2017 AMC 8 Problems/Problem 7

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 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 can see that numbers like 247247 can be written as ABCABC. We can see that the alternating sum of digits is C-B+A-C+B-A, which is 0, which means any number ABCABC is a multiple of 11.

By: Baolan (hi MVMS kids)

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

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2017 AMC 8 Problems/Problem 7

Contents

Problem 7

Solution 1

Solution 2

See Also