Topic

Number Theory

Digits, divisibility, factors and primes.

135 problems 📖 Read the lesson
Practice
Problem 1 · 2024 AMC 8 Easy
Number Theory last-digitmod-10

What is the ones digit of

222,222 − 22,222 − 2,222 − 222 − 22 − 2 ?
Show answer
Answer: B — The ones digit is 2.
Show hints
Hint 1 of 2
You don't need to do the whole subtraction. What part of the answer is the question actually asking about?
Still stuck? Show hint 2 →
Hint 2 of 2
Only the ones digits matter — and every number ends in 2. So skip the big subtraction entirely.
Show solution
  1. Only the ones digit matters. The five numbers being subtracted all end in 2, so their ones digits sum to 5 × 2 = 10 — together they take away something ending in 0.
  2. Subtracting a multiple of 10 from 222,222 doesn't touch its ones digit: it stays 2.
Another way — keep the intermediate positive (MAA):
  1. Look only at the last two digits so the running total never goes negative: 22 − 2 − 2 − 2 − 2 − 2.
  2. = 22 − 10 = 12. Ones digit: 2.
Mark: · log in to save
Problem 4 · 2024 AMC 8 Medium
Number Theory digit-sumperfect-squarework-backward

When Yunji added all the integers from 1 to 9, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?

Show answer
Answer: E — She left out 9.
Show hints
Hint 1 of 2
Start from the correct total 1+2+…+9. Taking one number out lands you near a special number — which one?
Still stuck? Show hint 2 →
Hint 2 of 2
Find the correct total of 1 through 9. Leaving out x makes the sum 45 − x. Which x makes that a perfect square?
Show solution
  1. 1 + 2 + 3 + … + 9 = 45.
  2. Leaving out x (from 1 to 9) gives 45 − x, which is between 36 and 44.
  3. The only perfect square in that range is 36 = 62.
  4. 45 − x = 36, so x = 9.
Mark: · log in to save
Problem 5 · 2024 AMC 8 Stretch
Number Theory divisibilityfactor-pairscasework

Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of 6. Which of the following integers cannot be the sum of the two numbers?

Show answer
Answer: B — The sum cannot be 6.
Show hints
Hint 1 of 2
The product is a multiple of 6 only when the two dice meet a special condition. Test each answer choice against it.
Still stuck? Show hint 2 →
Hint 2 of 2
A product is a multiple of 6 only if the pair contains a 3 or 6 (factor of 3) and an even number (factor of 2). Just test the answer choices against that.
Show solution
  1. The pair must include a multiple of 3 (a 3 or a 6) and an even number.
  2. Sum 6 comes only from (1,5), (2,4), (3,3) — products 5, 8, 9, none a multiple of 6. So 6 is impossible.
  3. Every other choice has a good pair: 5 = (2,3)→6, 7 = (1,6)→6, 8 = (2,6)→12, 9 = (3,6)→18.
Another way — list every valid pair (MAA):
  1. Pairs whose product is a multiple of 6 (need a multiple of 3 and an even number): (1,6), (2,3), (2,6), (3,6), (4,6), (5,6), (6,6).
  2. Their sums: 7, 5, 8, 9, 10, 11, 12. Among A–E, only 6 is missing.
Mark: · log in to save
Problem 4 · 2023 AMC 8 Stretch
Number Theory primesspiral-pattern
amc8-2023-04
Show answer
Answer: D — Three of them are prime.
Show hints
Hint 1 of 2
Forget the spiral pattern — what matters is which numbers end up on those four squares.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the four shaded numbers first (they sit on the diagonal through 7), then test each one for being prime.
Show solution
Approach: fill in the diagonal, then test each for primeness
  1. Continuing the spiral outward, the diagonal through 7 (going up-left and down-right) contains the four shaded numbers 19, 23, 39, 47.
  2. Test each: 19 prime, 23 prime, 47 prime; but 39 = 3 × 13 is composite.
  3. So 3 of the four shaded numbers are prime.
Another way — use perfect squares as landmarks (MAA):
  1. Without filling the whole grid: on an n×n spiral the number n2 sits in the upper-left (n even) or lower-right (n odd) corner. So 9 is at lower-right of the 3×3 block, 25 at lower-right of 5×5, 49 at lower-right of 7×7; 16 at upper-left of 4×4, 36 at upper-left of 6×6.
  2. Walking outward from those anchors locates the four shaded squares as 19, 23, 39, 47 — with 39 = 3 × 13 the only composite.
Mark: · log in to save
Problem 3 · 2022 AMC 8 Medium
Number Theory factorizationfactor-triplescasework

When three positive integers a, b, and c are multiplied together, their product is 100. Suppose a < b < c. In how many ways can the numbers be chosen?

Show answer
Answer: E — 4 ways.
Show hints
Hint 1 of 2
100 has only a handful of factor groupings. Be systematic so you don't miss any or double-count.
Still stuck? Show hint 2 →
Hint 2 of 2
List the ways to write 100 as a product of three different whole numbers, smallest to largest. (100 = 22 × 52.)
Show solution
  1. Try the smallest factor as 1: 1×2×50, 1×4×25, 1×5×20 — all have a < b < c.
  2. Now without a 1: 2×5×10 works (2 < 5 < 10).
  3. Others repeat a factor (like 1×10×10 or 2×2×25), so they don't count.
  4. That gives 4 valid choices.
Another way — bound the smallest factor first (MAA):
  1. Since a < b < c and abc = 100, we get a3 < 100, so a ≤ 4. And a must be a factor of 100, so a ∈ {1, 2, 4}.
  2. For each: a = 1 gives (1,2,50), (1,4,25), (1,5,20); a = 2 gives (2,5,10); a = 4 gives nothing (would need bc = 25 with 4 < b < c).
  3. Total: 4 ways.
Mark: · log in to save
Problem 5 · 2016 AMC 8 Medium
Number Theory divisibilitymod-10

The number N is a two-digit number.

  • When N is divided by 9, the remainder is 1.
  • When N is divided by 10, the remainder is 3.

What is the remainder when N is divided by 11?

Show answer
Answer: E — Remainder 7.
Show hints
Hint 1 of 2
Remainder 3 mod 10 means N ends in 3. So N ∈ {13, 23, 33, …, 93}.
Still stuck? Show hint 2 →
Hint 2 of 2
Which of those leaves remainder 1 when divided by 9? Sum of digits gives the answer fast (sum ≡ N mod 9).
Show solution
Approach: narrow to last-digit-3, then check mod 9
  1. Last digit 3, so N ∈ {13, 23, 33, 43, 53, 63, 73, 83, 93}.
  2. Sum-of-digits test for mod 9: only 73 has digit sum 10 ≡ 1 (mod 9). So N = 73.
  3. 73 mod 11: 11 × 6 = 66, remainder 7. Answer: 7.
Mark: · log in to save
Problem 4 · 2014 AMC 8 Easy
Number Theory parityprimes

The sum of two prime numbers is 85. What is the product of these two prime numbers?

Show answer
Answer: E — 166.
Show hint
Hint 1
Odd sum from two primes ⇒ one of them must be even. The only even prime is 2.
Show solution
Approach: the only even prime is 2
  1. 85 is odd, so one prime is even ⇒ it must be 2.
  2. The other prime: 85 − 2 = 83 (prime).
  3. Product: 2 × 83 = 166.
Mark: · log in to save
Problem 8 · 2014 AMC 8 Easy
Number Theory divisibility-by-11

Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?

Show answer
Answer: D — A = 3.
Show hints
Hint 1 of 2
The total is 11 × (integer), so 1A2 must be divisible by 11.
Still stuck? Show hint 2 →
Hint 2 of 2
Divisibility rule for 11: alternating sum of digits must be a multiple of 11.
Show solution
Approach: divisibility test for 11
  1. 1A2 divisible by 11 ⇒ alternating sum 1 − A + 2 = 3 − A is a multiple of 11.
  2. Single digit A ∈ {0, …, 9}: only A = 3 works (giving 0).
  3. Check: 132 = 11 × 12. ✓
Mark: · log in to save
Problem 1 · 2013 AMC 8 Easy
Number Theory next-multiple

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?

Show answer
Answer: A — 1 more car.
Show hint
Hint 1
Find the next multiple of 6 above 23.
Show solution
Approach: round up to a multiple of 6
  1. Next multiple of 6 after 23 is 24.
  2. She needs 24 − 23 = 1 more car.
Mark: · log in to save
Problem 15 · 2011 AMC 8 Easy
Number Theory exponent-rewrite

How many digits are in the product 45 · 510?

Show answer
Answer: D — 11 digits.
Show hint
Hint 1
Rewrite 45 as 210. Then 210 · 510 = 1010.
Show solution
Approach: rewrite into 10n
  1. 45 · 510 = (22)5 · 510 = 210 · 510 = 1010.
  2. 1010 = 1 followed by ten zeros ⇒ 11 digits.
Mark: · log in to save
Problem 14 · 2010 AMC 8 Easy
Number Theory prime-factorization

What is the sum of the prime factors of 2010?

Show answer
Answer: C — 77.
Show hint
Hint 1
Factor out small primes first: 2010 / 2 / 3 / 5 = ?
Show solution
Approach: factor primes step by step
  1. 2010 = 2 · 1005 = 2 · 3 · 335 = 2 · 3 · 5 · 67 (67 is prime).
  2. Sum: 2 + 3 + 5 + 67 = 77.
Mark: · log in to save
Problem 3 · 2007 AMC 8 Easy
Number Theory prime-factorization

What is the sum of the two smallest prime factors of 250?

Show answer
Answer: C — 7.
Show hint
Hint 1
250 = 2 · 53.
Show solution
Approach: factor 250
  1. 250 = 2 · 53. Two smallest (and only) primes: 2 and 5.
  2. Sum: 2 + 5 = 7.
Mark: · log in to save
Problem 10 · 2007 AMC 8 Easy
Number Theory sigma-function

For any positive integer n, define [n] to be the sum of the positive factors of n. For example, [6] = 1 + 2 + 3 + 6 = 12. Find [[11]].

Show answer
Answer: D — 28.
Show hint
Hint 1
[11] first; 11 is prime so [11] = 1 + 11 = 12. Then compute [12].
Show solution
Approach: apply twice
  1. [11] = 1 + 11 = 12 (since 11 is prime).
  2. [12] = 1 + 2 + 3 + 4 + 6 + 12 = 28.
Mark: · log in to save
Problem 8 · 2005 AMC 8 Easy
Number Theory parity-rules

Suppose m and n are positive odd integers. Which of the following must also be an odd integer?

Show answer
Answer: E — 3mn.
Show hint
Hint 1
odd × odd = odd; odd + odd = even; even + odd = odd.
Show solution
Approach: apply parity rules to each
  1. (A) odd + odd = even. (B) odd − odd = even. (C) odd + odd = even. (D) (odd + odd)2 = even2 = even.
  2. (E) odd · odd · odd = odd.
Mark: · log in to save
Problem 18 · 2005 AMC 8 Easy
Number Theory count-multiples

How many three-digit numbers are divisible by 13?

Show answer
Answer: C — 69.
Show hint
Hint 1
Three-digit multiples of 13: smallest is 13 · 8 = 104. Largest is 13 · 76 = 988. Count from 8 to 76.
Show solution
Approach: find smallest and largest multipliers
  1. Smallest 3-digit multiple: 13 · 8 = 104. Largest: 13 · 76 = 988.
  2. Multipliers 8 through 76: 76 − 8 + 1 = 69.
Mark: · log in to save
Problem 2 · 2003 AMC 8 Easy
Number Theory primes

Which of the following numbers has the smallest prime factor?

Show answer
Answer: C — 58 (its smallest prime factor is 2).
Show hint
Hint 1
No prime is smaller than 2 — which of these has 2 as a factor?
Show solution
Approach: the smallest prime is 2
  1. A prime factor can't be smaller than 2, and only even numbers have 2 as a factor.
  2. 58 is the only even choice, so its smallest prime factor is 2 — smaller than any odd number could manage. Answer 58.
Mark: · log in to save
Problem 4 · 2002 AMC 8 Easy
Number Theory place-value

The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?

Show answer
Answer: B — 4.
Show hints
Hint 1 of 2
Keep the thousands digit at 2, so the next palindrome looks like 2 _ _ 2.
Still stuck? Show hint 2 →
Hint 2 of 2
For that to mirror, the two middle digits must be equal — make them as small as you can while passing 2002.
Show solution
Approach: build the next palindrome from the outside in
  1. Don't change the leading 2, so the year stays of the form 2 _ _ 2 with equal middle digits.
  2. The smallest such year after 2002 is 2112, whose digit product is 2 × 1 × 1 × 2 = 4.
Mark: · log in to save
Problem 5 · 2002 AMC 8 Easy
Number Theory divisibility

Carlos Montado was born on Saturday, November 9, 2002. On what day of the week will Carlos be 706 days old?

Show answer
Answer: C — Friday.
Show hints
Hint 1 of 2
The day of the week repeats every 7 days, so only the remainder of 706 when divided by 7 matters.
Still stuck? Show hint 2 →
Hint 2 of 2
706 = 700 + 6, and 700 is a multiple of 7.
Show solution
Approach: strip off the whole weeks (multiples of 7)
  1. Since 700 is a multiple of 7, after 700 days the weekday is again Saturday.
  2. Six more days past Saturday lands on Friday.
Mark: · log in to save
Problem 4 · 2001 AMC 8 Easy
Number Theory place-value

The digits 1, 2, 3, 4, and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is

Show answer
Answer: E — 9.
Show hints
Hint 1 of 2
To make a number small, put the smallest digits on the left — but the units digit has to be even.
Still stuck? Show hint 2 →
Hint 2 of 2
Push the big digits as far right as the even-ending rule allows.
Show solution
Approach: small digits left, but keep the number even
  1. Smallest first three digits are 1, 2, 3. That leaves 4 and 9 for the tens and units.
  2. The units digit must be even, so 4 goes in the units and 9 in the tens: the number is 12394, and the tens digit is 9.
Mark: · log in to save
Problem 1 · 1996 AJHSME Easy
Number Theory divisor-counting

How many positive factors of 36 are also multiples of 4?

Show answer
Answer: B — 3.
Show hints
Hint 1 of 2
A factor of 36 that is a multiple of 4 is 4 times a factor of 36 ÷ 4 = 9.
Still stuck? Show hint 2 →
Hint 2 of 2
List the factors of 9.
Show solution
Approach: factor out the 4
  1. Such a number is 4 × (a factor of 9), and 9 has factors 1, 3, 9.
  2. That gives 4, 12, 36 — 3 numbers.
Mark: · log in to save
Problem 5 · 1986 AJHSME Easy
Number Theory time-arithmetic

A contest began at noon one day and ended 1000 minutes later. At what time did the contest end?

Show answer
Answer: D — 4:40 a.m.
Show hints
Hint 1 of 2
Convert 1000 minutes to hours and minutes.
Still stuck? Show hint 2 →
Hint 2 of 2
1000 = 16·60 + 40.
Show solution
Approach: convert to hh:mm and add
  1. 1000 minutes = 16 hours 40 minutes.
  2. Noon + 16 h 40 m = 4:40 a.m. (the next day).
Mark: · log in to save
Problem 16 · 2026 AMC 8 Hard
Number Theory divisibility-rule

Consider all positive four-digit integers whose digits are all even. What fraction of these integers are divisible by 4?

Show answer
Answer: D — 3/5.
Show hints
Hint 1 of 2
Divisibility by 4 depends only on the last two digits.
Still stuck? Show hint 2 →
Hint 2 of 2
With an even tens digit, the tens part is already a multiple of 4, so only the units digit decides it.
Show solution
Approach: reduce divisibility by 4 to the units digit
  1. A number is divisible by 4 exactly when its last two digits are. With an even tens digit, 10·(tens) is already a multiple of 4, so it comes down to the units digit being 0, 4, or 8.
  2. That's 3 of the 5 even units digits, and it holds for every choice of the other digits, so the fraction is 3/5.
Mark: · log in to save
Problem 18 · 2026 AMC 8 Hard
Number Theory consecutive-sums

In how many ways can 60 be written as the sum of two or more consecutive odd positive integers, arranged in increasing order?

Show answer
Answer: B — 2.
Show hints
Hint 1 of 2
A run of k consecutive odd numbers starting at a sums to k(a + k − 1).
Still stuck? Show hint 2 →
Hint 2 of 2
Test each k that divides 60 and keep the ones giving a positive odd starting value.
Show solution
Approach: write the run-sum and test divisors
  1. k consecutive odd numbers starting at a sum to k(a + k − 1) = 60.
  2. Checking divisors, only k = 2 (29 + 31) and k = 6 (5 + 7 + 9 + 11 + 13 + 15) give a positive odd start.
  3. So there are 2 ways.
Mark: · log in to save
Problem 13 · 2025 AMC 8 Medium
Number Theory mod-arithmeticcyclicity
amc8-2025-13
Show answer
Answer: A — Histogram (A).
Show hints
Hint 1 of 2
Don't compute 25 separate remainders — even numbers' remainders mod 7 repeat in a pattern.
Still stuck? Show hint 2 →
Hint 2 of 2
The cycle is 2, 4, 6, 1, 3, 5, 0 (period 7). 25 numbers = 3 full cycles + 4 extras.
Show solution
Approach: find the cycle, count full cycles and extras
  1. The remainders mod 7 of 2, 4, 6, 8, 10, 12, 14, … cycle through 2, 4, 6, 1, 3, 5, 0 with period 7.
  2. 25 even numbers = 3 full cycles (21 numbers, hitting each remainder 3 times) plus 4 extras: 2, 4, 6, 1.
  3. So remainders 1, 2, 4, 6 occur 4 times each; remainders 0, 3, 5 occur 3 times each — the pattern in choice A.
Mark: · log in to save
Problem 16 · 2025 AMC 8 Hard
Number Theory complementary-countingsum-constraint

Five distinct integers from 1 to 10 are chosen, and five distinct integers from 11 to 20 are chosen. No two numbers differ by exactly 10. What is the sum of the ten chosen numbers?

Show answer
Answer: C — 105.
Show hints
Hint 1 of 2
Each chosen low number blocks exactly one high number. How many highs are left to choose from?
Still stuck? Show hint 2 →
Hint 2 of 2
Five chosen lows block five highs — leaving exactly five unblocked highs, which are forced to be picked. They're the highs matching the unchosen lows + 10.
Show solution
Approach: the high choices are forced, and pair with the unchosen lows
  1. Each chosen low (say x) blocks the high x + 10. With 5 lows chosen, 5 highs are blocked — so the 5 chosen highs are exactly the 5 unblocked ones: those that are 10 more than the 5 unchosen lows.
  2. Sum of chosen highs = (sum of unchosen lows) + 5 × 10. So (chosen lows) + (chosen highs) = (chosen lows) + (unchosen lows) + 50 = (sum of 1 to 10) + 50.
  3. 1 + 2 + … + 10 = 55, so the total is 55 + 50 = 105.
Mark: · log in to save
Problem 15 · 2024 AMC 8 Hard
Number Theory factorizationcaseworkwork-backward

Let the letters F, L, Y, B, U, G represent distinct digits. Suppose FLYFLY is the greatest number that satisfies the equation

8 · FLYFLY = BUGBUG.

What is the value of FLY + BUG?

Show answer
Answer: C — 1107.
Show hints
Hint 1 of 2
A six-digit number that repeats a three-digit block (like ABCABC) has a hidden common factor.
Still stuck? Show hint 2 →
Hint 2 of 2
FLYFLY = 1001 · FLY and BUGBUG = 1001 · BUG, so the equation just says 8 · FLY = BUG.
Show solution
Approach: strip the repeat, then maximize digit-by-digit
  1. FLYFLY = 1001 · FLY and BUGBUG = 1001 · BUG, so the equation reduces to 8 · FLY = BUG.
  2. BUG is a 3-digit number, so 8 · FLY < 1000 → FLY ≤ 124. So F = 1, and L ≤ 2 (else 8 · FLY ≥ 1040).
  3. Maximize: L = 2 (can't be 1, F took it). Then 8 · 12Y needs all 6 digits distinct. Y = 4: 8 · 124 = 992 (two 9s, fails). Y = 3: 8 · 123 = 984 — digits {1, 2, 3, 9, 8, 4} all distinct ✓.
  4. FLY + BUG = 123 + 984 = 1107.
Mark: · log in to save
Problem 14 · 2023 AMC 8 Hard
Number Theory complementary-countingcareful-counting

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of 5-cent, 10-cent, and 25-cent stamps, with exactly 20 of each type. What is the greatest number of stamps Nicolas can use to make exactly $7.10 in postage?

Show answer
Answer: E — 55 stamps.
Show hints
Hint 1 of 2
Maximizing stamps used = minimizing stamps removed from his whole collection. What does the whole collection total?
Still stuck? Show hint 2 →
Hint 2 of 2
Total of all 60 stamps = $8. He needs to remove $0.90. Minimize the number of stamps that sum to $0.90.
Show solution
Approach: minimize stamps removed, not maximize stamps used
  1. Total value of all 60 stamps: 20·($0.05 + $0.10 + $0.25) = 20 · $0.40 = $8.00.
  2. He needs to make $7.10, so he removes $8.00 − $7.10 = $0.90 worth. Maximizing stamps used ≡ minimizing stamps removed.
  3. Minimum stamps summing to $0.90: three 25¢ (75¢) + one 10¢ + one 5¢ = $0.90 in 5 stamps.
  4. Stamps used = 60 − 5 = 55.
Mark: · log in to save
Problem 16 · 2023 AMC 8 Medium
Number Theory divisibilitysymmetry
amc8-2023-16
Show answer
Answer: C — 133 Ps, 134 Qs, 133 Rs.
Show hints
Hint 1 of 2
20 × 20 = 400 cells, with the letters cycling P, Q, R. How does 400 split among three letters?
Still stuck? Show hint 2 →
Hint 2 of 2
400 = 3 × 133 + 1, so one letter gets the extra 1. By the table's diagonals, P-count = R-count — so Q takes the extra.
Show solution
Approach: 400 cells split into thirds, with one letter winning the remainder
  1. The board has 20 × 20 = 400 cells, and the P/Q/R pattern repeats every 3 cells diagonally. 400 = 3 × 133 + 1, so the counts are 133, 133, 134 in some order.
  2. Look at the lower-left 2 × 2 corner of the pattern: Q R / P Q — one P, two Qs, one R. By the rest of the board's symmetry (P and R balance), the corner's extra Q is the surplus.
  3. Counts: 133 Ps, 134 Qs, 133 Rs.
Mark: · log in to save
Problem 18 · 2023 AMC 8 Hard
Number Theory divisibilitycasework

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump 5 pads to the right or 3 pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located 2023 pads to the right of her starting position?

Show answer
Answer: D — 411 jumps.
Show hints
Hint 1 of 2
Order doesn't matter — only the counts of right and left jumps. Set up an equation in R and L.
Still stuck? Show hint 2 →
Hint 2 of 2
5R − 3L = 2023. Need R ≥ 405 and 5R − 2023 divisible by 3.
Show solution
Approach: count-only equation, then mod constraint
  1. Order doesn't matter. Let R = right jumps, L = left jumps. Then 5R − 3L = 2023, with R, L ≥ 0.
  2. Minimum R: R ≥ 405 (else 5R < 2023, leaving negative L). And 5R − 2023 must be divisible by 3.
  3. Mod 3: 5R ≡ 2R, and 2023 ≡ 1, so 2R ≡ 1 ⇒ R ≡ 2 (mod 3). The smallest R ≥ 405 satisfying this is 407.
  4. Then L = (5·407 − 2023)/3 = 12/3 = 4. Total: 407 + 4 = 411 jumps.
Mark: · log in to save
Problem 17 · 2022 AMC 8 Medium
Number Theory last-digitmod-10

If n is an even positive integer, the double-factorial notation n!! represents the product of all the even integers from 2 to n. For example: 8!! = 2 × 4 × 6 × 8. What is the units digit of the following sum?

2!! + 4!! + 6!! + … + 2022!!
Show answer
Answer: B — Units digit 2.
Show hints
Hint 1 of 2
Once a double-factorial includes 10 as a factor, it ends in 0. Which terms in the sum still affect the ones digit?
Still stuck? Show hint 2 →
Hint 2 of 2
Only 2!!, 4!!, 6!!, 8!! contribute — everything from 10!! onward ends in 0.
Show solution
Approach: drop the 10!!-and-up terms (they end in 0)
  1. n!! for n ≥ 10 contains the factor 10, so its units digit is 0.
  2. Compute only the survivors: 2!! = 2, 4!! = 8, 6!! = 48, 8!! = 384. Their units digits: 2, 8, 8, 4.
  3. Sum of units digits: 2 + 8 + 8 + 4 = 22. Units digit: 2.
Mark: · log in to save
Problem 12 · 2020 AMC 8 Easy
Number Theory factorization

For a positive integer n, the factorial notation n! represents the product of the integers from n to 1. For example: 6! = 6 × 5 × 4 × 3 × 2 × 1. What value of N satisfies the following equation?

5! × 9! = 12 × N!
Show answer
Answer: A — N = 10.
Show hints
Hint 1 of 2
Compute 5! — can you factor 12 out of it neatly?
Still stuck? Show hint 2 →
Hint 2 of 2
5! = 120 = 12 × 10. So the equation becomes 12 × 10 × 9! = 12 × N!.
Show solution
Approach: factor 12 out of 5!
  1. 5! = 120 = 12 × 10.
  2. So 5! × 9! = 12 × 10 × 9! = 12 × (10 × 9!) = 12 × 10!.
  3. Therefore N! = 10!, so N = 10.
Mark: · log in to save
Problem 17 · 2020 AMC 8 Medium
Number Theory factorizationdivisibility

How many factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely 1, 2, 3, 4, 6, and 12.)

Show answer
Answer: B — 7 factors.
Show hints
Hint 1 of 2
Factor 2020 = 22 × 5 × 101. It has (2+1)(1+1)(1+1) = 12 factors total.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract those with at most 3 factors: 1 (one factor), primes 2/5/101 (two each), and squares of primes — here only 4 (three factors).
Show solution
Approach: subtract the easy-to-count low-factor cases
  1. 2020 = 22 · 5 · 101 has (2+1)(1+1)(1+1) = 12 factors.
  2. Factors with ≤ 3 factors: 1 (1 factor); 2, 5, 101 (2 factors each — primes); 4 (3 factors — prime squared). That's 5 of the 12.
  3. Remaining: 12 − 5 = 7 factors with more than 3 factors.
Mark: · log in to save
Problem 19 · 2020 AMC 8 Medium
Number Theory divisibilitydigit-sum

A number is called flippy if its digits alternate between two distinct digits. For example, 2020 and 37373 are flippy, but 3883 and 123123 are not. How many five-digit flippy numbers are divisible by 15?

Show answer
Answer: B — 4 numbers.
Show hints
Hint 1 of 2
Divisible by 15 = divisible by 5 AND by 3. The last digit must be 0 or 5; since the number is 5 digits, the first digit can't be 0.
Still stuck? Show hint 2 →
Hint 2 of 2
Five-digit flippy: pattern ababa. a ≠ 0 and ends in a, so a = 5. Digit sum = 15 + 2b ⇒ b ∈ {0, 3, 6, 9}.
Show solution
Approach: decompose div-by-15 into 5 and 3
  1. 5-digit flippy: ababa with a ≠ b and a ≠ 0.
  2. Div by 5 ⇒ last digit (= a) is 0 or 5. Since a ≠ 0, a = 5.
  3. Number is 5b5b5. Div by 3 ⇒ digit sum 15 + 2b div by 3 ⇒ b div by 3, so b ∈ {0, 3, 6, 9} (and all are ≠ 5).
  4. 4 flippy numbers.
Mark: · log in to save
Problem 13 · 2019 AMC 8 Medium
Number Theory divisibilitydigit-sum

A palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let N be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of N?

Show answer
Answer: A — 2.
Show hints
Hint 1 of 2
Two-digit palindromes (11, 22, 33, …) are all multiples of 11 — so any sum of three of them is a multiple of 11.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the smallest 3-digit multiple of 11 that isn't itself a palindrome. Can it be written as a sum of three distinct 2-digit palindromes?
Show solution
Approach: two-digit palindromes are multiples of 11
  1. Two-digit palindromes (11, 22, …, 99) are all multiples of 11; their sum is too. So N is a multiple of 11.
  2. Smallest 3-digit multiple of 11 that's NOT a palindrome: 110 (palindromes are 121, 131, … — 110 isn't one).
  3. Check: 110 = 11 + 22 + 77 ✓ (three distinct 2-digit palindromes).
  4. N = 110; digit sum = 1 + 1 + 0 = 2.
Mark: · log in to save
Problem 14 · 2019 AMC 8 Medium
Number Theory mod-10divisibility

Isabella has 6 coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every 10 days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the 6 dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?

Show answer
Answer: C — Wednesday.
Show hints
Hint 1 of 2
10 days ≡ 3 days mod 7. So the 6 redemption days are at day-of-week offsets 0, 3, 6, 2, 5, 1 from the starting day.
Still stuck? Show hint 2 →
Hint 2 of 2
Those 6 offsets cover 6 of 7 days — missing only offset 4. Sunday must be that missing day, so the start = Sunday − 4 days = Wednesday.
Show solution
Approach: compute the 6 days mod 7, find the missing one
  1. 10 days advances the day-of-week by 10 mod 7 = 3 days. So the six redemption days have day-of-week offsets {0, 3, 6, 2, 5, 1} mod 7 from the start.
  2. These 6 offsets cover everything except offset 4. Sunday must be that missing offset, so start day is Sunday − 4 days.
  3. Counting back from Sunday: Sat, Fri, Thu, Wed.
Mark: · log in to save
Problem 13 · 2018 AMC 8 Hard
Number Theory divisibilitycasework

Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?

Show answer
Answer: A — 4 values.
Show hints
Hint 1 of 2
Set 4f + l = 410, with f < l ≤ 100. Get bounds on l first.
Still stuck? Show hint 2 →
Hint 2 of 2
Since 4f is divisible by 4 and 410 ≡ 2 (mod 4), we need l ≡ 2 (mod 4).
Show solution
Approach: modular constraint + range
  1. Average 82 over 5 tests ⇒ total = 410. Let f = first-four score, l = last. Then 4f + l = 410 with f < l ≤ 100.
  2. l > 82 (else 4f + l would need f ≥ l).
  3. Mod 4: 4f ≡ 0, 410 ≡ 2 ⇒ l ≡ 2 (mod 4). In (82, 100]: l ∈ {86, 90, 94, 98}.
  4. 4 values.
Mark: · log in to save
Problem 14 · 2018 AMC 8 Medium
Number Theory factorizationdigit-sum

Let N be the greatest five-digit number whose digits have a product of 120. What is the sum of the digits of N?

Show answer
Answer: D — 18.
Show hints
Hint 1 of 2
Greedy: maximize the leftmost digit first — biggest single digit dividing 120.
Still stuck? Show hint 2 →
Hint 2 of 2
120 = 8 × 15. 15 = 5 × 3. 3 = 3 × 1. Remaining slots: 1, 1. Number: 85311.
Show solution
Approach: greedy left-to-right factorization
  1. Largest digit dividing 120 is 8 (since 9 doesn't divide 120). After taking out 8: leftover product 15.
  2. Largest digit dividing 15 is 5. Leftover: 3. Largest digit dividing 3 is 3. Leftover: 1 → fill remaining digits with 1.
  3. N = 85311. Digit sum: 8 + 5 + 3 + 1 + 1 = 18.
Mark: · log in to save
Problem 18 · 2018 AMC 8 Medium
Number Theory factorizationprime-test

How many positive factors does 23,232 have?

Show answer
Answer: E — 42 factors.
Show hints
Hint 1 of 2
Find the prime factorization of 23,232, then multiply (exponent + 1) for each prime.
Still stuck? Show hint 2 →
Hint 2 of 2
23,232 = 26 · 3 · 112.
Show solution
Approach: prime factorize, multiply (exponent + 1) per prime
  1. Factor: 23,232 = 2 · 11,616 = 22 · 5,808 = … = 26 · 363 = 26 · 3 · 121 = 26 · 3 · 112.
  2. Number of factors: (6+1)(1+1)(2+1) = 7 · 2 · 3 = 42.
Mark: · log in to save
Problem 12 · 2017 AMC 8 Easy
Number Theory divisibility

The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

Show answer
Answer: D — Between 60 and 79.
Show hint
Hint 1
Same remainder mod 4, 5, AND 6 means n − 1 is divisible by all three. So n − 1 = lcm(4, 5, 6).
Show solution
Approach: lcm shift
  1. n − 1 is divisible by 4, 5, and 6, so by lcm(4, 5, 6) = 60.
  2. Smallest n > 1: n = 61, which lies between 60 and 79.
Mark: · log in to save
Problem 19 · 2017 AMC 8 Hard
Number Theory factorizationdivisibility

For any positive integer M, the notation M! denotes the product of the integers 1 through M. What is the largest integer n for which 5n is a factor of the sum

98! + 99! + 100! ?
Show answer
Answer: D — 26.
Show hints
Hint 1 of 2
Factor out 98! from all three terms. The leftover is a clean number.
Still stuck? Show hint 2 →
Hint 2 of 2
98! + 99! + 100! = 98!(1 + 99 + 99·100) = 98! · 10,000. Count factors of 5 in each piece.
Show solution
Approach: factor out 98!, count 5s
  1. 98! + 99! + 100! = 98!(1 + 99 + 99·100) = 98!(100 + 9900) = 98! · 10,000.
  2. 10,000 = 104 = 24 · 54 → contributes 4 factors of 5.
  3. Factors of 5 in 98!: ⌊98/5⌋ + ⌊98/25⌋ = 19 + 3 = 22.
  4. Total: 22 + 4 = 26.
Mark: · log in to save
Problem 11 · 2016 AMC 8 Medium
Number Theory place-valuecareful-counting

Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is 132.

Show answer
Answer: B — 7 numbers.
Show hints
Hint 1 of 2
Two-digit + reversed = (10a + b) + (10b + a) = 11(a + b). So a + b = 12.
Still stuck? Show hint 2 →
Hint 2 of 2
Count digit pairs (a, b) with a + b = 12, a ∈ {1, …, 9}, b ∈ {0, …, 9}.
Show solution
Approach: factor out 11 from the sum
  1. (10a + b) + (10b + a) = 11(a + b) = 132 ⇒ a + b = 12.
  2. Valid pairs with a from 1–9 and b from 0–9: (3,9), (4,8), (5,7), (6,6), (7,5), (8,4), (9,3).
  3. 7 two-digit numbers.
Mark: · log in to save
Problem 15 · 2016 AMC 8 Medium
Number Theory difference-of-squaresfactorization

What is the largest power of 2 that is a divisor of 134 − 114?

Show answer
Answer: C — 32.
Show hint
Hint 1
Difference of squares twice: a4 − b4 = (a2 + b2)(a2 − b2).
Show solution
Approach: difference of squares + count factors of 2
  1. 134 − 114 = (132 + 112)(132 − 112) = 290 · 48.
  2. 290 = 2 · 145 (one factor of 2). 48 = 24 · 3 (four factors of 2).
  3. Total: 21+4 = 25 = 32.
Mark: · log in to save
Problem 20 · 2016 AMC 8 Medium
Number Theory divisibilityfactorization

The least common multiple of a and b is 12, and the least common multiple of b and c is 15. What is the least possible value of the least common multiple of a and c?

Show answer
Answer: A — 20.
Show hints
Hint 1 of 2
b divides both lcms — so b divides gcd(12, 15) = 3. Try b = 3.
Still stuck? Show hint 2 →
Hint 2 of 2
Then minimize a and c: smallest a giving lcm(a, 3) = 12 is 4, and smallest c giving lcm(c, 3) = 15 is 5.
Show solution
Approach: constrain b, then minimize a and c
  1. b must divide both 12 and 15, so b | gcd(12, 15) = 3.
  2. Take b = 3. Then smallest a with lcm(a, 3) = 12 is a = 4. Smallest c with lcm(c, 3) = 15 is c = 5.
  3. lcm(4, 5) = 20.
Mark: · log in to save
Problem 22 · 2015 AMC 8 Hard
Number Theory divisor-countinglcm

On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?

Show answer
Answer: C — 60 students.
Show hints
Hint 1 of 2
Each day's row count is a divisor of n, all different. Twelve days ⇒ n has at least 12 divisors.
Still stuck? Show hint 2 →
Hint 2 of 2
n is divisible by both 15 and 6, so by lcm(6, 15) = 30. Smallest multiple of 30 with exactly 12 divisors?
Show solution
Approach: translate to a divisor-counting problem
  1. Each day's number of students per row is a divisor of n, and all 12 are different ⇒ n has exactly 12 divisors (no 13th option exists).
  2. n is a multiple of 15 and 6, hence of lcm(6, 15) = 30 = 2 · 3 · 5. That has only (1+1)(1+1)(1+1) = 8 divisors.
  3. Double one prime exponent to get 12: 22 · 3 · 5 = 60 has (2+1)(1+1)(1+1) = 12 divisors. The other options (32·2·5 = 90, 2·3·52=150) are larger.
  4. Smallest n = 60.
Mark: · log in to save
Problem 24 · 2015 AMC 8 Hard
Number Theory linear-diophantinemod-3

A baseball league consists of two four-team divisions. Each team plays every other team in its division N games. Each team plays every team in the other division M games with N > 2M and M > 4. Each team plays a 76 game schedule. How many games does a team play within its own division?

Show answer
Answer: B — 48 games.
Show hints
Hint 1 of 2
A team faces 3 division rivals (N games each) and 4 outside teams (M games each): 3N + 4M = 76.
Still stuck? Show hint 2 →
Hint 2 of 2
Use the bounds M > 4 and N > 2M; also 4M ≡ 76 (mod 3) forces a residue on M.
Show solution
Approach: diophantine equation with inequalities
  1. 3N + 4M = 76.
  2. Mod 3: 4M ≡ 76 ⇒ M ≡ 1 (mod 3). So M ∈ {7, 10, 13, …}.
  3. N > 2M ⇒ 76 = 3N + 4M > 10M ⇒ M < 7.6.
  4. Combining M > 4 and M < 7.6 and M ≡ 1 (mod 3): M = 7. Then N = (76 − 28)/3 = 16.
  5. Games within division: 3N = 3 × 16 = 48.
Mark: · log in to save
Problem 21 · 2014 AMC 8 Hard
Number Theory divisibility-by-3mod-arithmetic

The 7-digit numbers 74A52B1 and 326AB4C are each multiples of 3. Which of the following could be the value of C?

Show answer
Answer: A — C = 1.
Show hints
Hint 1 of 2
Divisibility by 3: sum of digits divisible by 3. Write the digit-sum condition for each number, then subtract.
Still stuck? Show hint 2 →
Hint 2 of 2
Combine both conditions to find C's residue mod 3.
Show solution
Approach: two mod-3 conditions, subtract
  1. First number: 7+4+5+2+1 = 19, so 19 + A + B ≡ 0 (mod 3) ⇒ A + B ≡ 2 (mod 3).
  2. Second number: 3+2+6+4 = 15, so 15 + A + B + C ≡ 0 (mod 3) ⇒ A + B + C ≡ 0 (mod 3).
  3. Subtract: C ≡ −2 ≡ 1 (mod 3). Choices give C = 1.
Mark: · log in to save
Problem 22 · 2011 AMC 8 Hard
Number Theory tens-digit-cyclemod-100

What is the tens digit of 72011?

Show answer
Answer: D — Tens digit 4.
Show hints
Hint 1 of 2
Tens digit depends only on the value mod 100. Compute 7k mod 100 for small k and look for a cycle.
Still stuck? Show hint 2 →
Hint 2 of 2
71 = 07, 72 = 49, 73 = 343 (43), 74 = 2401 (01). Cycle length 4.
Show solution
Approach: powers of 7 mod 100 cycle in 4
  1. Last two digits cycle: 07, 49, 43, 01, repeating with period 4.
  2. 2011 = 4 · 502 + 3 ⇒ 72011 ≡ 73 ≡ 43 (mod 100).
  3. Tens digit = 4.
Mark: · log in to save
Problem 17 · 2009 AMC 8 Hard
Number Theory exponent-parity-mod

The positive integers x and y are the two smallest positive integers for which the product of 360 and x is a square and the product of 360 and y is a cube. What is the sum of x and y?

Show answer
Answer: B — 85.
Show hint
Hint 1
Prime-factor 360 = 23 · 32 · 5. Make every exponent even (square) or a multiple of 3 (cube).
Show solution
Approach: fix exponents to the right parity / multiple of 3
  1. 360 = 23 · 32 · 51.
  2. Square: bump 2's exponent to 4 and 5's to 2 ⇒ multiply by 2 · 5 = 10. So x = 10.
  3. Cube: bump 2's to 3 (already), 3's to 3 (add one 3), 5's to 3 (add two 5s) ⇒ multiply by 3 · 25 = 75. So y = 75.
  4. x + y = 85.
Mark: · log in to save
Problem 24 · 2006 AMC 8 Hard
Number Theory factor-1010cryptarithm

In the multiplication problem below, A, B, C, D are different digits. ABA × CD = CDCD. What is A + B?

Show answer
Answer: A — 1.
Show hint
Hint 1
CDCD = CD × 101. So ABA = 101.
Show solution
Approach: spot the 101 pattern
  1. CDCD = CD · 101 ⇒ ABA = 101.
  2. So A = 1, B = 0; A + B = 1.
Mark: · log in to save
Problem 25 · 2006 AMC 8 Hard
Number Theory parity-of-prime

Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? (Visible sides: 44, 59, 38.)

Show answer
Answer: B — 14.
Show hints
Hint 1 of 2
Even-visible cards (44, 38) need an odd prime to make an odd sum; the 59 card needs an even prime to make an odd sum. Equal sums ⇒ the 59 card pairs with the only even prime, 2.
Still stuck? Show hint 2 →
Hint 2 of 2
Common sum = 59 + 2 = 61.
Show solution
Approach: parity forces 2 behind 59
  1. Sum behind 59 + 59 = 61 (odd) only if hidden is even ⇒ hidden = 2 (only even prime).
  2. Common sum 61. Behind 44: 61 − 44 = 17 (prime ✓). Behind 38: 61 − 38 = 23 (prime ✓).
  3. Average: (2 + 17 + 23)/3 = 42/3 = 14.
Mark: · log in to save
Problem 20 · 2005 AMC 8 Hard
Number Theory modular-meeting

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

Show answer
Answer: A — 6 turns.
Show hint
Hint 1
After k turns, Alice is at +5k and Bob is at −9k (mod 12). They coincide when 5k ≡ −9k (mod 12) ⇒ 14k ≡ 0 (mod 12).
Show solution
Approach: modular equality
  1. 14k ≡ 0 (mod 12) ⇒ 7k ≡ 0 (mod 6) ⇒ k ≡ 0 (mod 6).
  2. Smallest positive: k = 6.
Mark: · log in to save
Problem 19 · 2003 AMC 8 Hard
Number Theory divisibilityfactorization

How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?

Show answer
Answer: C — 3 integers.
Show hints
Hint 1 of 2
A number divisible by all three is divisible by their least common multiple.
Still stuck? Show hint 2 →
Hint 2 of 2
LCM(15, 20, 25) = 300 — now count its multiples in range.
Show solution
Approach: reduce to multiples of the LCM
  1. Being divisible by 15, 20, and 25 is the same as being divisible by their LCM.
  2. LCM(15, 20, 25) = 300.
  3. Multiples of 300 between 1000 and 2000: 1200, 1500, 1800 — that's 3.
Mark: · log in to save
Problem 14 · 2000 AMC 8 Hard
Number Theory last-digitmod-10

What is the units digit of 1919 + 9999?

Show answer
Answer: D — 8.
Show hints
Hint 1 of 2
Only the units digit of the base matters, and powers of 9 cycle 9, 1, 9, 1, …
Still stuck? Show hint 2 →
Hint 2 of 2
9 raised to an odd power ends in 9.
Show solution
Approach: track only the units digit
  1. Powers of 9 end in 9 for odd exponents and 1 for even ones. Both 19 and 99 end in 9, and both exponents are odd, so each power ends in 9.
  2. 9 + 9 = 18, so the units digit is 8.
Mark: · log in to save
Problem 10 · 1998 AJHSME Hard
Number Theory casework

Each of the letters W, X, Y, and Z represents a different integer in the set {1, 2, 3, 4}, but not necessarily in that order. If WX − YZ = 1, then the sum of W and Y is

Show answer
Answer: E — 7.
Show hints
Hint 1 of 2
For the difference to be exactly the whole number 1, each fraction should itself be a whole number.
Still stuck? Show hint 2 →
Hint 2 of 2
Try making one fraction 3/1 and the other 4/2.
Show solution
Approach: make both fractions whole numbers
  1. To get a difference of exactly 1, both fractions should be integers: 3/1 = 3 and 4/2 = 2 work, since 3 − 2 = 1.
  2. Then W = 3 and Y = 4, so W + Y = 7.
Mark: · log in to save
Problem 11 · 1997 AJHSME Hard
Number Theory divisor-counting

Let [N] mean the number of whole number divisors of N. For example, [3] = 2 because 3 has two divisors, 1 and 3. Find the value of

[ [11] × [20] ].
Show answer
Answer: A — 6.
Show hints
Hint 1 of 2
The divisor count comes from the prime factorization: add 1 to each exponent and multiply.
Still stuck? Show hint 2 →
Hint 2 of 2
Work the brackets from the inside out.
Show solution
Approach: count divisors via prime factorization
  1. ⟨11⟩ = 2 (11 is prime) and ⟨20⟩ = (2+1)(1+1) = 6, so the inner value is 2 × 6 = 12.
  2. Then ⟨12⟩ = (2+1)(1+1) = 6.
Mark: · log in to save
Problem 15 · 1996 AJHSME Hard
Number Theory units-digitmod-arithmetic

The remainder when the product 1492 · 1776 · 1812 · 1996 is divided by 5 is

Show answer
Answer: E — 4.
Show hints
Hint 1 of 2
Only the units digit of each factor matters for the remainder mod 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Multiply the units digits and look at that product's units digit.
Show solution
Approach: work with units digits
  1. The units digits are 2, 6, 2, 6, and 2 · 6 · 2 · 6 = 144 ends in 4.
  2. A number ending in 4 leaves remainder 4 when divided by 5.
Mark: · log in to save
Problem 12 · 1995 AJHSME Hard
Number Theory factor-pairs

A lucky year is one in which at least one date, written as month/day/year, has the property that the month times the day equals the last two digits of the year. For example, 1956 is lucky because 7/8/56 has 7 × 8 = 56. Which of the following is NOT a lucky year?

Show answer
Answer: E — 1994.
Show hints
Hint 1 of 2
For each year, try to write its last two digits as month × day with a valid month (1–12) and day.
Still stuck? Show hint 2 →
Hint 2 of 2
94 has very few factor pairs.
Show solution
Approach: look for a valid month × day factor pair
  1. 90 = 9 × 10, 91 = 7 × 13, 92 = 4 × 23, 93 = 3 × 31 — all have a month from 1–12 with a real day.
  2. But 94 = 2 × 47 only (besides 1 × 94), and no month 1–12 pairs with a valid day, so 1994 is not lucky.
Mark: · log in to save
Problem 15 · 1995 AJHSME Hard
Number Theory repeating-decimalcyclicity

What is the 100th digit to the right of the decimal point in the decimal form of 4/37?

Show answer
Answer: B — 1.
Show hints
Hint 1 of 2
Write 4/37 as a repeating decimal and find the repeating block.
Still stuck? Show hint 2 →
Hint 2 of 2
The block has length 3, so reduce 100 by 3.
Show solution
Approach: find the repeating block, then index
  1. 4/37 = 0.108108… , repeating the block 108 of length 3.
  2. Since 100 = 3·33 + 1, the 100th digit is the 1st of the block: 1.
Mark: · log in to save
Problem 22 · 1995 AJHSME Hard
Number Theory factoring

The number 6545 can be written as a product of a pair of positive two-digit numbers. What is the sum of this pair of numbers?

Show answer
Answer: A — 162.
Show hints
Hint 1 of 2
Factor 6545 into primes first.
Still stuck? Show hint 2 →
Hint 2 of 2
Group the prime factors into two two-digit numbers.
Show solution
Approach: prime factor, then pair into two-digit numbers
  1. 6545 = 5 · 7 · 11 · 17. Grouping as (5·17)(7·11) = 85 × 77 gives two two-digit numbers.
  2. Their sum is 85 + 77 = 162.
Mark: · log in to save
Problem 10 · 1994 AJHSME Hard
Number Theory divisors

For how many positive integer values of N is the expression 36N + 2 an integer?

Show answer
Answer: A — 7.
Show hints
Hint 1 of 2
N + 2 must be a divisor of 36.
Still stuck? Show hint 2 →
Hint 2 of 2
Since N ≥ 1, only divisors that are at least 3 count.
Show solution
Approach: count valid divisors of 36
  1. N + 2 must divide 36, and N ≥ 1 means N + 2 ≥ 3.
  2. The divisors of 36 that are ≥ 3 are 3, 4, 6, 9, 12, 18, 36 — 7 of them.
Mark: · log in to save
Problem 15 · 1994 AJHSME Hard
Number Theory patternmod-arithmetic
ajhsme-1994-15
Show answer
Answer: A — Up arrow, then right arrow.
Show hints
Hint 1 of 2
The arrow pattern repeats every 4 steps.
Still stuck? Show hint 2 →
Hint 2 of 2
Find 425 and 426 modulo 4 to see which arrows apply.
Show solution
Approach: use the period-4 repetition
  1. The arrows repeat with period 4: from a number ≡ 0 (right), ≡ 1 (up), ≡ 2 (right), ≡ 3 (down).
  2. Since 425 ≡ 1 and 426 ≡ 2 (mod 4), the moves are up then right — choice A.
Mark: · log in to save
Problem 15 · 1992 AJHSME Hard
Number Theory periodic-sequencemod-arithmetic

What is the 1992nd letter in the sequence ABCDEDCBAABCDEDCBAABCDEDCBA… ?

Show answer
Answer: C — C.
Show hints
Hint 1 of 2
The block ABCDEDCBA repeats; count its length.
Still stuck? Show hint 2 →
Hint 2 of 2
Find 1992 modulo that length.
Show solution
Approach: use the repeating block
  1. The block ABCDEDCBA has 9 letters and repeats.
  2. 1992 = 9·221 + 3, so the 1992nd letter is the 3rd of the block: C.
Mark: · log in to save
Problem 9 · 1991 AJHSME Hard
Number Theory inclusion-exclusion

How many whole numbers from 1 through 46 are divisible by either 3 or 5 or both?

Show answer
Answer: B — 21.
Show hints
Hint 1 of 2
Count multiples of 3 and of 5 separately.
Still stuck? Show hint 2 →
Hint 2 of 2
Subtract the multiples of 15, which got counted twice.
Show solution
Approach: inclusion-exclusion
  1. Multiples of 3: ⌊46/3⌋ = 15; of 5: ⌊46/5⌋ = 9; of 15: ⌊46/15⌋ = 3.
  2. So 15 + 9 − 3 = 21.
Mark: · log in to save
Problem 13 · 1991 AJHSME Hard
Number Theory trailing-zerosfactor-2-and-5

How many zeros are at the end of the product 25 × 25 × 25 × 25 × 25 × 25 × 25 × 8 × 8 × 8?

Show answer
Answer: C — 9.
Show hints
Hint 1 of 2
Each trailing zero comes from a pair of factors 2 and 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the total 2's and 5's; the smaller count wins.
Show solution
Approach: match 2's with 5's
  1. The product is 25⁷ · 8³ = 5¹⁴ · 2⁹, giving fourteen 5's and nine 2's.
  2. Trailing zeros = the smaller count = 9.
Mark: · log in to save
Problem 14 · 1989 AJHSME Hard
Number Theory minimize-differenceplace-value

When placing each of the digits 2, 4, 5, 6, 9 in exactly one of the boxes of this subtraction problem, what is the smallest difference that is possible?

   
−     
 
Show answer
Answer: C — 149.
Show hints
Hint 1 of 2
To minimize the difference, make the 3-digit number as small as you can and the 2-digit number as large as you can.
Still stuck? Show hint 2 →
Hint 2 of 2
Pick the smallest hundreds digit first, then build each number from the digits you have left.
Show solution
Approach: smallest 3-digit minus largest 2-digit
  1. Smallest 3-digit using three of {2, 4, 5, 6, 9}: lead with 2, then take the next two smallest ascending → 245.
  2. Largest 2-digit from the remaining {6, 9} is 96. Difference: 245 − 96 = 149.
Mark: · log in to save
Problem 16 · 1989 AJHSME Hard
Number Theory parityprimes

In how many ways can 47 be written as the sum of two primes?

Show answer
Answer: A — 0.
Show hints
Hint 1 of 2
Odd = odd + even, so one of the two primes must be even.
Still stuck? Show hint 2 →
Hint 2 of 2
The only even prime is 2 — what does the other prime have to be?
Show solution
Approach: force one prime to be 2, then check
  1. 47 is odd, so one prime is even. The only even prime is 2, forcing the other prime to be 47 − 2 = 45.
  2. But 45 = 9 × 5 isn't prime, so there are 0 ways.
Mark: · log in to save
Problem 11 · 1988 AJHSME Hard
Number Theory bound-by-perfect-squares

√164 is

Show answer
Answer: E — between 12 and 13.
Show hints
Hint 1 of 2
Bound 164 between two consecutive perfect squares.
Still stuck? Show hint 2 →
Hint 2 of 2
12² = 144 and 13² = 169.
Show solution
Approach: squeeze between perfect squares
  1. 12² = 144 < 164 < 169 = 13².
  2. So √164 lies between 12 and 13.
Mark: · log in to save
Problem 14 · 1988 AJHSME Hard
Number Theory factor-pairsmax-sum

◇ and △ are whole numbers and ◇ × △ = 36. The largest possible value of ◇ + △ is

Show answer
Answer: E — 37.
Show hints
Hint 1 of 2
List all factor pairs of 36 and add each pair.
Still stuck? Show hint 2 →
Hint 2 of 2
Sums are biggest when the factors are most spread out.
Show solution
Approach: list factor pairs
  1. Factor pairs of 36: (1,36), (2,18), (3,12), (4,9), (6,6). Their sums: 37, 20, 15, 13, 12.
  2. The largest is 1 + 36 = 37.
Mark: · log in to save
Problem 20 · 1987 AJHSME Hard
Number Theory counterexample

"If a whole number n is not prime, then the whole number n − 2 is not prime." A value of n which shows this statement to be false is

Show answer
Answer: A — 9.
Show hints
Hint 1 of 2
Look for n that is itself not prime but n − 2 IS prime.
Still stuck? Show hint 2 →
Hint 2 of 2
Check each composite choice's n − 2.
Show solution
Approach: find a composite n whose n−2 is prime
  1. Try n = 9: 9 is not prime, and 9 − 2 = 7 is prime.
  2. That falsifies the claim, so n = 9.
Mark: · log in to save
Problem 17 · 1986 AJHSME Hard
Number Theory factor-outparity-of-product

Let o be an odd whole number and let n be any whole number. Which of the following statements about the whole number (o² + no) is always true?

Show answer
Answer: E — it is odd only if n is even.
Show hints
Hint 1 of 2
Factor: o² + no = o(o + n).
Still stuck? Show hint 2 →
Hint 2 of 2
o is odd, so the product is odd iff (o + n) is odd, which means n is even.
Show solution
Approach: factor and track parity
  1. o(o + n) is odd ⇔ both factors odd. o is already odd, so the product is odd iff o + n is odd, i.e. iff n is even.
  2. So the expression is odd only if n is even — answer E.
Mark: · log in to save
Problem 20 · 1985 AJHSME Hard
Number Theory 31-day-monthweekday-frequency

In a certain year, January had exactly four Tuesdays and four Saturdays. On what day did January 1 fall that year?

Show answer
Answer: C — Wednesday.
Show hints
Hint 1 of 2
31 days = 4 weeks + 3 days. Three weekdays appear 5 times, the other four appear 4 times.
Still stuck? Show hint 2 →
Hint 2 of 2
For Tuesday and Saturday both to land at 4 times, neither can be among the first 3 days of the month.
Show solution
Approach: find which start makes Tue and Sat both rare
  1. The three weekdays starting on Jan 1 each appear 5 times in a 31-day January. For Tue and Sat both to appear only 4 times, neither can be in those first three weekdays.
  2. Only starting on Wednesday (Wed, Thu, Fri) leaves both Tue and Sat out of that group.
Mark: · log in to save
Problem 6 · 2025 AMC 8 Medium
Number Theory divisibilitymod-10

Sekou writes down the numbers 15, 16, 17, 18, 19. After he erases one of his numbers, the sum of the remaining four numbers is a multiple of 4. Which number did he erase?

Show answer
Answer: C — 17.
Show hints
Hint 1 of 2
Adding the five up and testing each removal is slow. What does each number have in common with 4?
Still stuck? Show hint 2 →
Hint 2 of 2
Look at each number's remainder when divided by 4. The remainder of the whole sum tells you which one to erase.
Show solution
Approach: look at remainders mod 4
  1. Remainders mod 4: 15→3, 16→0, 17→1, 18→2, 19→3. Their sum is 9, which leaves remainder 1 mod 4.
  2. To make the remaining four sum divisible by 4, erase the one whose remainder is 1 — that's 17.
Mark: · log in to save
Problem 9 · 2024 AMC 8 Medium
Number Theory divisibilitysubstitution

All of the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?

Show answer
Answer: E — 28 marbles.
Show hints
Hint 1 of 2
Pick one color to stand for a variable, then write the others in terms of it. The total reveals a hidden factor.
Still stuck? Show hint 2 →
Hint 2 of 2
Let r = red count. Green = 2r, blue = 4r. Total = 7r — it must be a multiple of 7.
Show solution
Approach: find the hidden multiple
  1. Let r be the number of red marbles. Half as many red as green → green = 2r. Twice as many blue as green → blue = 4r.
  2. Total = r + 2r + 4r = 7r — always a multiple of 7.
  3. Among the choices, only 28 = 7 × 4 is a multiple of 7.
Mark: · log in to save
Problem 7 · 2018 AMC 8 Medium
Number Theory divisibilitydigit-sum

The 5-digit number 2018U is divisible by 9. What is the remainder when this number is divided by 8?

Show answer
Answer: B — Remainder 3.
Show hints
Hint 1 of 2
Div by 9 ⇔ digit sum div by 9. Find U.
Still stuck? Show hint 2 →
Hint 2 of 2
U = 7 (since 2+0+1+8 = 11; 11+U div by 9 ⇒ U = 7). Then 20187 mod 8.
Show solution
Approach: use the divisibility-by-9 rule
  1. Sum of digits 2+0+1+8+U = 11 + U must be a multiple of 9 (with 0 ≤ U ≤ 9). So U = 7.
  2. 20187 ÷ 8 = 2523 remainder 3 (since 2523 × 8 = 20184). Remainder: 3.
Mark: · log in to save
Problem 7 · 2017 AMC 8 Medium
Number Theory factorizationdivisibility

Let Z be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of Z?

Show answer
Answer: A — 11.
Show hint
Hint 1
Write Z = abcabc as a multiple of abc. The multiplier factors into nice primes.
Show solution
Approach: abcabc = 1001 · abc
  1. Z = abcabc = abc · 1001.
  2. 1001 = 7 · 11 · 13. So 11 is always a factor of Z.
Mark: · log in to save
Problem 8 · 2017 AMC 8 Medium
Number Theory divisibilitycasework

Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."

  1. It is prime.
  2. It is even.
  3. It is divisible by 7.
  4. One of its digits is 9.

This information allows Malcolm to determine Isabella's house number. What is its units digit?

Show answer
Answer: D — 8.
Show hints
Hint 1 of 2
Which two statements can't both be true together?
Still stuck? Show hint 2 →
Hint 2 of 2
A 2-digit number can't be both even and prime (the only even prime is 2). So either (1) or (2) is false — and since 3 of 4 are true, the false one must be (1) "prime".
Show solution
Approach: find the contradiction, then narrow
  1. (1) prime and (2) even can't both be true for a 2-digit number (the only even prime is 2). So the false statement is (1), making (2), (3), (4) all true: the number is even, div by 7, and has a 9 as a digit.
  2. Even + divisible by 7 ⇒ divisible by 14. Two-digit multiples of 14: 14, 28, 42, 56, 70, 84, 98. Only 98 has a 9 digit.
  3. Units digit: 8.
Mark: · log in to save
Problem 9 · 2017 AMC 8 Medium
Number Theory divisibilitycasework

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?

Show answer
Answer: D — 4 yellow marbles.
Show hints
Hint 1 of 2
The total must be divisible by both 3 and 4 — so by 12. Try 12 first; if it fails, try 24.
Still stuck? Show hint 2 →
Hint 2 of 2
12 total: 4 blue + 3 red + 6 green = 13 > 12. Doesn't work. Try 24.
Show solution
Approach: smallest total divisible by lcm(3, 4)
  1. Total must be a multiple of 12 (so that 1/3 and 1/4 are integers).
  2. Try 12: blue = 4, red = 3, green = 6 ⇒ total already 13 > 12. Doesn't fit.
  3. Try 24: blue = 8, red = 6, green = 6 ⇒ yellow = 24 − 8 − 6 − 6 = 4.
Mark: · log in to save
Problem 7 · 2016 AMC 8 Medium
Number Theory perfect-squarefactorization

Which of the following numbers is not a perfect square?

Show answer
Answer: B — 2^2017.
Show hint
Hint 1
Any prime raised to an EVEN power is a perfect square. Any perfect square (like 4) raised to any power is also a perfect square.
Show solution
Approach: even exponent OR squared base = perfect square
  1. 12016 = 1 (trivially a square).
  2. 32018 and 52020: even exponents on a prime → perfect squares.
  3. 42019 = (22)2019 = 24038: even exponent overall → perfect square.
  4. 22017: prime with odd exponent → NOT a perfect square.
Mark: · log in to save
Problem 9 · 2016 AMC 8 Easy
Number Theory factorizationprimes

What is the sum of the distinct prime integer divisors of 2016?

Show answer
Answer: B — 12.
Show hint
Hint 1
Factor 2016 into primes.
Show solution
Approach: prime-factorize and add
  1. 2016 = 25 · 32 · 7.
  2. Distinct primes: 2, 3, 7. Sum = 2 + 3 + 7 = 12.
Mark: · log in to save
Problem 14 · 2015 AMC 8 Medium
Number Theory divisibilityalgebra-from-pattern

Which of the following integers cannot be written as the sum of four consecutive odd integers?

Show answer
Answer: D — 100.
Show hints
Hint 1 of 2
Write the four consecutive odd integers as n, n+2, n+4, n+6 and add them up. Look for a divisibility pattern in the sum.
Still stuck? Show hint 2 →
Hint 2 of 2
The sum is 4n + 12 = 4(n + 3). Since n is odd, n + 3 is even, so the sum is divisible by 8.
Show solution
Approach: the sum must be a multiple of 8
  1. Four consecutive odd integers: n, n+2, n+4, n+6. Sum = 4n + 12 = 4(n + 3).
  2. n is odd, so n + 3 is even ⇒ the sum is a multiple of 8.
  3. Check each choice mod 8: 16, 40, 72, 200 are multiples of 8; 100 is not (100 = 8·12 + 4).
  4. Only 100 fails.
Mark: · log in to save
Problem 13 · 2014 AMC 8 Medium
Number Theory parity

If n and m are integers and n2 + m2 is even, which of the following is impossible?

Show answer
Answer: D — n + m cannot be odd.
Show hint
Hint 1
Squares preserve parity: k2 is even iff k is even. So n2 + m2 even ⇒ n, m have the same parity.
Show solution
Approach: parity of squares matches parity of base
  1. n2 + m2 even ⇒ n2 and m2 have the same parity ⇒ n and m have the same parity.
  2. Same parity ⇒ n + m is always even, never odd.
Mark: · log in to save
Problem 10 · 2013 AMC 8 Medium
Number Theory lcm-gcdprime-factorization

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

Show answer
Answer: C — 330.
Show hint
Hint 1
lcm(a, b) / gcd(a, b) = ab / gcd(a, b)2. Or factor and compare exponents.
Show solution
Approach: prime factor each number
  1. 180 = 22 · 32 · 5. 594 = 2 · 33 · 11.
  2. gcd: 2 · 32 = 18. (Take the minimum exponent of each shared prime.)
  3. lcm: 22 · 33 · 5 · 11 = 4 · 27 · 55 = 5940.
  4. Ratio: 5940 / 18 = 330.
Another way — shortcut:
  1. lcm(a, b) · gcd(a, b) = a · b, so lcm/gcd = ab/gcd2.
  2. 180 · 594 = 106,920. gcd = 18 ⇒ lcm/gcd = 106,920 / 324 = 330.
Mark: · log in to save
Problem 13 · 2013 AMC 8 Medium
Number Theory place-value-differencedivisibility-by-9

When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?

Show answer
Answer: A — 45.
Show hint
Hint 1
If a score's last two digits are 10a + b, reversing gives 10b + a. The difference is 9(a − b): always a multiple of 9.
Show solution
Approach: swap creates a multiple of 9 difference
  1. (10a + b) − (10b + a) = 9(a − b).
  2. The sum's error must therefore be a multiple of 9.
  3. Among the choices, only 45 = 9 · 5 qualifies.
Mark: · log in to save
Problem 12 · 2012 AMC 8 Medium
Number Theory units-digit-cycle

What is the units digit of 132012?

Show answer
Answer: A — 1.
Show hint
Hint 1
Only the units digit of 13 matters: same as 32012. Powers of 3 cycle: 3, 9, 7, 1, 3, 9, 7, 1, … (period 4).
Show solution
Approach: powers of 3 units-digit cycle (3, 9, 7, 1)
  1. Units digit of 13n equals units digit of 3n.
  2. Cycle length 4: 31→3, 32→9, 33→7, 34→1, then repeats.
  3. 2012 ≡ 0 (mod 4) ⇒ units digit = 1.
Mark: · log in to save
Problem 13 · 2012 AMC 8 Medium
Number Theory gcd

Jamar bought some pencils costing more than a penny each at the school bookstore and paid $1.43. Sharona bought some of the same pencils and paid $1.87. How many more pencils did Sharona buy than Jamar?

Show answer
Answer: C — 4 more pencils.
Show hint
Hint 1
Working in cents: the price (in cents) divides both 143 and 187. Take their gcd; since price > 1 cent, only one option survives.
Show solution
Approach: gcd of the two amounts
  1. gcd(143, 187) = 11 (since 143 = 11 · 13, 187 = 11 · 17). Price > 1 cent ⇒ price = 11 cents.
  2. Difference Sharona − Jamar = (187 − 143) / 11 = 44 / 11 = 4.
Mark: · log in to save
Problem 15 · 2012 AMC 8 Medium
Number Theory lcm

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

Show answer
Answer: D — Between 61 and 65.
Show hint
Hint 1
If x ≡ 2 (mod each of 3, 4, 5, 6), then x − 2 is a multiple of all of them, so a multiple of lcm(3, 4, 5, 6) = 60.
Show solution
Approach: shift to a multiple-of-LCM problem
  1. x − 2 divisible by 3, 4, 5, 6 ⇒ divisible by lcm = 60.
  2. Smallest such x > 2 is x = 60 + 2 = 62, which lies between 61 and 65.
Mark: · log in to save
Problem 18 · 2012 AMC 8 Medium
Number Theory product-of-distinct-primes

What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?

Show answer
Answer: A — 3127.
Show hints
Hint 1 of 2
Not prime & no factor < 50 ⇒ product of at least two primes, all ≥ 50. Not a square ⇒ the primes are distinct.
Still stuck? Show hint 2 →
Hint 2 of 2
Smallest two primes ≥ 50 are 53 and 59.
Show solution
Approach: smallest two primes ≥ 50
  1. First primes ≥ 50: 53, 59, 61, …
  2. Smallest product of two distinct such primes: 53 · 59 = 3127.
Mark: · log in to save
Problem 17 · 2011 AMC 8 Medium
Number Theory prime-factorization

Let w, x, y, and z be whole numbers. If 2w · 3x · 5y · 7z = 588, then what does 2w + 3x + 5y + 7z equal?

Show answer
Answer: A — 21.
Show hint
Hint 1
Factor 588 into primes: 588 = 4 · 147 = 4 · 3 · 49.
Show solution
Approach: prime factorization
  1. 588 = 22 · 31 · 50 · 72 ⇒ w = 2, x = 1, y = 0, z = 2.
  2. 2w + 3x + 5y + 7z = 4 + 3 + 0 + 14 = 21.
Mark: · log in to save
Problem 24 · 2011 AMC 8 Medium
Number Theory parityprimes

In how many ways can 10001 be written as the sum of two primes?

Show answer
Answer: A — 0 ways.
Show hint
Hint 1
Odd sum ⇒ one prime must be even ⇒ that prime is 2. Then check 10001 − 2 = 9999.
Show solution
Approach: force one prime to be 2 and check primality of 9999
  1. 10001 is odd; primes summing to odd require one to be even, so it must be 2.
  2. Then the other is 10001 − 2 = 9999 = 3 · 3333 (digit sum 36 divisible by 3). Not prime.
  3. So 0 representations.
Mark: · log in to save
Problem 11 · 2009 AMC 8 Medium
Number Theory gcd

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43. Some of the 30 sixth graders each bought a pencil, and they paid a total of $1.95. How many more sixth graders than seventh graders bought a pencil?

Show answer
Answer: D — 4 more.
Show hint
Hint 1
Working in cents, the price divides both 143 and 195. Compute gcd(143, 195).
Show solution
Approach: find the common price via gcd
  1. 143 = 11 · 13; 195 = 3 · 5 · 13. So price | 13 and price > 1 ⇒ price = 13¢.
  2. Seventh graders: 143/13 = 11. Sixth graders: 195/13 = 15.
  3. Difference: 15 − 11 = 4.
Mark: · log in to save
Problem 15 · 2008 AMC 8 Medium
Number Theory divisibilityaverage-as-integer

In Theresa's first 8 basketball games, she scored 7, 4, 3, 6, 8, 3, 1 and 5 points. In her ninth game, she scored fewer than 10 points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than 10 points and her points-per-game average for the 10 games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

Show answer
Answer: B — 40.
Show hints
Hint 1 of 2
Sum of first 8: 37. After game 9 (score < 10), total is between 37 and 47; must be a multiple of 9 (mean integer).
Still stuck? Show hint 2 →
Hint 2 of 2
Then after game 10 the total is < 56 and a multiple of 10.
Show solution
Approach: fit each total to the divisibility condition
  1. Sum after 8: 37.
  2. After 9: total in [38, 47], divisible by 9 ⇒ 45. Game 9 = 8.
  3. After 10: total in [46, 55], divisible by 10 ⇒ 50. Game 10 = 5.
  4. Product: 8 · 5 = 40.
Mark: · log in to save
Problem 22 · 2008 AMC 8 Medium
Number Theory range-of-integers

For how many positive integer values of n are both n3 and 3n three-digit whole numbers?

Show answer
Answer: A — 12.
Show hint
Hint 1
Let x = n/3 (so n = 3x). Then 3n = 9x. Both x and 9x are 3-digit.
Show solution
Approach: substitute and find the integer range
  1. 100 ≤ x ≤ 999 and 100 ≤ 9x ≤ 999.
  2. The second is the binding constraint: x ≤ 111.
  3. Combined with x ≥ 100: x ∈ {100, 101, …, 111} — 12 values.
Mark: · log in to save
Problem 18 · 2007 AMC 8 Medium
Number Theory last-digit

The product of the two 99-digit numbers 303,030,303,…,030,303 and 505,050,505,…,050,505 has thousands digit A and units digit B. What is the sum of A and B?

Show answer
Answer: D — 8.
Show hint
Hint 1
The last 4 digits of each factor are 0303 and 0505. Multiply those mod 10000.
Show solution
Approach: compute the last four digits
  1. 303 · 505 = 153015.
  2. Last 4 digits: 3015 ⇒ thousands digit A = 3, units digit B = 5.
  3. A + B = 8.
Mark: · log in to save
Problem 11 · 2006 AMC 8 Medium
Number Theory digit-sumcasework

How many two-digit numbers have digits whose sum is a perfect square?

Show answer
Answer: C — 17.
Show hints
Hint 1 of 2
Digit sums of two-digit numbers range 1–18. Perfect squares in range: 1, 4, 9, 16.
Still stuck? Show hint 2 →
Hint 2 of 2
Count two-digit numbers with each of these digit sums.
Show solution
Approach: case on perfect-square digit sum
  1. Sum = 1: {10} ⇒ 1.
  2. Sum = 4: {13, 22, 31, 40} ⇒ 4.
  3. Sum = 9: {18, 27, 36, 45, 54, 63, 72, 81, 90} ⇒ 9.
  4. Sum = 16: {79, 88, 97} ⇒ 3.
  5. Total: 1 + 4 + 9 + 3 = 17.
Mark: · log in to save
Problem 23 · 2006 AMC 8 Medium
Number Theory chinese-remainder-by-listing

A box contains gold coins. If the coins are equally divided among six people, four coins are left over. If the coins are equally divided among five people, three coins are left over. If the box holds the smallest number of coins that meets these two conditions, how many coins are left when equally divided among seven people?

Show answer
Answer: A — 0.
Show hint
Hint 1
List numbers ≡ 4 (mod 6) and find the first one that's also ≡ 3 (mod 5).
Show solution
Approach: intersect two congruences
  1. ≡ 4 (mod 6): 4, 10, 16, 22, 28, …
  2. ≡ 3 (mod 5): 3, 8, 13, 18, 23, 28, …
  3. First common: 28. 28 / 7 = 4 remainder 0.
Mark: · log in to save
Problem 19 · 2004 AMC 8 Medium
Number Theory lcm

A whole number larger than 2 leaves a remainder of 2 when divided by each of the numbers 3, 4, 5, and 6. The smallest such number lies between which two numbers?

Show answer
Answer: B — Between 60 and 79.
Show hint
Hint 1
x − 2 is divisible by lcm(3, 4, 5, 6) = 60.
Show solution
Approach: shift by 2 and use LCM
  1. Smallest x > 2 with x − 2 = 60 ⇒ x = 62.
  2. 62 lies between 60 and 79.
Mark: · log in to save
Problem 9 · 2000 AMC 8 Medium
Number Theory powerscareful-counting
amc8-2000-09
Show answer
Answer: D — 6.
Show hints
Hint 1 of 2
List all the three-digit powers of 5, and of 2.
Still stuck? Show hint 2 →
Hint 2 of 2
The shared crossing digit forces which power of 2 fits.
Show solution
Approach: only a few three-digit powers exist
  1. Three-digit powers of 5 are only 125 and 625; three-digit powers of 2 are 128, 256, and 512.
  2. The crossing forces the power of 2 to be 256, so the outlined square holds its units digit, 6.
Mark: · log in to save
Problem 11 · 2000 AMC 8 Medium
Number Theory divisibilitycasework

The number 64 has the property that it is divisible by its units digit. How many whole numbers between 10 and 50 have this property?

Show answer
Answer: C — 17.
Show hints
Hint 1 of 2
Group the numbers by their units digit and test divisibility.
Still stuck? Show hint 2 →
Hint 2 of 2
Units digit 1, 2, or 5 always works; 0 never does (no division by 0).
Show solution
Approach: casework on the units digit
  1. Units digit 1, 2, and 5 each give 4 working numbers (12 total). Then 33 (digit 3), 24 and 44 (digit 4), 36 (digit 6), and 48 (digit 8) work; digits 0, 7, 9 give none.
  2. Total = 12 + 1 + 2 + 1 + 1 = 17.
Mark: · log in to save
Problem 5 · 1997 AJHSME Medium
Number Theory digit-sumlisting

There are many two-digit multiples of 7, but only two of them have a digit sum of 10. The sum of these two multiples of 7 is

Show answer
Answer: A — 119.
Show hints
Hint 1 of 2
List the two-digit multiples of 7 and check each digit sum.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the two whose digits add to 10.
Show solution
Approach: scan multiples of 7 for digit sum 10
  1. Among 14, 21, 28, …, 98, the ones with digit sum 10 are 28 and 91.
  2. Their sum is 28 + 91 = 119.
Mark: · log in to save
Problem 7 · 1996 AJHSME Medium
Number Theory powersexponential

Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has 4 goldfish at the same time that Gretel has 128 goldfish, in how many months from that time will they have the same number of goldfish?

Show answer
Answer: B — 5 months.
Show hints
Hint 1 of 2
Write both counts as powers of 2 and match the exponents.
Still stuck? Show hint 2 →
Hint 2 of 2
Brent: 4 · 4^m; Gretel: 128 · 2^m.
Show solution
Approach: equate powers of 2
  1. After m months Brent has 4 · 4^m = 2^(2m+2) and Gretel has 128 · 2^m = 2^(m+7).
  2. Setting 2m + 2 = m + 7 gives m = 5.
Mark: · log in to save
Problem 6 · 1994 AJHSME Medium
Number Theory divisibility

The units digit (one's digit) of the product of any six consecutive positive whole numbers is

Show answer
Answer: A — 0.
Show hints
Hint 1 of 2
Among six consecutive numbers, what's guaranteed to appear?
Still stuck? Show hint 2 →
Hint 2 of 2
A multiple of 5 and an even number together make a multiple of 10.
Show solution
Approach: the product is a multiple of 10
  1. Any six consecutive numbers include a multiple of 5 and at least one even number.
  2. Their product is therefore a multiple of 10, so its units digit is 0.
Mark: · log in to save
Problem 3 · 1993 AJHSME Medium
Number Theory prime-factorization

Which of the following numbers has the largest prime factor?

Show answer
Answer: B — 51.
Show hints
Hint 1 of 2
Factor each number into primes.
Still stuck? Show hint 2 →
Hint 2 of 2
Then compare the biggest prime in each.
Show solution
Approach: factor and compare largest primes
  1. 39 = 3·13, 51 = 3·17, 77 = 7·11, 91 = 7·13, 121 = 11².
  2. The largest prime factor among these is 17, from 51.
Mark: · log in to save
Problem 7 · 1993 AJHSME Medium
Number Theory exponent-rules

33 + 33 + 33 =

Show answer
Answer: A — 3⁴.
Show hints
Hint 1 of 2
Three copies of the same thing is 3 times it.
Still stuck? Show hint 2 →
Hint 2 of 2
3 × 3³ = 3¹ × 3³.
Show solution
Approach: factor out the common power
  1. 3³ + 3³ + 3³ = 3 × 3³.
  2. Adding exponents, 3 × 3³ = 3⁴.
Mark: · log in to save
Problem 1 · 1990 AJHSME Medium
Number Theory place-valueminimize

What is the smallest sum of two 3-digit numbers that can be obtained by placing each of the six digits 4, 5, 6, 7, 8, 9 in one of the six boxes in this addition problem?

   
+     
 
Show answer
Answer: C — 1047.
Show hints
Hint 1 of 2
The hundreds digits matter most, so put the two smallest there.
Still stuck? Show hint 2 →
Hint 2 of 2
Then the next-smallest in the tens, and the largest in the units.
Show solution
Approach: small digits into the big place values
  1. Put 4 and 5 in the hundreds (900), 6 and 7 in the tens (130), and 8 and 9 in the units (17).
  2. The smallest sum is 900 + 130 + 17 = 1047.
Mark: · log in to save
Problem 4 · 1990 AJHSME Medium
Number Theory units-digit-of-squares

Which of the following could not be the units digit (one's digit) of the square of a whole number?

Show answer
Answer: E — 8.
Show hint
Hint 1
The last digit of n² depends only on the last digit of n — there are just 10 cases to check.
Show solution
Approach: list the possible last digits of squares
  1. Squares end only in 0, 1, 4, 5, 6, or 9.
  2. 8 is not on that list, so a square can never end in 8.
Mark: · log in to save
Problem 10 · 1988 AJHSME Medium
Number Theory mod-7

Chris's birthday is on a Thursday this year. What day of the week will it be 60 days after her birthday?

Show answer
Answer: A — Monday.
Show hints
Hint 1 of 2
Days of the week repeat every 7 days, so reduce 60 modulo 7.
Still stuck? Show hint 2 →
Hint 2 of 2
60 = 8·7 + 4.
Show solution
Approach: reduce mod 7
  1. 60 days = 8 full weeks plus 4 extra days.
  2. 4 days after Thursday → Friday, Saturday, Sunday, Monday.
Mark: · log in to save
Problem 8 · 1987 AJHSME Medium
Number Theory bound-the-sum
ajhsme-1987-08
Show answer
Answer: B — 5.
Show hints
Hint 1 of 2
Check the smallest and largest possible sums to see whether the digit count can change.
Still stuck? Show hint 2 →
Hint 2 of 2
Smallest: A = B = 1. Largest: A = B = 9.
Show solution
Approach: bracket the sum
  1. Smallest: 9876 + 132 + 11 = 10019. Largest: 9876 + 932 + 91 = 10899.
  2. Both have 5 digits, so the sum always does.
Mark: · log in to save
Problem 9 · 1987 AJHSME Medium
Number Theory lcm-from-prime-factors

When finding the sum 1⁄2 + 1⁄3 + 1⁄4 + 1⁄5 + 1⁄6 + 1⁄7, the least common denominator used is

Show answer
Answer: C — 420.
Show hints
Hint 1 of 2
Take the highest power of each prime in 2, 3, 4, 5, 6, 7.
Still stuck? Show hint 2 →
Hint 2 of 2
4 contributes 2², the rest contribute 3, 5, and 7 once.
Show solution
Approach: take the highest power of each prime
  1. Primes appearing: 2² (from 4), 3, 5, 7.
  2. LCM = 4 × 3 × 5 × 7 = 420.
Mark: · log in to save
Problem 7 · 1986 AJHSME Medium
Number Theory bound-square-roots

How many whole numbers are between √8 and √80?

Show answer
Answer: B — 6.
Show hints
Hint 1 of 2
√8 sits between 2 and 3; √80 sits between 8 and 9.
Still stuck? Show hint 2 →
Hint 2 of 2
Count the integers strictly between.
Show solution
Approach: bound each square root by the nearest perfect squares
  1. 4 < 8 < 9, so 2 < √8 < 3. 64 < 80 < 81, so 8 < √80 < 9.
  2. Whole numbers strictly between √8 and √80 are 3, 4, 5, 6, 7, 8 — that's 6.
Mark: · log in to save
Problem 8 · 1986 AJHSME Medium
Number Theory last-digitguess-and-check
ajhsme-1986-08
Show answer
Answer: E — 8.
Show hints
Hint 1 of 2
The ones digit of the product is 2 · B, which must end in 6.
Still stuck? Show hint 2 →
Hint 2 of 2
Then check that B2 × 7B actually equals 6396.
Show solution
Approach: ones-digit clue then verify
  1. Product ends in 6, and 2 × B ends in 6 only for B = 3 or B = 8.
  2. Test: 82 × 78 = 6396. ✓ So B = 8.
Mark: · log in to save
Problem 24 · 2026 AMC 8 Stretch
Number Theory legendre-formula

The notation n! is the product of the first n positive integers. Define the superfactorial of n to be the product of the factorials 1! · 2! · 3! · … · n! (so the superfactorial of 3 is 1! · 2! · 3! = 12). How many factors of 7 appear in the prime factorization of the superfactorial of 51?

Show answer
Answer: E — 171.
Show hints
Hint 1 of 2
The 7-count of the superfactorial is the sum of the 7-counts of 1! through 51!.
Still stuck? Show hint 2 →
Hint 2 of 2
Each k! holds ⌊k/7⌋ + ⌊k/49⌋ sevens; group the values of k by that count.
Show solution
Approach: sum Legendre's 7-count over every factorial
  1. The number of 7s is the sum of v₇(k!) for k = 1 to 51, where v₇(k!) = ⌊k/7⌋ + ⌊k/49⌋.
  2. Seven values of k each contribute 1, 2, 3, 4, 5, and 6, totaling 7(1+2+3+4+5+6) = 147; and k = 49, 50, 51 contribute 8 each, adding 24.
  3. The total is 147 + 24 = 171.
Mark: · log in to save
Problem 22 · 2025 AMC 8 Hard
Number Theory factorizationfactor-pairs
amc8-2025-22
Show answer
Answer: D — 7 different coat counts.
Show hints
Hint 1 of 2
Imagine adding a phantom coat after the last one. Now the row is a clean repeating pattern.
Still stuck? Show hint 2 →
Hint 2 of 2
After the phantom, you have 36 hooks split into d identical blocks of b hooks each, with b ≥ 2 and d ≥ 2. How many (b, d) factorizations of 36 are there?
Show solution
Approach: add a phantom coat, count factor pairs
  1. Add a phantom coat after the last real coat. Now 36 hooks form a perfectly repeating pattern: each block is (some empty hooks) + (one coat), of length b ≥ 2.
  2. If there are d blocks, then bd = 36, and the real coat count = d − 1 (we added one).
  3. Both b ≥ 2 (each block has ≥ 1 empty + 1 coat) and d ≥ 2 (at least 1 original coat + the phantom).
  4. 36 = 22 × 32 has (2+1)(2+1) = 9 divisors. Removing the two ordered factorizations with a 1 (1×36 and 36×1) leaves 9 − 2 = 7.
Mark: · log in to save
Problem 23 · 2025 AMC 8 Hard
Number Theory primesdifference-of-squaresprime-test

How many four-digit numbers have all three of the following properties?

  1. The tens digit and ones digit are both 9.
  2. The number is 1 less than a perfect square.
  3. The number is the product of exactly two prime numbers.
Show answer
Answer: B — Exactly 1.
Show hints
Hint 1 of 2
The number ends in 99, so the perfect square just above it ends in 00. What does that say about its square root?
Still stuck? Show hint 2 →
Hint 2 of 2
Use a2 − 1 = (a − 1)(a + 1). For the number to be a product of exactly two primes, both factors must be prime — twin primes.
Show solution
Approach: a² ends in 00, then look for twin primes around a
  1. A 4-digit number ending in 99 plus 1 ends in 00. So that perfect square = (10k)2, and our number is (10k)2 − 1 = (10k − 1)(10k + 1).
  2. Four-digit range gives 10k ∈ {40, 50, 60, 70, 80, 90, 100}.
  3. We need both 10k − 1 and 10k + 1 prime. Check: 39×41 (39 = 3×13, no), 49×51 (49 = 72, no), 59×61 (both prime ✓), 69×71 (69 = 3×23, no), 79×81 (81 = 34, no), 89×91 (91 = 7×13, no), 99×101 (99 = 9×11, no).
  4. Only 59 × 61 = 3599 works. Exactly 1.
Mark: · log in to save
Problem 23 · 2024 AMC 8 Hard
Number Theory factorizationgridgrid-counting
amc8-2024-23
Show answer
Answer: C — 7000 cells.
Show hints
Hint 1 of 2
The example (0,4) to (2,0) crosses 4 cells. Notice: 2 + 4 − gcd(2,4) = 4. That's a general formula.
Still stuck? Show hint 2 →
Hint 2 of 2
Apply the formula to (3000, 5000) horizontal/vertical offsets: 3000 + 5000 − gcd(3000, 5000).
Show solution
Approach: use the lattice-line cell-count formula
  1. A line segment whose horizontal offset is a and vertical offset is b crosses a + b − gcd(a, b) grid cells. (You'd cross a + b cells if the line never hit a grid corner; each lattice-point crossing collapses two cell-entries into one, saving 1 per shared factor.)
  2. From (2000, 3000) to (5000, 8000): horizontal offset 3000, vertical offset 5000.
  3. gcd(3000, 5000) = 1000. Cells = 3000 + 5000 − 1000 = 7000.
Another way — scale down (MAA):
  1. The slope from (2000, 3000) to (5000, 8000) is 5/3. The segment is equivalent to 1000 copies of a primitive (0,0)→(3,5) piece, since gcd(3000, 5000) = 1000.
  2. Each primitive (3,5) segment crosses 7 cells. Total: 7 × 1000 = 7000.
Mark: · log in to save
Problem 22 · 2023 AMC 8 Hard
Number Theory factorizationsubstitution

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term in the sequence is 4000. What is the first term?

Show answer
Answer: D — 5.
Show hints
Hint 1 of 2
Write the first six terms as a, b, and powers of a and b. The 6th term will be a3b5.
Still stuck? Show hint 2 →
Hint 2 of 2
Factor 4000: 4000 = 53 × 25. Match exponents.
Show solution
Approach: track exponents of a and b through the sequence
  1. Let the first two terms be a, b. Then the next four are ab, ab2, a2b3, a3b5. (Each term sums the exponents of the previous two.)
  2. a3b5 = 4000 = 53 × 25. So a = 5, b = 2.
  3. First term: 5.
Mark: · log in to save
Problem 22 · 2020 AMC 8 Hard
Number Theory work-backwardcasework

When a positive integer N is fed into a machine, the output is calculated by the rule: if N is even, output N/2; if N is odd, output 3N + 1. Example: 7 → 22 → 11 → 34 → 17 → 52 → 26. When the same 6-step process is applied to a different starting N, the final output is 1. What is the sum of all such integers N?

Show answer
Answer: E — Sum is 83.
Show hints
Hint 1 of 2
Work backward from 1. The predecessors of any k are: 2k (always), and (k − 1)/3 (only if that's an odd integer).
Still stuck? Show hint 2 →
Hint 2 of 2
Build the inverse-tree six levels up from 1. The level-6 leaves are the valid starting values.
Show solution
Approach: invert the machine, walk backward 6 steps from 1
  1. Forward: even → halve, odd → 3n+1. Inverting from a value k: predecessors are 2k (always), and (k−1)/3 (only if that's an odd integer).
  2. Walk backward from 1: 1 → {2} → {4} → {8, 1} → {16, 2} → {32, 5, 4} → {64, 10, 8, 1}.
  3. Sum: 64 + 10 + 8 + 1 = 83.
Mark: · log in to save
Problem 21 · 2018 AMC 8 Hard
Number Theory divisibilitycasework

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

Show answer
Answer: E — 5 integers.
Show hints
Hint 1 of 2
Each remainder is exactly 4 less than the divisor (2 = 6−4, 5 = 9−4, 7 = 11−4). So x ≡ −4 mod each of them.
Still stuck? Show hint 2 →
Hint 2 of 2
x + 4 is divisible by lcm(6, 9, 11) = 198. Count multiples of 198 with x + 4 in [104, 1003].
Show solution
Approach: spot the common shift, then use lcm
  1. 2 = 6 − 4, 5 = 9 − 4, 7 = 11 − 4 ⇒ x ≡ −4 mod 6, 9, 11 simultaneously.
  2. So x + 4 is a multiple of lcm(6, 9, 11) = 2 · 32 · 11 = 198.
  3. 100 ≤ x ≤ 999 ⇒ 104 ≤ x + 4 ≤ 1003. Multiples of 198 in that range: 198, 396, 594, 792, 990. 5 values.
Mark: · log in to save
Problem 25 · 2018 AMC 8 Hard
Number Theory perfect-cubeestimate-and-pick

How many perfect cubes lie between 28 + 1 and 218 + 1, inclusive?

Show answer
Answer: E — 58 cubes.
Show hints
Hint 1 of 2
218 = (26)3 = 643. So cubes ≤ 218 + 1 means base ≤ 64.
Still stuck? Show hint 2 →
Hint 2 of 2
28 = 256, 28 + 1 = 257. The next cube above 257 is 73 = 343 (since 63 = 216).
Show solution
Approach: bracket the cube-root bounds
  1. Upper end: 218 = (26)3 = 643, so 643 ≤ 218 + 1 ✓. Cubes at base ≥ 65 exceed the range.
  2. Lower end: 63 = 216 < 257 = 28 + 1, but 73 = 343 ≥ 257 ✓. Smallest valid base: 7.
  3. Count integers from 7 to 64 inclusive: 64 − 7 + 1 = 58.
Mark: · log in to save
Problem 23 · 2017 AMC 8 Hard
Number Theory factorizationarithmetic-sequence

Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?

Show answer
Answer: C — 25 miles.
Show hints
Hint 1 of 2
Each day's min-per-mile must divide 60 (so that 60 min yields an integer mile count). Find four divisors of 60 in arithmetic progression with common difference 5.
Still stuck? Show hint 2 →
Hint 2 of 2
Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. The only arithmetic-progression-5 of length 4 is 5, 10, 15, 20.
Show solution
Approach: min-per-mile must divide 60
  1. 60 minutes total per day; for integer miles, the min-per-mile divides 60.
  2. Four divisors of 60 forming an AP with common difference 5: 5, 10, 15, 20.
  3. Miles each day: 60/5 + 60/10 + 60/15 + 60/20 = 12 + 6 + 4 + 3 = 25.
Mark: · log in to save
Problem 24 · 2017 AMC 8 Hard
Number Theory complementary-countingdivisibility

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

Show answer
Answer: D — 146 days.
Show hints
Hint 1 of 2
The pattern repeats every lcm(3, 4, 5) = 60 days. Count call-days in 60, then scale to 365 (with the leftover 5 days handled).
Still stuck? Show hint 2 →
Hint 2 of 2
Inclusion–exclusion in a 60-day block: 20 + 15 + 12 − 5 − 4 − 3 + 1 = 36 call days. So 60 − 36 = 24 no-call days.
Show solution
Approach: inclusion-exclusion over a 60-day cycle, then handle leftovers
  1. lcm(3, 4, 5) = 60, so the call pattern repeats every 60 days. In one 60-day block:
  2. Calls by inclusion-exclusion: 20 (every 3) + 15 (every 4) + 12 (every 5) − 5 (every 12) − 4 (every 15) − 3 (every 20) + 1 (every 60) = 36 call days. So 60 − 36 = 24 no-call days per block.
  3. 365 days = 6 full 60-day blocks + 5 extra days. No-call days from blocks: 6 × 24 = 144.
  4. In the last 5 days (days 361–365 ≡ days 1–5 of a new cycle): day 1 no call, day 2 no call, day 3 call, day 4 call, day 5 call. So 2 more no-call days.
  5. Total: 144 + 2 = 146.
Mark: · log in to save
Problem 24 · 2016 AMC 8 Hard
Number Theory divisibilitycasework

The digits 1, 2, 3, 4, and 5 are each used once to write a five-digit number PQRST. The three-digit number PQR is divisible by 4, the three-digit number QRS is divisible by 5, and the three-digit number RST is divisible by 3. What is P?

Show answer
Answer: A — P = 1.
Show hints
Hint 1 of 2
QRS div by 5 means S ends in 0 or 5; the only digit available is 5. So S = 5.
Still stuck? Show hint 2 →
Hint 2 of 2
PQR div by 4 means QR is divisible by 4. With S = 5 used, QR is a 2-digit from {1, 2, 3, 4}. Options: 12, 24, 32.
Show solution
Approach: narrow each divisibility constraint in order
  1. S = 5 (the only digit from {1,2,3,4,5} that makes QRS end in 0 or 5).
  2. QR must be a 2-digit number from {1, 2, 3, 4} divisible by 4: {12, 24, 32}.
  3. Test each. QR = 12: leftover {3, 4} for P, T. RST = 25T, digit sum 7 + T; T ∈ {3, 4} gives 10 or 11, neither div by 3.
  4. QR = 24: leftover {1, 3}. RST = 45T, digit sum 9 + T; T = 3 gives 12 (div 3) ✓. So Q = 2, R = 4, T = 3, P = 1. Number: 12435.
  5. QR = 32: leftover {1, 4}. RST = 25T sum 7 + T; T ∈ {1, 4} gives 8 or 11, neither div by 3.
Mark: · log in to save
Problem 25 · 2001 AMC 8 Stretch
Number Theory divisibilitycareful-counting

There are 24 four-digit whole numbers that use each of the four digits 2, 4, 5, and 7 exactly once. Only one of these four-digit numbers is a multiple of another one. Which of the following is it?

Show answer
Answer: D — 7425.
Show hints
Hint 1 of 2
The multiple still uses the digits 2, 4, 5, 7, so it stays between 2457 and 7542.
Still stuck? Show hint 2 →
Hint 2 of 2
That keeps the factor tiny: 2457 × 4 is already too big, so test multiplying by 3.
Show solution
Approach: the only feasible factor is 3
  1. Any such multiple uses the same four digits, so it lies between 2457 and 7542; since 2457 × 4 ≈ 9828 overshoots, the factor can only be 2 or 3.
  2. No doubling of a number in the set lands back in the set, but 2475 × 3 = 7425, which uses exactly 2, 4, 5, 7.
  3. So the number is 7425 (= 3 × 2475).
Mark: · log in to save
Problem 24 · 1999 AMC 8 Stretch
Number Theory units-digitcyclicity

When 19992000 is divided by 5, the remainder is

Show answer
Answer: D — 1.
Show hints
Hint 1 of 2
Only the units digit of 1999 matters for the units digit of the power.
Still stuck? Show hint 2 →
Hint 2 of 2
Powers of 9 end in 9, 1, 9, 1, … — and an even exponent lands on 1.
Show solution
Approach: track the units digit's cycle
  1. Powers of a number ending in 9 cycle 9, 1, 9, 1, …; since 2000 is even, 1999²⁰⁰⁰ ends in 1.
  2. A number ending in 1 leaves remainder 1 when divided by 5.
Mark: · log in to save
Problem 24 · 1998 AJHSME Stretch
Number Theory triangular-numbersmod-arithmetic

A board of 8 columns has squares numbered left to right, top to bottom (row one is 1–8, row two is 9–16, and so on). A student shades square 1, then skips one and shades square 3, skips two and shades square 6, skips three and shades square 10, and continues this way until every column has at least one shaded square. What is the number of the shaded square that first achieves this?

Show answer
Answer: E — 120.
Show hints
Hint 1 of 2
The shaded squares are the triangular numbers 1, 3, 6, 10, 15, … .
Still stuck? Show hint 2 →
Hint 2 of 2
A square's column is set by its value mod 8 — you need all 8 remainders to appear.
Show solution
Approach: triangular numbers mod 8; wait for remainder 0
  1. The shaded squares are triangular numbers, and a square's column depends on its value mod 8.
  2. Every remainder shows up early except 0; the first triangular number divisible by 8 is 120, so the last column fills at square 120.
Mark: · log in to save
Problem 23 · 1997 AJHSME Stretch
Number Theory boundingcasework

Some positive integers have both properties: (I) the sum of the squares of their digits is 50, and (II) each digit is larger than the one to its left. The product of the digits of the largest such integer is

Show answer
Answer: C — 36.
Show hints
Hint 1 of 2
The digits strictly increase, so they're distinct; squaring shows there can be at most four of them.
Still stuck? Show hint 2 →
Hint 2 of 2
To make the number large, push the leading digits up while keeping the square-sum at 50.
Show solution
Approach: bound the digit count, then maximize
  1. Five increasing digits would have squares summing to at least 1+4+9+16+25 = 55 > 50, so at most four digits.
  2. The digits 1, 2, 3, 6 give 1 + 4 + 9 + 36 = 50 and form the largest such number, 1236.
  3. Its digit product is 1 × 2 × 3 × 6 = 36.
Mark: · log in to save
Problem 25 · 1997 AJHSME Stretch
Number Theory units-digitcyclicity

All the even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the units digit of the product?

Show answer
Answer: D — 6.
Show hints
Hint 1 of 2
Only units digits matter; each block of ten contributes the units 2, 4, 6, 8.
Still stuck? Show hint 2 →
Hint 2 of 2
2 × 4 × 6 × 8 ends in 4, and there are ten blocks — find the units digit of 4¹⁰.
Show solution
Approach: group by tens, then find the units of a power
  1. Each block of ten (like 2, 4, 6, 8) multiplies to a units digit of 4, and there are 10 such blocks, so we need the units digit of 4¹⁰.
  2. 4² ends in 6, and any power of 6 ends in 6, so 4¹⁰ = (4²)⁵ ends in 6.
Mark: · log in to save
Problem 23 · 1994 AJHSME Stretch
Number Theory place-valuemaximize
ajhsme-1994-23
Show answer
Answer: D — Form YYZ.
Show hints
Hint 1 of 2
The sum equals 113·X + 10·Y; keep it to three digits while making it large.
Still stuck? Show hint 2 →
Hint 2 of 2
Find the digits of that largest sum and match them to X, Y, Z.
Show solution
Approach: express the sum, then maximize within three digits
  1. Adding XXX + YX + X gives 113·X + 10·Y. To stay three digits, X ≤ 8; taking X = 8, Y = 9 gives 904 + 90 = 994.
  2. 994 reads as 9, 9, 4 = Y, Y, (a new digit Z), so the form is YYZ.
Mark: · log in to save
Problem 25 · 1994 AJHSME Stretch
Number Theory small-casespattern

Find the sum of the digits in the answer to

9999…9994 nines×4444…4494 fours

where a string of 94 nines is multiplied by a string of 94 fours.

Show answer
Answer: A — 846.
Show hints
Hint 1 of 2
Try small versions: 9 × 4, 99 × 44, 999 × 444, and find the digit-sum pattern.
Still stuck? Show hint 2 →
Hint 2 of 2
The digit sum turns out to be 9 times the number of nines.
Show solution
Approach: spot the pattern from small cases
  1. 9·4 = 36 (digit sum 9), 99·44 = 4356 (18), 999·444 = 443556 (27): the digit sum is 9 × (number of nines).
  2. With 94 nines, the digit sum is 9 × 94 = 846.
Mark: · log in to save
Problem 20 · 1993 AJHSME Stretch
Number Theory borrowing-patterndigit-sum

When 1093 − 93 is expressed as a single whole number, the sum of the digits is

Show answer
Answer: D — 826.
Show hints
Hint 1 of 2
Try a small case like 10⁴ − 93 = 9907 to see the shape.
Still stuck? Show hint 2 →
Hint 2 of 2
10ⁿ − 93 is a string of nines ending in 07.
Show solution
Approach: find the digit pattern from a small case
  1. 10⁴ − 93 = 9907, 10⁵ − 93 = 99907 — so 10ⁿ − 93 is (n − 2) nines followed by 07.
  2. For n = 93 that's 91 nines and a 0 and 7: 91 × 9 + 7 = 826.
Mark: · log in to save
Problem 24 · 1993 AJHSME Stretch
Number Theory patternrows

The figure below shows a triangular ‘staircase’ array of numbers. The first row has 1 number, the second row has 3, the third row has 5, and so on (the kth row has 2k−1 numbers, in order).

1
2  3  4
5  6  7  8  9
10  11  12  13  14  15  16
…

What number is directly above 142 in this array of numbers?

Show answer
Answer: C — 120.
Show hints
Hint 1 of 2
Row k ends at the perfect square k², and holds 2k − 1 numbers.
Still stuck? Show hint 2 →
Hint 2 of 2
Find which row holds 142, then the number sitting one row up and aligned with it.
Show solution
Approach: use the row structure of the triangle
  1. Rows end at 1, 4, 9, 16, …, k², so 142 is in row 12 (122–144), as its 21st of 23 entries.
  2. Row 11 (101–121) sits centered above, so directly above the 21st entry is the 20th entry of row 11: 101 + 19 = 120.
Mark: · log in to save
Problem 20 · 1991 AJHSME Stretch
Number Theory cryptarithmplace-value
ajhsme-1991-20
Show answer
Answer: A — 1.
Show hints
Hint 1 of 2
Add the column-place values: ABC + AB + A = 111·A + 11·B + C.
Still stuck? Show hint 2 →
Hint 2 of 2
Set that equal to 300 and find digits A, B, C.
Show solution
Approach: expand by place value and solve
  1. 111A + 11B + C = 300. Taking A = 2 gives 11B + C = 78, so B = 7 and C = 1 (271 + 27 + 2 = 300).
  2. All digits differ, so C = 1.
Mark: · log in to save
Problem 11 · 1990 AJHSME Stretch
Number Theory consecutiveopposite-pairs
ajhsme-1990-11
Show answer
Answer: E — 81.
Show hints
Hint 1 of 2
The six faces are six consecutive numbers including the visible 11, 14, 15.
Still stuck? Show hint 2 →
Hint 2 of 2
The visible faces meet at a corner, so they can't be opposite each other — that pins which six numbers.
Show solution
Approach: use the equal-opposite-sums to fix the six numbers
  1. The visible 11, 14, 15 are mutually adjacent, so they aren't opposite pairs. The six consecutive numbers must be 11–16, pairing as 11+16, 12+15, 13+14 (each summing to 27).
  2. Their total is 11 + 12 + 13 + 14 + 15 + 16 = 81.
Mark: · log in to save
Problem 22 · 1990 AJHSME Stretch
Number Theory mod-arithmeticdivisors

Several students are seated at a large circular table. They pass around a bag of 100 pieces of candy. Each person takes one piece and passes the bag to the next person. If Chris takes the first and the last piece of candy, then the number of students at the table could be

Show answer
Answer: B — 11.
Show hints
Hint 1 of 2
Chris takes pieces 1, n+1, 2n+1, … where n is the number of students.
Still stuck? Show hint 2 →
Hint 2 of 2
Taking the 100th piece too means 99 is a multiple of n.
Show solution
Approach: the gap between Chris's pieces must divide 99
  1. Chris takes the 1st and every nth piece after, so for the 100th to be his, n must divide 100 − 1 = 99.
  2. Of the choices, only 11 divides 99, so there could be 11 students.
Mark: · log in to save
Problem 22 · 1989 AJHSME Stretch
Number Theory lcm-of-cycle-lengths

The letters A, J, H, S, M, E and the digits 1, 9, 8, 9 are "cycled" separately as follows and put together in a numbered list:

      AJHSME  1989
  1.  JHSMEA  9891
  2.  HSMEAJ  8919
  3.  SMEAJH  9198
      .........

What is the number of the line on which AJHSME 1989 will appear for the first time?

Show answer
Answer: C — 12.
Show hints
Hint 1 of 2
The letters return to AJHSME every 6 lines; the digits return to 1989 every 4 lines.
Still stuck? Show hint 2 →
Hint 2 of 2
The two parts line up again the first time both cycles finish together.
Show solution
Approach: LCM of the two cycle lengths
  1. The 6 letters come back after 6 cycles; the 4 digits come back after 4 cycles.
  2. Both line up together at the smallest common multiple: LCM(6, 4) = 12.
Mark: · log in to save
Problem 25 · 1986 AJHSME Stretch
Number Theory average-of-arithmetic-progression

Which of the following sets of whole numbers has the largest average?

Show answer
Answer: D — multiples of 5 between 1 and 101.
Show hints
Hint 1 of 2
For an arithmetic progression, the average is just (first + last)⁄2.
Still stuck? Show hint 2 →
Hint 2 of 2
Compare each set's first and last terms.
Show solution
Approach: (first + last) ⁄ 2
  1. Averages: multiples of 2 → (2 + 100)⁄2 = 51; of 3 → (3 + 99)⁄2 = 51; of 4 → (4 + 100)⁄2 = 52; of 5 → (5 + 100)⁄2 = 52.5; of 6 → (6 + 96)⁄2 = 51.
  2. Largest is 52.5 — multiples of 5.
Mark: · log in to save