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 and so we get <math>\boxed{\textbf{(A)}\ 11}</math>.
  
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.
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==Video Solution==
 
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https://youtu.be/7an5wU9Q5hk?t=647
By: Baolan (hi MVMS kids)
 
  
 
==See Also==
 
==See Also==

Revision as of 17:19, 13 February 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 and so we get $\boxed{\textbf{(A)}\ 11}$.

Video Solution

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

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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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