2017 AMC 8 Problem 7

Attempt Problem 7 of the 2017 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

Let ZZ be a 66-digit positive integer, such as 247247,247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must be a factor of Z?Z?
11 11
19 19
101 101
111 111
1111 1111
Answer: A
Concepts:prime factorizationplace value
Small hint:
Let nn be the three-digit block that is repeated. Express ZZ in terms of nn.
Big hint:
Factor 10011001.
Video solution:
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Written solution:
Let nn be the three-digit number formed by either repeated block. Then Z=1000n+n=1001n=7⋅11⋅13⋅n.\begin{aligned} Z&=1000n+n\\ &=1001n\\ &=7\cdot11\cdot13\cdot n. \end{aligned} Therefore, 1111 must be a factor of Z.Z.
Thus, A is the correct answer.
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