AMC 10 · 2017 · #23
Grade 6 number-theoryPick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Dividing a 79-digit number by 45 head-on is hopeless. But 45 = 9 x 5, and 9 and 5 have no common factor, so the remainder mod 45 is locked in once we know the remainder mod 9 and the remainder mod 5. Each of those is easy: mod 5 depends only on the last digit, and mod 9 depends only on the digit sum. Find the two small remainders, then test the few possibilities that fit both.
Split 45 into 9 and 5
Factor 45 = 9 × 5. Since 9 and 5 are coprime, N mod 45 is fixed by N mod 9 and N mod 5 — two easy divisions replace one hard one.
Breaking 45 into the coprime pieces 9 and 5 turns one impossible division into two easy ones.
6.NS.B.4Identify SubproblemsRemainder when divided by 5
A number's remainder mod 5 comes from its last digit alone. N's last digit is 4, so N leaves remainder 4 mod 5.
Tens, hundreds, and beyond are all multiples of 5, so only the last digit can leave a remainder.
4.NBT.B.6Look For A PatternRemainder when divided by 9
N's remainder mod 9 equals its digit sum's, since every power of 10 is 1 mod 9. Here 1+2+…+44 = 990 = 9 × 110, so N mod 9 = 0.
Every power of 10 is one more than a multiple of 9, so a number and its digit sum always land on the same remainder.
6.EE.A.3Look For A PatternCombine the two clues
We need a multiple of 9 that also leaves remainder 4 mod 5. Among 0, 9, 18, 27, 36, only 9 works, since 9 = 5 + 4. So the answer is (C).
Only one multiple of 9 under 45 also lands 4 past a multiple of 5, so it must be the answer.
4.OA.B.4Eliminate PossibilitiesTo divide a giant number by 45, split 45 into 9 and 5: the last digit handles the 5 and the digit sum handles the 9, then find the one remainder that fits both.
- Split 45 into 9 and 5
- Remainder when divided by 5
- Remainder when divided by 9
- Combine the two clues