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

Revision as of 13:00, 18 January 2021

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 ans so we get $\boxed{\textbf{(A)}\ 11}$ .

Video Solution

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

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 5: / Line 5: @@
 <math>\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111</math>
-==Solutions==
+==Solution 1==
-===Solution 1===
 Let <math>Z = \overline{ABCABC} = 1001 \cdot \overline{ABC} = 7 \cdot 11 \cdot 13 \cdot \overline{ABC}.</math> Clearly, <math>Z</math> is divisible by <math>\boxed{\textbf{(A)}\ 11}</math>.
-===Solution 2===
+==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 ans so we get <math>\boxed{\textbf{(A)}\ 11}</math>.
-===Video Solution===
+==Video Solution==
 https://youtu.be/7an5wU9Q5hk?t=647

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