Difference between revisions of "2017 AMC 8 Problems/Problem 7"

Revision as of 15:10, 5 October 2018

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 $0$ . Because $0$ is a multiple of $11$ , any number $ABCABC$ is a multiple of $11$ , so the answer is $A$

-Baolan :P WVHS '22

Solution 3

The most important step is to realize that any number in the form $\overline{ABCABC} = \overline{ABC000}+\overline{ABC} = 1000\overline{ABC}+\overline{ABC} = 1001\overline{ABC}$ . Thus every number in this form is divisible by $1001$ , and the answer is $\textbf{(A) }11$ , because it is the only choice that is a factor of $1001$ .

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

@@ Line 13: / Line 13: @@
 We can see that numbers like <math>247247</math> can be written as <math>ABCABC</math>. We can see that the alternating sum of digits is <math>C-B+A-C+B-A</math>, which is <math>0</math>. Because <math>0</math> is a multiple of <math>11</math>, any number <math>ABCABC</math> is a multiple of <math>11</math>, so the answer is <math>A</math>
--Baolan (hi MVMS)
+-Baolan :P WVHS '22
 ==Solution 3==

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