AMC 8 · 2017 · #7

Grade 4 number-theory
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
place-valueprime-factorizationdivisibility-rulesfactors identify-subproblemspattern-recognition ↑ Prerequisites: multi-digit-arithmeticfactors
📏 Medium solution 💡 3 insights
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Problem
A 6-digit number Z has the special form where its first three digits repeat as its last three digits, like 247247. The problem asks which of the five choices — 11, 19, 101, 111, 1111 — must always be a factor of Z, no matter what those three repeated digits are.

Pick an answer.

(A)
11
(B)
19
(C)
101
(D)
111
(E)
1111

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

Instead of jumping into algebra, we use Tool #9 (Easier Problem) by working with the concrete example 247247 the problem hands us, plus one or two more abcabc numbers. Tool #5 (Pattern) helps us notice that every abcabc number is exactly abc × 1001 — once we see that pattern, the question reduces to "which choices divide 1001?". Finally Tool #3 (Eliminate Possibilities) checks the five choices against 1001 to pick the winner. We deliberately avoid Tool #13 (Algebra) — the place-value pattern is easier to discover by playing with one example than by setting up variables first.

1STEP 1

Start with the given example: divide 247247 by its repeated block 247 to get 1001.

247247 ÷ 247 = 1001
2STEP 2

Confirm it's no fluke: 315315 ÷ 315 also gives 1001.

315315 ÷ 315 = 1001
3STEP 3

By place value the front abc is worth 1000 copies, so abcabc = abc × 1001.

abcabc = abc × 1000 + abc = abc × 1001
4STEP 4

Any factor of 1001 divides Z, so factor it: 1001 = 7 × 11 × 13.

1001 = 7 × 11 × 13
5STEP 5

Test each choice against 1001: only 11 divides it (1001 ÷ 11 = 91); the rest leave remainders.

1001 ÷ 11 = 91 ✓ others: not whole
Answer
11
Spot-check with two different abcabc numbers. 247247 ÷ 11 = 22477 (whole) and 123123 ÷ 11 = 11193 (whole) — 11 works for both. Now try a choice that shouldn't always work, like 19: 123123 ÷ 19 ≈ 6480.16 is not whole, so 19 is not a guaranteed factor. Only 11 — sitting inside 1001 = 7 × 11 × 13 — survives for every abcabc. Answer (A) 11 holds up.
💡Key takeaway

This AMC 8 problem only needs Grade 4 place value and factor pairs you already know — once you see that abcabc = abc × 1001, it's just asking which choice divides 1001!

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