Sarah has a bag of 18 balls numbered from 1 to 18. What is the smallest number of balls Sarah must remove from the bag to be sure that she has removed at least three prime numbers?
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Answer: D — 14
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Hint 1 of 2
Ask how many balls are NOT prime — those could all come out before any prime.
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Hint 2 of 2
After removing every non-prime, the next few must be primes; worst case removes all non-primes first.
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Approach: worst-case (pigeonhole) counting
Primes from 1–18: 2,3,5,7,11,13,17 — seven of them; the other 11 numbers are non-prime.
Worst case she draws all 11 non-primes first, then needs 3 more to guarantee three primes.
The circumference of the large wheel of a bicycle measures 4·2 m, and that of the small wheel 0·9 m. To begin with, the valves on both wheels are at the lowest point; then the bicycle moves off. After a few metres both valves are again at the lowest point at the same time. After how many metres does this happen for the first time?
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Answer: C — 12·6 m
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Hint 1 of 2
Both valves return to the bottom after a whole number of turns of each wheel.
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Hint 2 of 2
Find the least common multiple of the two circumferences.
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Approach: least common multiple of the wheel circumferences
The distance must be a whole-number multiple of both 4.2 m and 0.9 m.
Work in tenths of a metre: lcm(42, 9) = 126, so 12.6 m.
A palindrome is a number that reads the same forwards and backwards, for example 121 or 444. What is the sum of the digits of the largest three-digit palindrome that is also a multiple of 6?
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Answer: E — 24
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Hint 1 of 2
A palindrome aba is a multiple of 6 when it is even and its digit sum is divisible by 3.
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Hint 2 of 2
Even means the outer digit a is even; push a as high as possible, then the middle digit.
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Approach: build the largest even palindrome with digit sum divisible by 3
For aba to be even, a must be even, so the largest choice is a = 8.
Then 8b8 needs digit sum 16 + b divisible by 3; the largest such b is 8, giving 888.
888 is even and divisible by 3, so it is a multiple of 6, with digit sum 24.
The daughter of Mary’s daughter was born today. In two years’ time the product of Mary’s age, her daughter’s age and her granddaughter’s age will be exactly 2024, and each of the three ages will be an even number. How old is Mary today?
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Answer: B — 44
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Hint 1 of 2
The granddaughter is 0 today, so in two years she is 2; that 2 is a factor of 2024.
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Hint 2 of 2
Factor 2024/2 into two even numbers that are sensible ages for a mother and her daughter.
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Approach: factor 2024 with the granddaughter's age fixed
In two years the ages are M+2, D+2 and 2, with product 2024, so (M+2)(D+2) = 1012.
All three ages are even, so M+2 and D+2 are both even; 1012 = 46 x 22 fits.
A number is written in each of the twelve circles shown. The number in each square is the product of the four numbers at the corners of that square. What is the product of the numbers in the eight bold circles?
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Answer: B — 40
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Hint 1 of 2
Multiply the four outer squares' numbers together and watch which circles get used and how many times.
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Hint 2 of 2
The centre square's number is the product of the four inner circles, and each inner circle sits in exactly two outer squares.
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Approach: multiply the four outer squares and divide out the inner circles
Multiplying the four outer (arm) squares \(10\times4\times6\times24=5760\) uses each of the eight bold outer circles once and each of the four inner circles twice.
The four inner circles multiply to the centre square's value 12, so their squared contribution is \(12^2=144\).
Therefore the product of the eight bold circles is \(5760\div144=40\), answer B.
We call a positive integer powerfree if none of its digits can be written as a power of an integer with an exponent bigger than 1. For example, the number 53 is powerfree, but the number 54 is not powerfree since \(4 = 2^2\). Which one of the following numbers is the difference between the biggest and the smallest two-digit powerfree numbers?
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Answer: B — 55
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Hint 1 of 2
A digit is “bad” if it is a perfect power: 0, 1, 4, 8, 9.
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Hint 2 of 2
Find the largest and smallest two-digit numbers whose digits are all allowed, then subtract.
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Approach: restrict to allowed digits
Digits that are powers (exponent>1) are 0=0², 1=1², 4=2², 8=2³, 9=3².
Allowed digits are 2, 3, 5, 6, 7.
The biggest two-digit powerfree number is 77 and the smallest is 22.
Marta observed that the number 2020 has the following property: the number formed by the two digits on the left is equal to the number formed by the two digits on the right. How many four-digit numbers, including 2020, have this same property?
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Answer: D — 90
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Hint 1 of 2
The property says the front two digits equal the back two digits.
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Hint 2 of 2
Such a number is just a two-digit block repeated; count the allowed blocks.
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Approach: count the repeated two-digit blocks
A number with this property looks like the block XY repeated, e.g. 2020 = 20|20.
The block must be a two-digit number from 10 to 99 (it can’t start with 0).
If one of the digits of a two-digit number is deleted, the result in both cases is a factor of the original number. How many two-digit numbers have this property?
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Answer: C — 14
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Hint 1 of 3
Deleting a digit leaves a single digit; the original must be divisible by each remaining digit.
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Hint 2 of 3
Require the number to be divisible by both its tens digit and its units digit.
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Hint 3 of 3
Every multiple of 11 works, plus a few others — count them all.
Show solution
Approach: test divisibility by each single digit
For a number with tens t and units u, both u and t must divide it (u ≠ 0).
Through 10–99 the ones that pass are 11, 12, 15, 22, 24, 33, 36, 44, 48, 55, 66, 77, 88, 99.
60 apples and 60 pears in total are shared out among several boxes. There should be the same number of apples in each box, but no two boxes should contain the same number of pears. Each box contains both fruits. What is the maximum number of boxes that can be filled in this way?
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Answer: D — 10
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Hint 1 of 3
Equal apples per box means the number of boxes divides 60.
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Hint 2 of 3
Distinct positive pear counts summing to 60 need at least 1+2+…+k; combine both limits.
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Hint 3 of 3
Find the largest k with 60 divisible by k and 1+2+…+k ≤ 60.
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Approach: combine the apple and pear constraints
With k boxes, k must divide 60 (equal apples) and 1+2+…+k ≤ 60 (distinct pear counts).
1+…+10 = 55 ≤ 60 and 10 divides 60, so k = 10 works.
k = 12 fails (1+…+12 = 78 > 60), so the maximum is 10 boxes.
Maria wants to share 42 apples, 60 peaches and 90 cherries fairly. She puts them into baskets that each hold the same number of apples, the same number of peaches and the same number of cherries, and gives one basket to each friend. At most, how many baskets can she fill?
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Answer: B — 6
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Hint 1 of 2
The number of baskets must divide each of 42, 60 and 90 exactly.
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Hint 2 of 2
Find the greatest common divisor of 42, 60 and 90.
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Approach: greatest common divisor
Each fruit count must split evenly among the baskets, so the number of baskets divides 42, 60 and 90.
105 numbers are written in a row: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, … where each number n is written exactly n times. How many of those numbers are divisible by 3?
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Answer: D — 30
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Hint 1 of 2
1 + 2 + 3 + … + 14 = 105, so the row runs from 1 up to 14.
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Hint 2 of 2
Only the multiples of 3 (namely 3, 6, 9, 12) count, and each appears as many times as its value.
Show solution
Approach: sum the counts of the multiples of 3 up to 14
Since 1 + 2 + … + 14 = 105, the numbers used run from 1 to 14.
The multiples of 3 in that range are 3, 6, 9 and 12, each written that many times.
How many three-digit numbers have the property that the two-digit number obtained by deleting the middle digit is exactly one ninth of the original number?
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Answer: D — 4
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Hint 1 of 2
Write the number as \(100a+10b+c\); deleting the middle digit gives \(10a+c\).
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Hint 2 of 2
The condition \(100a+10b+c = 9(10a+c)\) simplifies to a relation between the digits.
Show solution
Approach: set up the place-value equation and count solutions
Let the number be \(100a+10b+c\); deleting the middle digit gives \(10a+c\).
The condition \(100a+10b+c = 9(10a+c)\) becomes \(10a+10b = 8c\), i.e. \(5(a+b)=4c\).
So \(c\) is a multiple of 5, and \(c\neq 0\) forces \(c=5\), giving \(a+b=4\).
Number TheoryAlgebra & Patternssum-constraintcasework
Seven positive whole numbers a, b, c, d, e, f, g are written down next to each other in this order. The sum of all seven numbers is 2017. Every two adjacent numbers always differ by 1. Which number can be equal to 286?
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Answer: A — only a or g
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Hint 1 of 2
Neighbours differ by 1, so the seven numbers alternate even, odd, even, … — and the odd total 2017 fixes which positions are even.
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Hint 2 of 2
286 is below the average of about 288, so ask which position can dip the lowest.
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Approach: parity rules out three spots, then check which can reach the smallest value
Neighbours differ by 1, so the numbers alternate in parity. For the total 2017 to be odd, the four outer-pattern places a, c, e, g must be even and b, d, f odd.
286 is even, so it can only sit at one of a, c, e, g — that already rules out choices about b, d, f.
The average is 2017 ÷ 7 ≈ 288, and since steps are only ±1, the smallest a number can be is 286, reached only by going straight down from an end. An inner even place (c or e) has values on both sides pulling it up, so it cannot get below 288.
Hence 286 can occur only at an endpoint: only a or g, choice A.
Eight cards with the numbers 1, 2, 4, 8, 16, 32, 64, 128 are each in an unmarked envelope. Eva randomly chooses some of these eight envelopes. Ali takes the remaining ones. Both add their numbers together. They find out that Eva’s sum is 31 bigger than Ali’s sum. How many envelopes has Eva chosen?
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Answer: D — 5
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Hint 1 of 2
The eight cards are the powers of two from 1 to 128; their total is 255.
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Hint 2 of 2
If Eva's sum beats Ali's by 31 out of 255, find Eva's exact total, then read off its binary digits.
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Approach: binary split
The cards total 1+2+…+128 = 255.
Eva − Ali = 31 and Eva + Ali = 255, so Eva = (255+31)/2 = 143.
Tom has written down a few different positive whole numbers, all smaller than 101. The product of the numbers is not divisible by 18. At most how many numbers could he have written down?
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Answer: C — 68
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Hint 1 of 2
18 = 2 × 3², so the product fails only if it lacks a 2 or lacks two 3s.
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Hint 2 of 2
Keep as many numbers as possible while keeping the power of 3 below 2.
Show solution
Approach: control the factors of 3 to dodge 18
Keep all 67 numbers from 1–100 that are not multiples of 3.
Add one multiple of 3 that is not a multiple of 9 (one factor of 3 only); the product then isn't divisible by 9, hence not by 18.
Given a six-digit number whose digit sum is even and whose digit product is odd. Which of the following statements is true for this number?
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Answer: E — None of the statements (A)–(D) are correct.
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Hint 1 of 2
A product of digits is odd only when every single digit is odd.
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Hint 2 of 2
With all six digits odd, check each statement — and remember there are only five odd digits to choose from.
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Approach: deduce all digits are odd, then test each claim
An odd digit product forces all six digits to be odd (any even digit would make the product even).
Six odd digits sum to an even number, matching the condition — so such numbers exist (e.g. 111111).
Now: (A) zero even digits, false; (C) six odd digits is an even count, false; (D) only five distinct odd digits exist, so six different is impossible, false.
Along each side of a pentagon a positive integer is written. Numbers on adjacent sides never have a common factor bigger than 1, while numbers on non-adjacent sides always have a common factor bigger than 1. There are several possibilities for this situation, but one of the following numbers can never be on a side of the pentagon. Which one?
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Answer: C — 19
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Hint 1 of 2
Each side must share a factor with its two non-adjacent sides but none with its neighbours.
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Hint 2 of 2
A prime number on a side forces both its partner sides to be its multiples - and those partners are neighbours of each other.
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Approach: why a prime side is impossible
A side's two non-adjacent partners must share its factor; but in a pentagon those two partners are next to each other.
If the side were a prime, both partners would be multiples of that prime and so share it - yet as neighbours they must be coprime.
The only prime offered is 19, so 19 can never be used.
Friday writes several different positive whole numbers, all less than 11, next to each other in the sand. Robinson Crusoe looks at the sequence and notices with amusement that adjacent numbers are always divisible by each other. What is the maximum amount of numbers he could possibly have written in the sand?
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Answer: D — 9
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Hint 1 of 2
Among 1–10, each neighbour pair must have one number dividing the other.
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Hint 2 of 2
Treat numbers as dots and 'divides' as links, then find the longest chain of distinct numbers.
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Approach: build the longest chain of distinct numbers where each neighbour pair divides
Among 1–10, the only number that 7 divides or that divides 7 is 1 (since 14 is too big), so 7 can sit next to 1 only — meaning 7 must be an end of the row, touching 1.
If all ten numbers were used, 7 would need that single end spot, but then the rest of 1–10 still cannot all be chained, so 10 numbers is impossible.
Drop 7 and a valid row of 9 exists: 9, 3, 6, 2, 4, 8, 1, 5, 10, where every neighbour pair has one dividing the other.
The four-digit number 80?? is missing its last two digits. We know that this number is divisible by 8 and 9. What is the product of the last two digits?
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Answer: D — 24
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Hint 1 of 2
Divisible by both 8 and 9 means divisible by 72.
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Hint 2 of 2
Find the multiple of 72 between 8000 and 8099, then multiply its last two digits.
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Approach: use divisibility by 72
A number divisible by 8 and 9 is divisible by 72.
The multiple of 72 of the form 80__ is 8064 (= 72×112).
There are some cards on a table with various different positive integers written on them. All of these are smaller than 20 and their product is 2025. What is the maximum number of cards on the table?
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Answer: D — 5
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Hint 1 of 2
Factor 2025 and split it into as many factors below 20 as possible.
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Hint 2 of 2
2025 = 3^4 × 5^2; including the card '1' adds one more card for free.
Show solution
Approach: maximize the number of small factors
2025 = 3⁴ × 5²; each card value is below 20.
Break it up as 1 × 3 × 5 × 9 × 15 = 2025, all distinct and under 20.
Ardal fences a rectangular plot of land using 40 m of fence. Every side length of the rectangle is a prime number. What is the largest possible area of the plot?
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Answer: B — 91 m²
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Hint 1 of 2
The two side lengths add to half the perimeter; both must be prime.
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Hint 2 of 2
For a fixed sum, the product (area) is largest when the two numbers are closest together.
Show solution
Approach: find the prime pair summing to 20 with largest product
Perimeter 40 means the two sides add to 20, and both are prime.
Prime pairs adding to 20: (3,17), (7,13); their products are 51 and 91.
Karo has a box of matches with 30 matches. Using some of the matches she forms the number 2022. She has already formed the first two digits (see picture). How many matches will be left in the box when she has finished the number?
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Answer: B — 9
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Hint 1 of 2
Work out how many matches each digit needs, then add up what is still missing.
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Hint 2 of 2
She has already built '20'; only '22' remains, and each '2' costs 5 matches.
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Approach: count matches used, subtract from 30
A '2' is built from 5 matches and a '0' from 6 matches.
Forming '20' already used 5 + 6 = 11 matches.
The remaining digits '22' need 5 + 5 = 10 matches.
In the multiplication grid displayed, each white cell should show the product of the numbers in the grey cells that are in the same row and column respectively. One number is already entered. The integer x is bigger than the positive integer y. What is the value of y?
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Answer: A — 6
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Hint 1 of 2
The bottom-right white cell equals (y+1) times (x+1).
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Hint 2 of 2
Factor 77 and use that x and y are positive integers with x > y.
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Approach: factor the product in the corner cell
The 77 cell is the product of its grey row label (y+1) and column label (x+1): (y+1)(x+1) = 77.
77 = 7 * 11, and since x > y we take x+1 = 11, y+1 = 7.
In the addition shown, different letters represent different digits and equal letters represent equal digits. The resulting sum is a four-digit number, with B different from zero. What is the sum of the digits of this number?
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Answer: B — BB
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Hint 1 of 2
Write A + AB + ABA as an expression in the digit values of A and B.
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Hint 2 of 2
Match it to the four-digit result BEBA and pin down each letter.
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Approach: convert the cryptarithm to place value and solve for the digits
A + AB + ABA = 112A + 11B, and BEBA = 1010B + 100E + A.
Setting them equal forces A = 9, B = 1, E = 0, so the sum is 1019.
Three five-digit numbers are written onto three separate pieces of paper, as shown. Three of the digits in the picture are hidden. The sum of the three numbers is 57263. Which are the hidden digits?
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Answer: B — 1, 2 and 9
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Hint 1 of 2
The three five-digit numbers add to 57263; match digits column by column.
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Hint 2 of 2
Work through the addition with carries to pin down each hidden digit.
Show solution
Approach: column addition with the known total
Line up the three five-digit numbers so their sum is 57263.
Filling the columns with the right carries forces the three covered digits.
If the whole number age of a father is multiplied by the whole number age of his son, one obtains 2015. Both are born in the 20th century. How big is the age gap between father and son?
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Answer: D — 34
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Hint 1 of 2
Factor 2015 into a product of two whole numbers, both ages of people born in the 1900s.
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Hint 2 of 2
2015 = 5 × 13 × 31; find the factor pair that gives believable father/son ages.
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Approach: match a factor pair to plausible ages
2015 = 5 × 13 × 31.
The factor pair fitting a father and son both born in the 20th century is 65 × 31.
Which value of the variable n is a counterexample to the statement “If n is a prime number, then exactly one of the two numbers n − 2 and n + 2 is a prime number.”?
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Answer: E — 37
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Hint 1 of 2
You want a prime n for which the rule fails: both n−2 and n+2 prime, or neither.
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Hint 2 of 2
Check each option: 37 is prime but 35 and 39 are both composite, so neither n−2 nor n+2 is prime.
Show solution
Approach: find the prime that breaks the claim
The statement fails if, for a prime n, the count of primes among n−2, n+2 is 0 or 2.
For 37 (prime): 35 = 5·7 and 39 = 3·13 are both composite → neither is prime.
A grandmother, her daughter and her granddaughter each have their birthday in February. Together they are 100 years old, and each person’s age is a power of 2. In which year was the granddaughter born?
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Answer: C — 2010
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Hint 1 of 2
Three ages, each a power of 2, add to 100.
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Hint 2 of 2
Try the largest powers of 2 first and see what is left.
Show solution
Approach: find three powers of 2 summing to 100
Powers of 2 available: 1, 2, 4, 8, 16, 32, 64.
64 + 32 + 4 = 100, so the ages are 64, 32 and 4 (grandmother, mother, granddaughter).
If you add the numbers on opposite faces of this special die, you get the same total three times. The numbers on the hidden faces are prime numbers. Which number is on the face opposite to 14?
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Answer: E — 23
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Hint 1 of 2
Opposite faces all add to the same total; the three hidden faces are primes.
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Hint 2 of 2
Use the three visible numbers 18, 35, 14 and make each opposite a prime with one common sum.
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Approach: find the common opposite-sum that makes all hidden faces prime
Let every opposite pair add to S. Then the hidden faces are S−18, S−35 and S−14.
S = 37 works: hidden faces 19, 2 and 23 are all prime.
The number n is the biggest natural number for which \(4n\) is three digits long, and m is the smallest natural number for which \(4m\) is three digits long. What value does \(4n - 4m\) have?
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Answer: C — 896
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Hint 1 of 2
A three-digit multiple of 4 ranges from 100 to 999; find the largest and smallest n that land in that range.
A long number is built by writing the number 2009 in a row 2009 times. What is the sum of all the odd digits in this number that are immediately to the left of an even digit?
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Answer: D — 18072
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Hint 1 of 2
Inside the repeated block 2009, which digit sits just before an even digit?
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Hint 2 of 2
Each copy of 2009 contributes the same fixed amount; count how many junctions there are.
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Approach: find the repeating pattern, then multiply
The string is 2009 2009 2009 … (2009 copies).
Within one block 2-0-0-9, the only odd digit followed by an even one is the 9 before the next block's 2.
That happens at each junction between consecutive blocks: 2008 junctions, each adding 9.
A four-digit number ABCD is multiplied by its units digit D. The result is a different four-digit number DXYA, whose units and thousands digits are swapped compared to the original number (that is, ABCD × D = DXYA). Different letters can stand for the same digits. How many four-digit numbers ABCD have this property?
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Answer: E — 11
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Hint 1 of 3
The product is still a 4-digit number whose thousands digit is the original units digit \(D\) and whose units digit is the original thousands digit \(A\).
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Hint 2 of 3
First nail down \(A\) and \(D\) from the size and the last-digit rules, then see how free the middle digits are.
Still stuck? Show hint 3 →
Hint 3 of 3
Since multiplying by \(D\) only just stays 4-digit, \(A\) must be very small; the units digit of \(A\times D\) must equal \(A\).
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Approach: pin the end digits, then count the free middle
The result \(DXYA\) starts with \(D\) and ends with \(A\); for \(ABCD\times D\) to stay 4-digit while its leading digit jumps to \(D\), the only fit is \(A=1\) and \(D=9\) (then \(1BC9\times9\) lands in the 9000s and ends in \(\dots1\), since \(9\times9=81\)).
With \(A=1,\,D=9\) the number is \(\overline{1BC9}\); checking shows every such number works because \(\overline{1BC9}\times9\) automatically becomes \(\overline{9XY1}\).
So \(B\) and \(C\) are free over \(1009,1019,\dots,1109\) — that is 11 numbers, answer E.
The six-digit number ABCDEF consists of the digits 1, 2, 3, 4, 5 and 6, with each digit occurring exactly once. The number AB consisting of the first two digits is a multiple of 2. The number ABC is a multiple of 3. The number ABCD is a multiple of 4. The number ABCDE is a multiple of 5, and the entire number ABCDEF is a multiple of 6. What values can the digit F take?
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Answer: B — only 4
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Hint 1 of 3
Apply the rules in order: \(AB\) even, \(ABC\) a multiple of 3, \(ABCD\) of 4, \(ABCDE\) of 5, \(ABCDEF\) of 6.
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Hint 2 of 3
Divisible-by-5 with digits 1–6 forces \(E=5\); divisible-by-2 and divisible-by-6 force \(B\), \(D\), \(F\) to be the three even digits.
Still stuck? Show hint 3 →
Hint 3 of 3
So \(\{B,D,F\}=\{2,4,6\}\) and \(\{A,C\}=\{1,3\}\); pin them down with the multiple-of-3 and multiple-of-4 rules.
Show solution
Approach: apply the divisibility conditions step by step
Since \(ABCDE\) is a multiple of 5, \(E=5\); \(B\) (and \(F\)) are even and \(D\) is even (for the 4-rule), so the even digits 2, 4, 6 fill \(B,D,F\) and the odd digits 1, 3 fill \(A,C\).
\(ABC\) divisible by 3 needs \(A+B+C\) divisible by 3; with \(A+C=1+3=4\), \(B\) must be 2, leaving \(\{D,F\}=\{4,6\}\). The 4-rule on \(\overline{CD}\) (with \(C\in\{1,3\}\)) forces \(D=6\).
That leaves \(F=4\), and indeed 123654 and 321654 satisfy every rule, so \(F\) can only be 4, answer B.
Cristina has 12 cards numbered 1 to 12. She places eight of them in a circle so that the sum of any two adjacent numbers is a multiple of 3. Which numbers does Cristina not use?
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Answer: E — 3, 6, 9, 12
Show hints
Hint 1 of 2
Sort 1-12 by remainder when divided by 3: three groups of four.
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Hint 2 of 2
Two adjacent numbers sum to a multiple of 3 only if both are multiples of 3, or one is a 'remainder 1' and the other a 'remainder 2'.
Show solution
Approach: classify by residue mod 3 and build the circle
Mod 3 the numbers split into {3,6,9,12}, {1,4,7,10}, {2,5,8,11}.
Adjacent sums divisible by 3 force the circle to alternate the remainder-1 and remainder-2 groups, using all eight of those numbers.
That leaves the multiples of 3 unused, so Christina does not use 3, 6, 9, 12.
The sum of the digits of N is three times the sum of the digits of \(N+1\). What is the smallest possible sum of the digits of N?
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Answer: C — 12
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Hint 1 of 2
Adding 1 only changes the digit sum a lot when N ends in some nines (each trailing 9 turns into a 0).
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Hint 2 of 2
If N ends in exactly k nines, write the digit sum of N+1 in terms of the digit sum of N and solve.
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Approach: use trailing nines to relate the two digit sums, then minimise
If N ends in exactly \(k\) nines, then \(N+1\) turns those \(k\) nines into zeros and raises the next digit by 1, so \(S(N+1)=S(N)-9k+1\).
The condition \(S(N)=3\,S(N+1)\) becomes \(S(N)=3(S(N)-9k+1)\), which simplifies to \(2S(N)=27k-3\); the smallest valid case is \(k=1\), giving \(S(N)=12\).
This is achievable, e.g. \(N=39\) has digit sum 12 and \(N+1=40\) has digit sum 4, with \(12=3\times4\); so the smallest digit sum of N is 12, answer C.
Seven pairwise different single-digit numbers are distributed among the circles shown so that the product of the three numbers that are connected by a straight line is the same in all three cases. Which number is written in the circle with the question mark?
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Answer: A — 2
Show hints
Hint 1 of 2
The centre number is in all three products, so the two outer numbers on each line have equal products.
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Hint 2 of 2
Find three disjoint pairs of distinct digits with the same product, and see which centre value makes it work.
Show solution
Approach: equal products force matching pairs
All seven digits must be different and non-zero (a zero would force every product to be 0).
Trying the common product 72 = 8·9·1 = 6·4·3, the two horizontal lines can use {8,9,1} and {6,4,3}.
The vertical line shares the two centres (9 and 4 in one valid layout), so its third number is 72 ÷ (9·4) = 2.
Checking all arrangements, the question-mark circle always holds 2.
Once I met six sisters whose ages were six consecutive integers. I asked each one of them: How old is the oldest of your sisters? Which of the following numbers cannot be the sum of the six answers?
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Answer: D — 205
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Hint 1 of 2
Each sister names her oldest sister: five of them name the same person, but the oldest names someone different.
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Hint 2 of 2
Write the total of the six answers as one expression in the youngest age, then check it against each option.
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Approach: build the sum of the six answers, then test the fixed remainder
Call the ages \(n, n+1, \dots, n+5\); the oldest is \(n+5\), and her own oldest sister is \(n+4\).
Five sisters answer \(n+5\) and the oldest answers \(n+4\), so the six answers add to \(5(n+5)+(n+4)=6n+29\).
So any valid sum minus 29 must be a multiple of 6; checking the options, only \(205-29=176\) is not, so 205 cannot occur.
Julia wrote four positive integers, one at each vertex of a triangular-based pyramid. She calculated the sum of the numbers written on the vertices of one face and the products of the numbers written on the vertices of the other two faces, obtaining 15, 20 and 30, respectively. What is the highest possible value of the product of the four numbers?
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Answer: E — 120
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Hint 1 of 2
Each face uses three of the four vertices; two faces share an edge (two vertices).
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Hint 2 of 2
Pick vertex values matching the two products that share a pair, then check the sum-15 face.
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Approach: match shared-edge factor triples and maximise the product
The faces with products 20 and 30 share two vertices; try the shared pair {1,5}.
Then the third vertices are 4 (for 20) and 6 (for 30), giving values 1, 5, 4, 6.
The remaining face 5+4+6 = 15 matches the required sum.
The numbers a, b and c are three-digit numbers, and in each number the first digit is equal to the last one. Furthermore \(b = 2a + 1\) and \(c = 2b + 1\). How many possible values are there for the number a?
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Answer: C — 2
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Hint 1 of 2
“First digit equals last” means each of a, b, c is a 3-digit number of the form x?x.
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Hint 2 of 2
Use b = 2a + 1 and c = 2b + 1 and test which starting a keep all three in that form.
Show solution
Approach: chase the doubling chain through 3-digit ‘x?x’ numbers
a, b, c each read x?x (first digit = last). With b = 2a + 1 and c = 2b + 1, only a few a work.
a = 181 gives b = 363, c = 727; a = 191 gives b = 383, c = 767 — both valid.
What is the minimum number of elements of the set \(\{10, 20, 30, 40, 50, 60, 70, 80, 90\}\) that have to be removed so that the product of the remaining elements is a square number?
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Answer: B — 2
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Hint 1 of 3
A product is a square exactly when every prime appears an even number of times.
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Hint 2 of 3
Only one element, 70, carries the prime 7, so 70 must go — then check what its removal does to the other primes.
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Hint 3 of 3
Removing 70 throws the counts of 2 and 5 off, so a second element is needed to fix them.
Show solution
Approach: make every prime exponent even
In the full product \(2^{16}\cdot 3^{4}\cdot 5^{10}\cdot 7^{1}\) the only odd exponent is the lone 7, which comes solely from 70.
So 70 must be removed; but 70 = \(2\cdot 5\cdot 7\), and dropping it turns the exponents of 2 and 5 odd.
Removing one more element that supplies an odd 2 and an odd 5 (for example 40 = \(2^{3}\cdot 5\)) makes everything even, so the minimum is 2.
Some whole numbers are written on a board, among them the number 2018. The sum of all of them is 2018, and the product of all of them is also 2018. Which of the following could be how many numbers are on the board?
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Answer: B — 2017
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Hint 1 of 2
The product is already 2018 from that one number, so every other number must be \(\pm 1\).
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Hint 2 of 2
The sum is already 2018, so the extra \(+1\)'s and \(-1\)'s must cancel; that forces an exact count.
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Approach: every other number is +1 or -1, then match the sum and product constraints
The list already contains 2018, whose value equals both the required sum and the required product, so every other number must be a \(+1\) or a \(-1\) (these don't change a product and are the only integers that can).
For the product to stay \(+2018\), the number of \(-1\)'s must be even; for the sum to stay 2018, the \(+1\)'s and \(-1\)'s must cancel, so there are equally many of each, say \(k\) of each.
Then the board holds \(1 + k + k = 2k+1\) numbers, which is always odd, so among the options only 2017 is possible (take \(k = 1008\)).
Ed builds a big cube from several identical small white dice and colours some of the big cube's faces red. His sister Nicole drops it and it breaks back into the small dice. 45 of them have no red face. How many faces of the big cube did Ed colour red?
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Answer: C — 4
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Hint 1 of 2
A small die stays all-white only if it touches none of the coloured big-cube faces.
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Hint 2 of 2
Find a cube size where exactly 45 small dice can avoid the coloured faces.
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Approach: count the small dice that avoid every coloured face
Take the big cube as \(5\times5\times5\) (125 small dice) and colour the 4 side faces, leaving the top and bottom uncoloured.
The dice touching no red face are the inner \(3\times3\) of each of the 5 layers: \(3\times 3\times 5 = 45\), exactly as stated.
In the primate enclosure in a zoo there are four gorillas. They are all younger than 18 years old. No two have the same age, and all their ages are whole numbers. The product of their ages is 882. How big is the sum of their ages?
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Answer: D — 31
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Hint 1 of 2
Factor 882 and split it into four different whole numbers, all under 18.
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Hint 2 of 2
Use the prime factorisation 882 = 2 · 3² · 7².
Show solution
Approach: factor 882 into four distinct factors below 18
882 = 2 · 3² · 7².
The only way to write it as four distinct factors each under 18 is 1 · 7 · 9 · 14 = 882.
The numbers −3, −2, −1, 0, 1, 2 are written on the six faces of a die. The die is rolled twice. The numbers that were rolled are multiplied. How big is the probability that this product is negative?
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Answer: E — 13
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Hint 1 of 2
A product is negative only when one factor is positive and the other is negative.
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Hint 2 of 2
Count the positive faces and the negative faces; zero never helps.
Show solution
Approach: count favourable ordered rolls over all 36 outcomes
The faces are −3, −2, −1, 0, 1, 2: three negatives and two positives (0 gives product 0).
Negative product needs one of each sign; ordered, that is 3·2 + 2·3 = 12 of the 36 equally likely pairs.
A popular two-digit number is made up of the digits a and b. If the number pair is written down three times one after the other, a six-digit number is obtained. This new number is always divisible by
Show answer
Answer: C — 7
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Hint 1 of 2
Writing the two-digit number three times equals it times some fixed number.
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Hint 2 of 2
Factor that fixed multiplier 10101.
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Approach: factor the repetition multiplier
Writing the two-digit number ab three times gives ababab = ab × 10101.
Factor 10101 = 3 · 7 · 13 · 37, so 7 always divides the result.
My friend Heinz wants to use a special password that is made up of seven digits. Each digit used in the password appears as many times in the password as is the value of the digit. Additionally, equal digits are always next to each other. Therefore he can for example use 4444333 or 1666666 as passwords. How many possible passwords can he choose from?
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Answer: E — 13
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Hint 1 of 2
Each chosen digit takes up as many of the 7 slots as its value, so the values must add to 7.
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Hint 2 of 2
Equal digits stay together as one block, so count orderings of the blocks.
Show solution
Approach: partition 7 into distinct digit-values, then order the blocks
Since a digit d fills d slots and equal digits are adjacent, choose distinct digit-values that sum to 7: {7}, {1,6}, {2,5}, {3,4}, {1,2,4}.
Each choice of k blocks can be arranged in k! orders: 1 + 2 + 2 + 2 + 6 = 13.
Logic & Word ProblemsNumber Theorycaseworkcareful-counting
Paul wants to write a positive whole number onto every tile in the number wall shown, so that every number is equal to the sum of the two numbers on the tiles that are directly below. What is the maximum number of odd numbers he can write on the tiles?
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Answer: B — 14
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Hint 1 of 2
Work in parity (odd/even): a tile is odd exactly when the two below it differ in parity.
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Hint 2 of 2
Search the bottom row patterns of the six-row wall to maximise odd tiles.
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Approach: reduce to parity and optimise the bottom row of the wall
Only odd/even matters: a tile is odd exactly when the two tiles below it have different parities, so the whole wall is fixed once the bottom row's pattern of odds and evens is chosen.
For the six-row (21-tile) wall, testing the bottom-row patterns, the best choice (such as even, odd, odd, even, odd, odd) makes 14 of the 21 tiles odd.
So the maximum number of odd tiles is 14, choice B.
Lisa places some points on a circle and then connects them in sequence to make a polygon. She adds up the interior angles of the polygon. By mistake she misses out one angle and obtains the sum 2017. How big is the angle that she has overlooked?
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Answer: E — 143°
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Hint 1 of 2
The true angle sum of a polygon is a multiple of 180°.
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Hint 2 of 2
The missed angle is the gap up to the next multiple of 180 above 2017.
Show solution
Approach: round up to the nearest valid polygon angle sum
The interior angles of an n-gon sum to (n−2)·180°, a multiple of 180.
The smallest multiple of 180 above 2017 is 2160 = 12·180.
The overlooked angle is 2160 − 2017 = 143°, choice E.
The points A and B lie on a circle with centre M. The point P lies on the straight line through A and M. PB touches the circle in B. The lengths of the segments PA and MB are whole numbers, and PB = PA + 6. How many possible values for MB are there?
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Answer: D — 6
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Hint 1 of 2
PB is tangent, so its square equals the product of the whole secant and its external part (power of the point P).
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Hint 2 of 2
Turn the relation into MB = 6 + 18/PA and require whole numbers.
Show solution
Approach: use the tangent-secant power of a point, then count integer solutions
Power of the point P: \(PB^2 = PA \cdot (PA + 2\,MB)\), since the secant through A and M has external part PA and crosses the circle again a diameter (2·MB) further on.
With PB = PA + 6: \((PA+6)^2 = PA^2 + 2\,PA\cdot MB\) gives \(12\,PA + 36 = 2\,PA\cdot MB\), so \(MB = 6 + \dfrac{18}{PA}\).
MB is a whole number when PA divides 18: PA ∈ {1, 2, 3, 6, 9, 18}, giving 6 distinct values of MB, choice D.
In the Kangaroo Republic, every month has 40 days, which are numbered through from 1 to 40. Every day with a number that is divisible by 6 is a public holiday, and likewise every day with a prime number. How often per month does it occur that there is exactly one working day between two public holidays?
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Answer: A — 1
Show hints
Hint 1 of 2
Mark every day that is a multiple of 6 or a prime as a holiday.
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Hint 2 of 2
Look for a holiday, then a single working day, then another holiday.
Show solution
Approach: list and scan
Holidays are the primes and the multiples of 6 up to 40.
Scanning for the pattern holiday–working–holiday, only day 4 sits alone between holidays 3 and 5.
Jakob writes down four consecutive positive whole numbers. He calculates all possible sums of three of those numbers and realises that none of those sums is a prime number. What is the smallest number that Jakob could have written down?
Show answer
Answer: C — 7
Show hints
Hint 1 of 2
Adding three of four consecutive numbers leaves out one; write the four sums.
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Hint 2 of 2
Two of the sums are automatically multiples of 3; make the other two composite too.
Show solution
Approach: search the smallest start
For a start n the four sums are 3n+3, 3n+4, 3n+5, 3n+6; the first and last are multiples of 3.
Need 3n+4 and 3n+5 composite as well; the first n that works is n = 7 (25 and 26).
A date can be written in the form DD.MM.YYYY; e.g. today’s date is 17.03.2016. We call a date “surprising” if all 8 digits used in this notation are different. In which month does the next surprising date occur?
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Answer: B — June
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Hint 1 of 3
All eight digits must differ, so the year YYYY itself must already use four distinct digits.
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Hint 2 of 3
A day's first digit is 0-3 and a month's first digit is 0 or 1, which sharply limits which years can work.
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Hint 3 of 3
Step forward from 2016 to the first year whose digits leave a legal day and month with no repeats.
Show solution
Approach: find the first year that leaves room for a valid day and month
Any year from 2017 onward that starts 20.. reuses the 0 (months and small days also need a 0 or repeat), so no surprising date appears in the 2000s.
Checking the 2100s, 2200s and early 2300s, the leading digits keep colliding with the only small digits a valid month (01-12) and day (01-31) can use, so none works.
The first year that frees up enough distinct small digits is 2345, and its earliest surprising date is 17.06.2345 (digits 1,7,0,6,2,3,4,5 all different).
The number 2013 is made up of the four consecutive digits 0, 1, 2, 3. How many years before the year 2013 was the date last made up of four consecutive digits?
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Answer: C — 581
Show hints
Hint 1 of 2
Search backward from 2013 for a year whose four digits are four consecutive values in some order.
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Hint 2 of 2
Try the digit-set {1,2,3,4} arranged to make the largest year below 2013.
Show solution
Approach: search backward for four-consecutive-digit years
Just below 2013, the most recent year using four consecutive digits is 1432 (digits 1, 2, 3, 4).
Let Q be the number of square numbers among the natural numbers from 1 to \(2013^{6}\), and K the number of cube numbers among the natural numbers from 1 to \(2013^{6}\). Which of the following holds true?
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Answer: A — \(Q = 2013\times K\)
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Hint 1 of 2
The count of perfect squares up to a number is the square root of that number (rounded down); the count of cubes is the cube root.
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Hint 2 of 2
Take the square root and cube root of 2013⁶ exactly.
Using the numbers 1, 2, 3, …, 22, eleven fractions \(\frac{a}{b}\) are formed where each number is used exactly once. What is the maximum number of these fractions that can have whole-number values?
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Answer: B — 10
Show hints
Hint 1 of 2
A fraction is a whole number when the numerator is a multiple of the denominator; pair the numbers 1–22 to make as many such pairs as possible.
Still stuck? Show hint 2 →
Hint 2 of 2
Big primes like 13, 17, 19 are hard to use as denominators — one pair is forced to be non-integer.
Show solution
Approach: pair numbers so one divides the other, maximising integer pairs
A fraction \(\frac{a}{b}\) is a whole number when \(b\) divides \(a\), so pair each number with one it divides into.
The large primes 13, 17 and 19 have no other multiple in 1–22, so each can only sit on top of 1 to give an integer — but there is just one number 1, so at least two of them are forced into a non-integer fraction.
Matching the rest greedily (e.g. \(\frac{22}{11}, \frac{20}{10}, \frac{18}{9}, \dots\)) lets 10 fractions come out whole while one stays non-integer.
A positive integer N is smaller than the sum of its three biggest proper factors (N itself is not a proper factor of N). Which of the following statements is true?
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Answer: B — All such numbers N are divisible by 6.
Show hints
Hint 1 of 2
The biggest proper factors of N are N/2, N/3, N/4, ... — write the sum of the top three this way.
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Hint 2 of 2
For that sum to exceed N, the small divisors 2 and 3 must both divide N.
Show solution
Approach: express the largest proper factors as N over its smallest divisors
The three biggest proper factors are N divided by its three smallest divisors (other than 1).
Their sum can exceed N only when both 2 and 3 divide N (for example N/2 + N/3 + N/4 = 13N/12 > N).
For a positive whole number \(n \ge 2\), let \(\langle n\rangle\) denote the largest prime number less than or equal to n. How many positive whole numbers k satisfy the condition \(\langle k+1\rangle + \langle k+2\rangle = \langle 2k+3\rangle\)?
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Answer: B — 1
Show hints
Hint 1 of 2
Both sides are at most 2k+3, and the left side hits that maximum only when k+1 and k+2 are both prime.
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Hint 2 of 2
Test small k directly; the equality is very restrictive.
Show solution
Approach: bound both sides, then test small k
The left side is at most (k+1)+(k+2) = 2k+3, and the right side ⟨2k+3⟩ is at most 2k+3.
Equality forces a tight prime arrangement; checking k = 1 gives 2 + 3 = 5 = ⟨5⟩, which works.
For k = 2, 3, 4, ... the largest primes drop below the needed totals, so none work.
All factors of a number N (with the exception of 1 and N itself) are written down one after another. It turns out that the biggest of these factors is 45 times as big as the smallest. For how many numbers N is this true?
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Answer: C — 2
Show hints
Hint 1 of 2
The smallest proper factor is the least prime p; the largest proper factor is N/p.
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Hint 2 of 2
Set N/p = 45p, so N = 45p², and require p to really be the smallest prime factor.
Show solution
Approach: express N through its smallest and largest proper factors
The smallest proper factor is the least prime p of N and the largest is N/p.
'Largest = 45 × smallest' means N/p = 45p, so N = 45p².
Since 45 = 3²·5, p must be 2 or 3 for p to stay the smallest prime: N = 180 or N = 405.
Both work (factors of 180 run 2…90, of 405 run 3…135), so there are 2 such numbers.
A single-digit prime number is called “strange.” A prime number with more than one digit is called “strange” if the numbers obtained by deleting its first digit and by deleting its last digit are both strange prime numbers again. How many strange prime numbers are there?
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Answer: D — 9
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Hint 1 of 2
Start from the single-digit 'strange' primes 2, 3, 5, 7 and build longer ones digit by digit.
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Hint 2 of 2
A longer prime is strange only if dropping its first digit AND dropping its last digit each give a strange prime.
Show solution
Approach: build strange primes upward by length
The single-digit strange primes are 2, 3, 5, 7.
A two-digit prime is strange if removing either end digit leaves a strange prime: 23, 37, 53, 73 qualify.
A three-digit prime needs both trimmed numbers strange: only 373 works (37 and 73 are strange).
No four-digit prime qualifies, so the count is 4 + 4 + 1 = 9.
