AMC 10 · 2007 · #11

Grade 6 number-theory
Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery
place-valuedigit-decompositiondivisibility-rulesprime-factorization identify-subproblems ↑ Prerequisites: place-value
📏 Medium solution 💡 3 insights
Problem
Three-digit numbers form a loop, each one's last two digits opening the next. Find the largest prime that must divide the sum of every such loop.

Pick an answer.

(A)
3
(B)
7
(C)
13
(D)
37
(E)
43
How to solve
Strategy Introduce a Variable

The overlap rule means each digit is shared by three consecutive terms, so the loop is really one ring of digits d₁, d₂, …, d_n written around a circle. Tool #4 (Introduce a Variable) names those digits and writes each term as 100 d_k + 10 d_k+1 + d_k+2. Tool #5 (Look for a Pattern) spots that every digit lands in the hundreds place once, the tens place once, and the units place once as you slide around the ring. Tool #15 (Organize Information in More Ways) adds the terms column by column instead of term by term, which collapses the sum to 111 times the digit total. Then Tool #3 (Eliminate Possibilities) factors 111 and uses the answer choices plus one tiny loop to pin down the largest prime that is forced every time.

1STEP 1

Name the ring of digits

Naming the ring of digits is simpler than naming the numbers.

term_k = 100 d_k + 10 d_k+1 + d_k+2, k = 1, 2, …, n (indices mod n)
2STEP 2

Each digit visits every place once

Each digit visits every place exactly once.

d_k appears as: 100 d_k (term k), 10 d_k (term k-1), 1 · d_k (term k-2)
3STEP 3

Add by place value, not term by term

Adding by place value factors the sum into 111 times the digit total.

S = 100T + 10T + T = (100 + 10 + 1) T = 111 T, T = d₁ + … + d_n
4STEP 4

Factor 111 and pin the largest forced prime

Factoring shows nothing larger is forced, so the answer is 37, choice (D).

111 = 3 × 37; S = 111 T → 37 ∣ S always; S = 111 shows nothing larger is forced → (D)
Answer
37
Test the given example digits. The ring 2,4,7,5,6,… makes terms 247, 475, 756, …; whatever the full ring is, the claim says S must be 111 times the digit total. Check a complete small loop directly: 247 + 475 + 752 uses the ring 2,4,7,5 closing back, and 247+475+752+524 = 1998 = 111 × 18, and 2+4+7+5 = 18 — the 111 T rule holds. Both 3 and 37 divide 1998. The trap answers are the other primes near 37: 7, 13, and 43 each divide some loops but not the plain loop 111, so none is guaranteed; and 3 works always but is smaller than 37. So 37 being the largest forced prime is consistent.
💡Key takeaway

Because the loop makes every digit land in the hundreds, tens, and units place exactly once, the whole sum is always 111 = 3 × 37 times the digit total, so 37 is the biggest prime guaranteed to divide it.

  • Name the ring of digits
  • Each digit visits every place once
  • Add by place value, not term by term
  • Factor 111 and pin the largest forced prime

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