AMC 10 · 2007 · #22

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
A finite sequence of three-digit integers has this property: the tens and units digits of each term are, in order, the hundreds and tens digits of the next term, and the tens and units digits of the last term are the hundreds and tens digits of the first term, so the sequence closes into a loop. For example, such a sequence might begin 247, 475, 756 and end with 824. Let S be the sum of all the terms in the sequence. What is the largest prime that always divides S?

Pick an answer.

(A)
$\ 3$
(B)
$\ 7$
(C)
$\ 13$
(D)
$\ 37$
(E)
$\ 43$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Consecutive terms share digits, so name the whole ring d₁, d₂, …, d_n around a circle: term k is then 100 d_k + 10 d_k+1 + d_k+2.

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 is the hundreds digit of its term, the tens digit of the prior term, and the units digit of the one two back — every place 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

So add by columns, not by terms: with T the total of all the digits, S = 100T + 10T + T = 111 T.

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

Factor 111 and pin the largest forced prime

S = 111 T always, and 111 = 3 × 37; the one-term loop 111 shows nothing larger is forced, so (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

More like this

Same archetype — closest grade level first.