2017 AMC 8 Problems/Problem 7

Revision as of 15:50, 22 November 2017 by Nukelauncher (talk | contribs) (Solution)

Problem 7

Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$?

$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Solution

Let $Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.$ Clearly, $Z$ is divisible by $\boxed{\textbf{(A)}\ 11}$.

~nukelauncher

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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