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

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==Solution 2==
 
==Solution 2==
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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>.
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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>\boxed{\textbf{(A)}\ 11}</math>.
 
  
 
==See Also==
 
==See Also==

Revision as of 16:29, 9 November 2020

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 $Z$?

$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Video Solution

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

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

See Also

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

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