Number Theory — What whole numbers are made of.

Quick: is 3,471 a multiple of 3? You did not reach for long division — you can't, in your head, that fast. But there IS a way to know in two seconds. Add the digits: 3+4+7+1 = 15. That's a multiple of 3, so 3,471 is too. No dividing.

That's the whole game of this chapter. Divisible means “splits evenly, no leftover.” 12 is divisible by 3 because 12 ÷ 3 = 4 exactly; 13 is not, because 13 ÷ 3 leaves 1 over. You COULD test by dividing — but for every common divisor (2, 3, 4, 5, 6, 8, 9, 10, 11) there's a shortcut rule that reads the answer straight off the digits. Learn these cold. They show up on almost every contest.

The weird rules — divisibility by 7 and 13

The rules for 7 and 13 don't look anything like the others. They feel like party tricks. They are party tricks — but they work, and competition kids who know them save real time.

Examples.

• Is 161 divisible by 7? Chop the 1: leaves 16. Double the 1: 2. Subtract: 16 − 2 = 14. Yes, 14 = 7·2 ✓. (And 161 = 7·23.)
• Is 203 divisible by 7? Chop 3, double = 6, subtract from 20: 20 − 6 = 14 ✓.
• Is 1,729 divisible by 7? Chop 9, double = 18. 172 − 18 = 154. Still big — repeat. Chop 4, double = 8. 15 − 8 = 7 ✓. So 1729 is divisible by 7.

Examples.

• Is 169 divisible by 13? Chop 9, times 4 = 36. 16 + 36 = 52 = 13·4 ✓. (And 169 = 13².)
• Is 1001 divisible by 13? Chop 1, times 4 = 4. 100 + 4 = 104 = 13·8 ✓.

The 1001 magic — one trick for 7, 11, AND 13

Here's a fact every contest kid should memorize:

1001 = 7 × 11 × 13

This single identity unlocks a SHARED divisibility test for all three of those primes:

Example. Is 247,247 divisible by 7, 11, and 13?

• Groups from the right: 247 and 247.
• Alternating sum: 247 − 247 = 0.
• 0 is divisible by everything, so 247,247 is divisible by all of 7, 11, and 13 ✓.

This is exactly the famous “6-digit ABCABC trick”: any number that repeats a 3-digit block — like 247247 or 854854 — equals abc · 1001, so it's automatically divisible by 7, 11, AND 13.

Be honest with yourself. If you're going to memorize ONE “weird” fact, make it 1001 = 7·11·13. The chop-double-subtract rule for 7 and the chop-times-4-add rule for 13 are bonus tricks for showing off — useful sometimes, but the 1001 identity is the one that pays off on real contest problems.

Why does the rule work? Watch the pattern.

Here are the first 10 multiples of 9. Add the digits of each one:

Always 9. Not a trick — a pattern.

For bigger numbers, the digit sum might be 18 or 27 or 36 instead — but those are all multiples of 9 too. So the rule still works:

• 144 → 1+4+4 = 9 ✓ (144 = 9×16)
• 198 → 1+9+8 = 18 ✓ (18 is 9×2)
• 999 → 9+9+9 = 27 ✓ (27 is 9×3)

Why the LAST-DIGITS rules work — the big part falls away

The ÷4 rule (look at the last two digits) and the ÷8 rule (last three) feel arbitrary. They're not. The reason is one tiny fact: 4 divides 100, and 8 divides 1000.

Take any number and chop it at the hundreds: 7,328 is 73×100 + 28. The chunk 73×100 is built entirely out of 100s — and every 100 is 4×25, so that whole chunk is already a stack of 4s. It can't tip the answer. Only the leftover 28 decides it.

Same story for 8: every 1000 is 8×125, so the thousands-and-up chunk is a stack of 8s and falls away — only the last three digits get a vote. That's the whole reason the rule stops where it does.

Why the ÷3 and ÷9 rule works — every 10 is “one extra”

You saw the pattern for 9 above. Here's the engine behind it in one line: 10 leaves a leftover of 1 when you divide by 9 (10 = 9 + 1). So does 100 (= 99 + 1), and 1000, and every power of 10.

So when you build 471 as 4×100 + 7×10 + 1, each hundred contributes “1 leftover,” each ten contributes “1 leftover,” and the ones contribute themselves. Sweep up all the leftovers and you get exactly 4 + 7 + 1 — the digit sum. The digit sum IS the leftover (mod 9). Since 3 also divides 9-and-friends the same way, the rule works for 3 too.

Why the ÷11 rule alternates + − + −

Multiply any four-digit number ABCD by 11 and watch the digits. Each digit lands in TWO neighboring columns at once (because 11 = 10 + 1), so the columns become:

Add them with alternating signs and every letter cancels: A − (A+B) + (B+C) − (C+D) + D = 0. That cancellation is exactly why the alternating digit sum of any multiple of 11 comes out to 0 (or a multiple of 11). The deeper reason: powers of 10 bounce +1, −1, +1, −1 mod 11 (since 10 = 11 − 1), which is where the alternating signs are born.

The one engine behind EVERY rule — multiples add and subtract

Every rule above is really the same trick wearing different costumes. Here is the engine they all run on:

Picture three baskets, each holding exactly 7 eggs. Pour two together: 14 eggs, still a clean stack of 7s. Pour one out: 7 eggs, still clean. You can never get a leftover by combining whole baskets — a leftover can only come from a basket that wasn’t full to begin with.

That is exactly why we keep saying “the big chunk falls away.” Take 7328 tested for 4: we wrote it as 7300 + 28. The 7300 is a stack of 4s (a multiple), so by the engine it can’t change the answer — only the leftover 28 votes. And if 28 had NOT been a multiple of 4, adding the clean 7300 could never rescue it. The digit-sum rule, the last-digits rule, the alternating-sum rule — each one peels off a big multiple-of-c chunk and leaves a tiny leftover that decides everything.

Testing a COMPOSITE — split it, but only the RIGHT way

There’s no special digit rule for 36. So how do you test it fast? Break 36 into a product and test each piece. We already do this for 6 (“divisible by 2 AND 3”). The catch most people miss: it only works when the two pieces share no common factor.

The good split. 36 = 4 × 9, and 4 and 9 share nothing (their GCD is 1 — they’re relatively prime). So a number is divisible by 36 exactly when it passes the rule for 4 AND the rule for 9, separately.

• Is 612 divisible by 36? By 4: last two digits 12 = 4×3 ✓. By 9: digit sum 6+1+2 = 9 ✓. Both pass ⇒ yes. (Check: 612 = 36×17.)

Why it breaks: 3 and 12 share a factor of 3, so “passes 3 and passes 12” only guarantees divisibility by lcm(3,12) = 12, not 36. The overlap gets counted once instead of twice. The fix: split into pieces with no shared factor (relatively prime), and their product gives the real test. 4 × 9 works because lcm(4,9) = 36; 3 × 12 fails because lcm(3,12) = 12.

Framing inspired by AoPS Prealgebra.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

Take 12 marbles. You can lay them in a neat rectangle a few ways: 2 rows of 6, 3 rows of 4. Now try 7 marbles. One row of 7 — and that's it. No fatter rectangle works. Some numbers refuse to be split into a rectangle. Those are the special ones.

A prime number is a whole number bigger than 1 that you CAN’T break into a multiplication of smaller whole numbers. Its only divisors are 1 and itself.

The opposite is composite — it splits. 6 = 2 × 3, so 6 is composite. 7 won’t split, so 7 is prime. Primes are the atoms of the number world: every other number is built by multiplying primes together.

Here’s the picture: imagine arranging n dots into a rectangle. If you can ONLY form a 1-by-n strip, the number is prime. If you can form a fatter rectangle, it’s composite.

How to FIND every prime — the Sieve of Eratosthenes

A Greek named Eratosthenes found a way to catch every prime under 100 with no dividing at all — only crossing out. The idea: primes are whatever is LEFT once you remove every multiple.

Write 1 to 100 in a grid. Then:

1. Cross out 1 (not prime). Circle 2, then cross out every other multiple of 2 (4, 6, 8, …).
2. Circle 3 (still standing), then cross out every multiple of 3 (6, 9, 12, …).
3. Circle 5, cross out its multiples. Then 7, cross out its multiples.
4. Stop. You only need to sieve by primes up to √100 = 10 — that's only 2, 3, 5, 7. Everything still uncrossed is prime.

Why stop at 7? Same reason as the prime-test below: any composite under 100 must have a factor at or below √100 = 10, so the strikers 2, 3, 5, 7 catch all of them. The 25 survivors are exactly the primes under 100.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

The first 15 primes — memorize on sight

Below 50 there are exactly 15 primes. Color them red on the 1–50 grid:

Three facts you’ll use again and again

How to test if n is prime — the √n shortcut

You don’t need to try every divisor up to n − 1. Just primes up to √n. Why? Because if n = a × b with both factors bigger than √n, the product would overshoot n — impossible. So at least one factor must be small.

Hand 60 to ten different people and ask them to break it into primes. Someone starts 60 = 6×10, someone else 60 = 4×15, someone else 60 = 2×30. Different first steps — but keep going, and EVERY one of them lands on the exact same pile of primes: 2, 2, 3, 5. Always. That's not a coincidence; it's a law.

Every whole number ≥ 2 can be written as a product of primes in exactly one way (order doesn't matter). This is the Fundamental Theorem of Arithmetic — the most important fact in number theory.

So 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5, and that's the ONLY prime recipe for 60. Think of it as the number's fingerprint: no two numbers share one.

How to find the prime factorization (the “factor tree”): divide by the smallest prime that fits, write the factor, repeat with the quotient. Stop when you reach 1.

Read the leaves of the tree (the red primes at the bottom): 2, 2, 3, 7. So 84 = 2² × 3 × 7. You can also build a different-looking tree (split 84 as 4×21 first), but you'll always get the same primes at the leaves. That's the uniqueness theorem.

Why is this important? Because once you have the factorization, almost everything about the number becomes easy:

• Is it a perfect square? Yes if every prime's exponent is even. (36 = 2²·3² → yes.)
• How many divisors does it have? See chapter 4.
• Is it divisible by some target? Check if all the target's primes are in the fingerprint.
• What's the GCD or LCM with another number? See chapter 5.

How many divisors does 1,000,000 have? Listing them would take all afternoon. But once you know its prime factorization, you can name the count — 49 — in one line, without writing a single divisor down. Here's the formula that does it.

Where does the +1 come from? Watch a divisor get BUILT. Every divisor is built by choosing how many copies of each prime to include. Take 40 = 2³ · 5. To build a divisor, you make two choices: how many 2s (you can take 0, 1, 2, or 3 — four options), and how many 5s (0 or 1 — two options).

Pick one box from the top row and one from the bottom row, and you've built a divisor. 2¹ · 5⁰ = 2, 2³ · 5¹ = 40, 2⁰ · 5⁰ = 1 — every combination is different, and there are 4 × 2 = 8 of them. The “+1” means “don't forget the option of taking ZERO of that prime.” That's why you count (3+1) twos and (1+1) fives, then multiply.

Example with 36. 36 = 2² · 3². For a divisor of 36, the exponent of 2 can be 0, 1, or 2 (three choices). The exponent of 3 can also be 0, 1, or 2 (three choices). Total divisors = 3 × 3 = 9.

Let's list them in a 3×3 grid to check:

3 rows × 3 columns = 9 divisors. Matches the (2+1)(2+1) formula.

Another example. How many divisors does 360 have? From chapter 3, 360 = 2³ × 3² × 5. So divisor count = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24 divisors.

Two patterns to spot in the divisor list

Pattern 1 — divisors come in pairs. Every divisor d of n has a partner n / d, which is also a divisor. The pair multiplies to n.

Try it. The divisors of 12 are 1, 2, 3, 4, 6, 12. Pair them up:

So the 6 divisors of 12 form 3 pairs. This is true in general: if n is NOT a perfect square, divisors pair up cleanly (count is always even).

See it with 40. List the pairs from the outside in — small factor on the left, its partner on the right — and they march toward the middle:

Now watch what's different about a perfect square. Do the same for 36, and the innermost pair turns out to be 6 × 6 — the two halves are the SAME number, so they fold into one:

That lone middle number is √36 = 6, pairing with itself. Every other divisor still comes in a clean pair, but that one stands alone — so a perfect square always has an ODD count, and every other number has an even count.

Pattern 2 — counting perfect-square divisors. A divisor of n is itself a perfect square exactly when each prime’s exponent (in the divisor) is even.

Example. How many perfect-square divisors does 360 = 2³ · 3² · 5 have? For each prime, count even-exponent choices:

• Exponent of 2 in the divisor: even choices are {0, 2}. 2 options.
• Exponent of 3: even choices are {0, 2}. 2 options.
• Exponent of 5: even choice is {0}. 1 option.

Total: 2 · 2 · 1 = 4 perfect-square divisors (they are 1, 4, 9, 36).

Adding up the divisors — multiply the prime-power sums

Same idea as counting, but now you want the TOTAL of all the divisors, not only how many. Listing and adding works for small numbers, but it gets slow fast.

Slow way for 20. Divisors are 1, 2, 4, 5, 10, 20. Add: 1 + 2 + 4 + 5 + 10 + 20 = 42. Fine for 20. Painful for 360.

Here’s the shortcut. Factor: 20 = 2² × 5. Write a little sum for each prime — all its powers from p0 up to the top — then MULTIPLY the sums:

(1 + 2 + 4) × (1 + 5) = 7 × 6 = 42

Same 42, no listing. Why does multiplying work? When you expand (1 + 2 + 4)(1 + 5) the regular way, every term is one power of 2 times one power of 5 — and that’s exactly the recipe for a divisor. The grid below shows the expansion landing on all six divisors:

Read the cells: that's 1, 2, 4 across the top of each column times 1 or 5 — every divisor of 20 appears once, so adding them all is the same as multiplying the two bracket-sums.

Bigger example. Sum of the divisors of 72 = 2³ · 3²: (1 + 2 + 4 + 8)(1 + 3 + 9) = 15 × 13 = 195. Try listing all twelve divisors of 72 and adding — you’ll get 195 too, the slow way.

Framing inspired by Competition Math for Middle School (AoPS).

Counting only the ODD divisors (or only the EVEN ones)

Sometimes a problem only wants the odd factors, or only the even ones. Don’t list and sort — the prime factorization splits them for you.

The key fact: a divisor is odd exactly when it has NO factor of 2. So to count odd divisors, cover up the power of 2 in the factorization and count the divisors of what’s left.

Example. How many odd divisors does 120 = 2³ · 3 · 5 have? Ignore the 2³ — the odd divisors come only from 3 · 5. Count those: (1+1)(1+1) = 4 (they are 1, 3, 5, 15).

And the EVEN divisors? Don’t count them one by one — subtract. Total divisors of 120: (3+1)(1+1)(1+1) = 16. Even ones = total − odd = 16 − 4 = 12. That’s complementary counting: count the easy set, subtract from the whole.

Framing inspired by Competition Math for Middle School (AoPS).

Two more questions you can answer instantly from prime factorizations.

• The GCD (greatest common divisor) of two numbers is the biggest number that divides BOTH. GCD(12, 18) = 6.
• The LCM (least common multiple) of two numbers is the smallest number that’s a multiple of BOTH. LCM(12, 18) = 36.

Recipe (two rules)

Write each number’s prime factorization. For each prime, look at the two exponents:

The Venn picture — one diagram, two answers

Put 12’s primes in the left circle (2, 2, 3). Put 18’s primes in the right circle (2, 3, 3). The shared primes go in the MIDDLE.

Same two circles, two different questions. GCD is the shared overlap; LCM is everything together.

Where LCM matters most: when two cycles need to line up again. Bells at 4 and 6 minutes both ring together every LCM(4, 6) = 12 minutes. Lights flashing every 8 and 10 seconds sync every LCM(8, 10) = 40 seconds.

The Euclidean algorithm — GCD without factoring

What if the two numbers are HUGE and you don’t want to factor them? There’s a 2300-year-old shortcut (Euclid wrote it down). Replace the bigger number with its remainder when divided by the smaller. Repeat until one number is zero. The other is the GCD.

Try it. What’s gcd(252, 105)?

• 252 ÷ 105 = 2 remainder 42. Now gcd(105, 42).
• 105 ÷ 42 = 2 remainder 21. Now gcd(42, 21).
• 42 ÷ 21 = 2 remainder 0. Stop.

The last non-zero number is 21. So gcd(252, 105) = 21. (Check: 252 = 21 · 12, 105 = 21 · 5 ✓.)

This is much faster than factoring 252 and 105 when the numbers get big. Most contest problems use small numbers where factoring is fine, but the algorithm is good to have in your back pocket.

Two remainders at once — find where they line up

LCM tells you when two cycles sync up exactly. Sometimes they line up at an offset instead. Here's the classic version:

Find the smallest positive whole number that leaves remainder 1 when divided by 6, and remainder 6 when divided by 11.

Don't set up equations. List each cycle and look for the first match.

• Remainder 1 when divided by 6: 1, 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, …
• Remainder 6 when divided by 11: 6, 17, 28, 39, 50, 61, …

The first number in both lists is 61. Done.

Here's the payoff: once you have one answer, the next ones come every LCM(6, 11) = 66. So the whole family is 61, 127, 193, … — start at 61, step by 66.

List the cycles, grab the first match, then step by the LCM.

The “one short” shortcut — when every remainder is divisor − 1

Some remainder problems collapse to one line if you spot the pattern. Here’s the classic:

Find the smallest whole number bigger than 1 that leaves remainder 5 when divided by 6, remainder 6 when divided by 7, and remainder 7 when divided by 8.

Don’t list anything. Notice each remainder is exactly one less than its divisor: 5 = 6−1, 6 = 7−1, 7 = 8−1. “One short of a clean multiple” is the same as saying n + 1 is a clean multiple. So n + 1 is divisible by 6, 7, AND 8 at once — that means n + 1 is a multiple of LCM(6, 7, 8).

LCM(6, 7, 8) = 168 (take 2³ from the 8, 3 from the 6, 7 from the 7). The smallest such n + 1 is 168, so n = 167. Check: 167 = 27×6 + 5 ✓, 167 = 23×7 + 6 ✓, 167 = 20×8 + 7 ✓.

Scaling — factor out the common piece

When two numbers share an obvious common chunk, pull it out FIRST. GCD and LCM both scale with it:

See it small. 24 and 36 are both 12 times something: 24 = 12×2, 36 = 12×3. So gcd(24, 36) = 12 · gcd(2, 3) = 12 · 1 = 12, and lcm[24, 36] = 12 · lcm[2, 3] = 12 · 6 = 72. No factor tree needed.

See it big. What is lcm[606060, 707070]? Both are 10101 times something: 606060 = 10101 × 60, 707070 = 10101 × 70. Pull out the 10101: lcm = 10101 · lcm[60, 70]. Now 60 = 2²·3·5 and 70 = 2·5·7, so lcm[60, 70] = 2²·3·5·7 = 420. Thus lcm = 10101 × 420 = 4,242,420 — and you never factored the six-digit monsters.

Coprime shortcut — LCM is just the product

When two numbers share NO common factor (GCD 1, relatively prime), there’s no overlap to worry about, so their LCM is the whole product:

if gcd(a, b) = 1, then lcm[a, b] = a × b

For example lcm[8, 15] = 8 × 15 = 120 (8 = 2³, 15 = 3·5 — nothing shared). This also drops straight out of the identity gcd × lcm = a × b when the GCD is 1.

(Remainder problem adapted from Problem Solving via the AMC, Australian Maths Trust.)

The digit sum of a number is what you get when you add its digits.

• Digit sum of 1234 = 1+2+3+4 = 10.
• Digit sum of 999 = 9+9+9 = 27.
• Digit sum of 50 = 5+0 = 5.

Already we used this in chapter 1: for divisibility by 3 or 9, the digit sum tells you the answer. But there's a deeper fact:

This is why the divisibility rule for 9 works. And it gives a quick trick called casting out nines: if you keep adding the digits of the digit sum, you eventually get a single digit — that's the original number's remainder mod 9 (or 9 itself if the original was a multiple of 9). Adults used this for centuries before calculators to check their arithmetic.

Example. What's 1234 mod 9? Digit sum: 1+2+3+4 = 10. Digit sum again: 1+0 = 1. So 1234 ≡ 1 (mod 9). Check: 1234 = 137·9 + 1 ✓.

Casting out nines for an arithmetic check. If 472 + 351 = 823, the digit sums should match too: 4+7+2 = 13 → 4; 3+5+1 = 9 → 9 (which is 0 mod 9); 4 + 0 = 4. And 8+2+3 = 13 → 4. Both sides give 4 ✓. (If they hadn't matched, you'd know you had an arithmetic mistake.)

Place value — what each digit is really worth

A digit sum throws away WHERE each digit sits. But a lot of contest problems are about exactly that — the position. The number 358 is not “3 and 5 and 8.” It’s

That “break it into place pieces” move is the whole idea. A two-digit number with tens digit a and ones digit b is 10a + b. A three-digit number is 100a + 10b + c. Write it that way and the position does the work.

Biggest number? Biggest digit in the highest seat.

You’re handed some digits and asked for the LARGEST number you can build. A spot worth 100 beats a spot worth 10, which beats a spot worth 1. So the biggest digit should grab the most valuable seat.

Palindromes and reversing digits

A palindrome reads the same forwards and backwards: 121, 3443, 7. Build one from the outside in — the first and last digit must match, then the next pair in, and so on.

And here’s a slick fact about reversing a two-digit number. The number is 10a + b; its reversal is 10b + a. Add them: 10a + b + 10b + a = 11(a + b). Reversal-sums are always multiples of 11. Subtract instead and you get 9(a − b) — always a multiple of 9. Place value turns “reverse the digits” into clean algebra.

Place value means each seat has a weight. Put your biggest digit where the weight is biggest.

Cryptarithms — when letters stand for digits

Some puzzles hide the digits behind letters or boxes: ABC + CBA = DDDD, or ?? × ? = 176. The rule is always the same — different letters are different digits, the same letter is the same digit every time. Your instinct is to try random digits. Resist it. Use place value plus one toehold fact.

The toehold: a repeated-digit number is a multiple. dd = 11d (so 44 = 11×4). aba-style and abab-style numbers hide a factor too (abab = ab × 101). Spotting that one factor usually cracks the puzzle.

Warm-up. The SAME digit fills every box: ?? × ? = 176. Call it d. The two-digit box ?? is dd = 11d, so 11d × d = 176, i.e. 11d² = 176. Divide: d² = 16, so d = 4. Check: 44 × 4 = 176 ✓. The hidden factor 11 cracked it instantly.

Different letters = different digits. Pin the units column first, then let place value finish it.

What's the last digit of 2 to the 100th power? That number has 31 digits — no calculator you own will show it. And yet you can name its final digit in about five seconds, because the last digit of a power doesn't wander. It marches in a tiny loop.

Watch the powers of 2 and keep only the LAST digit:

Only the last digit matters: 2, 4, 8, 6, 2, 4, 8, 6, … repeating forever in groups of 4. So we say “powers of 2 have units-digit period 4.”

The strip above is really a loop bent straight. Coil it back up and you get a little wheel — each time you raise the power by one, you take one step clockwise:

So “what does 2ⁿ end in?” becomes “where do you land after n steps around a 4-spot wheel?” — and that's exactly n mod 4. The wheel is why the exponent only matters as its leftover.

The other cycles you should know cold:

How to find the units digit of an:

1. Keep only the units digit of a.
2. Look up its cycle (above).
3. Find which spot in the cycle n lands on by computing n mod (cycle length).

Example. Units digit of 3⁷? Cycle of 3 has period 4. 7 mod 4 = 3. So we want the 3rd entry of the cycle 3, 9, 7, 1 — that's 7. (Check: 3⁷ = 2187 ✓.)

Example. Units digit of 7¹⁰⁰? Cycle of 7 has period 4. 100 mod 4 = 0, which means the last entry of the cycle (the 4th). Cycle of 7 is 7, 9, 3, 1. Last entry is 1.

Parity is one word for “is this number even or odd?” The simplest math fact in the universe — but on a contest, it’s a superpower.

The combination rules — two tables

When you ADD or MULTIPLY two numbers, the result’s parity is determined by the parities of the pieces. Memorize:

Where parity wins on a contest

• “Is X even or odd?” Track parity through the operations — you don’t need the actual value.
• “Odd number = sum of two primes”: one prime MUST be 2 (the only even prime).
• Counting moves on a grid. Each step flips parity. A square at distance 5 requires 5, 7, 9, … moves — never an even count.
• “Sum of n consecutive integers is even/odd?” Just count how many odd terms are in the list.

Whole-number equations — isolate, then list

Some problems hand you an equation like 3x + 5y = 31 and tell you x and y are whole numbers (often counts of things, so they can't be negative). One equation, two unknowns — in regular algebra that has infinitely many answers. But “whole numbers only” changes everything. Now there are only a handful.

Your instinct is to guess pairs at random. Resist it. Here's the clean method.

1. Isolate one variable. Solve for whichever one is easier: 5y = 31 − 3x, so y = (31 − 3x) ÷ 5.
2. Walk x up from 0 and keep only the x-values that make the top a clean multiple of 5 (so y comes out whole).
3. Stop the moment y would go negative.

Walk it: x = 0 → 31÷5, not whole. x = 1 → 28÷5, no. x = 2 → 25÷5 = 5 ✓. x = 3 → 22÷5, no. x = 4 → 19÷5, no. x = 5 → 16÷5, no. x = 6 → 13÷5, no. x = 7 → 10÷5 = 2 ✓. x = 8 → 7÷5, no. x = 9 → 4÷5, no. x = 10 → 3x = 30, only 1 left, not a multiple of 5; x = 11 makes 3x = 33 > 31, so y goes negative — stop.

Two solutions: (x, y) = (2, 5) and (7, 2). Notice the winning x-values jumped by 5 each time (2, then 7). That's not luck: once you find one, the next comes every 5 steps in x (the coefficient of y). Isolate one variable, list the whole-number hits, stop when it goes negative.

(Listing method adapted from Problem Solving via the AMC, Australian Maths Trust.)

Coin and ticket problems — build the equation first

Coin and ticket word problems all hide the same shape: so many things, this much total value. The hard part isn't the math — it's writing the equation.

Watch how a count condition and a value condition combine. You have 12 coins, all dimes (10¢) and quarters (25¢), worth $1.80 in all. How many of each?

Let d = dimes. Then quarters = 12 − d (the count uses up all 12 coins — that's the trick beginners miss). Now the value, in cents:

10d + 25(12 − d) = 180

Expand: 10d + 300 − 25d = 180, so −15d = −120, giving d = 8. So 8 dimes and 4 quarters. Check: 8×10 + 4×25 = 80 + 100 = 180¢ = $1.80 ✓.

The whole game: name ONE unknown, write the other count as “total minus that,” then turn the money into one equation. Count condition fixes the other variable; value condition gives the equation.

(Coin-problem setup adapted from Problem Solving via the AMC, Australian Maths Trust.)

“How many solutions?” — count without listing

Sometimes the question isn't “find x and y” — it's “how MANY whole-number pairs work?” If the count is big, listing every one is slow and you'll lose track. There's a faster way.

How many pairs of positive whole numbers (x, y) satisfy 2x + 7y = 100?

Isolate the variable with the BIGGER coefficient — it grows fastest, so it has the fewest legal values. Solve for y: 7y = 100 − 2x. For y to be a whole number, 100 − 2x must be a multiple of 7.

Find the first hit: x = 1 gives 100 − 2 = 98 = 7×14 ✓, so (1, 14) works. After that, x jumps by 7 each time (the y-coefficient) and y drops by 2: (1,14), (8,12), (15,10), (22,8), (29,6), (36,4), (43,2). The next, x = 50, forces y = 0 — not positive, so stop.

That's 7 solutions — and you found them by stepping, not by checking all 50 values of x. Once the first solution and the step size are known, you can even count by arithmetic: x runs 1, 8, …, 43, which is (43 − 1) ÷ 7 + 1 = 7 values. Find one solution, learn the step, then count — don't grind out the whole list.

(Solution-counting adapted from Problem Solving via the AMC, Australian Maths Trust.)

Modular arithmetic is the math of leftovers. Instead of asking “what’s the actual answer?” you ask “what’s left over after dividing?”

You already know it from clocks. If it’s 9:00 and you wait 5 hours, the clock reads 2:00, not 14:00. The hour hand doesn’t care that 9 + 5 = 14 — it only cares about the leftover after 12.

9 + 5 = 14, but the clock loops back at 12. So 14 mod 12 = 2. The clock IS modular arithmetic.

Days of the week work the same way, but with cycle 7. Today + 7 days = same day. Today + 30 days = same day shifted by 30 mod 7 = 2.

Reading the notation “a ≡ b (mod n)”

The superpower: do arithmetic with leftovers

Once you switch into leftover-mode, you can replace huge numbers with tiny ones before adding or multiplying. The leftover of the final answer comes out the same.

The fencepost trap — counting a range

“How many whole numbers from 4 to 20?” Your instinct says 20 − 4 = 16. Resist it — that’s the classic off-by-one mistake. Subtracting counts the GAPS between numbers, not the numbers themselves.

Think of fence posts and the panels between them. A fence with 5 posts has only 4 panels. Posts > panels, always by one.

Same trap shows up as cuts on a rope (n cuts make n+1 pieces) and posts on a track. Whenever you count things in a line, ask: am I counting the posts, or the gaps? For an inclusive count, it’s last − first plus one.