2017 AMC 8 Problems/Problem 7

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 $\boxed{\textbf{(A)}\ 11}$ .

Solution 3

To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, $+2-4+7-2+4-7=0$ . Because the result is 0, the number 247247 is divisible by 11 and so we get $\boxed{\textbf{(A)}\ 11}$ . ---LarryFlora

Solution 4

Similar to solution 1, let Z=ABCABC. To prove it is divisible by 11, we can compute its alternating sum, which is A-B+C-A+B-C=0 which is divisible by 11. Therefore the answer is $\boxed{\textbf{(A)}\ 11}$ ~PEKKA

Video Solution

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

https://youtu.be/-saKu3lLU0U

~savannahsolver

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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