2017 AMC 8 Problems/Problem 7

Revision as of 17:19, 13 February 2021 by Serenaxu (talk | contribs) (Solution 2)

Problem

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 1

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

Solution 2

We are given one of the numbers Z can be so we can just try out the options to see which one is divisible by 247247 and so we get $\boxed{\textbf{(A)}\ 11}$.

Video Solution

https://youtu.be/7an5wU9Q5hk?t=647

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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