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

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==Problem 7==
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==Problem==
  
 
Let <math>Z</math> 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 <math>Z</math>?
 
Let <math>Z</math> 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 <math>Z</math>?
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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 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.
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==Video Solution==
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https://youtu.be/7an5wU9Q5hk?t=647
  
 
==See Also==
 
==See Also==

Revision as of 12:07, 19 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

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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