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 <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>.
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==Solution 3==
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To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, <math>+2-4+7-2+4-7=0</math>. Because the result is 0, the number 247247 is divisible by 11 and so we get <math>\boxed{\textbf{(A)}\ 11}</math>. ---LarryFlora
  
-Baolan
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==Solution 4==
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Similar to solution 1, let Z=ABCABC. To prove it is divisible by 11, we can compute its alternating sum, which is A-B+C-A+B-C=0 which is divisible by 11. Therefore the answer is <math>\boxed{\textbf{(A)}\ 11}</math>
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~PEKKA
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==Video Solution==
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https://youtu.be/7an5wU9Q5hk?t=647
  
 
==See Also==
 
==See Also==

Revision as of 16:30, 31 December 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 $\boxed{\textbf{(A)}\ 11}$.

Solution 3

To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, $+2-4+7-2+4-7=0$. Because the result is 0, the number 247247 is divisible by 11 and so we get $\boxed{\textbf{(A)}\ 11}$. ---LarryFlora

Solution 4

Similar to solution 1, let Z=ABCABC. To prove it is divisible by 11, we can compute its alternating sum, which is A-B+C-A+B-C=0 which is divisible by 11. Therefore the answer is $\boxed{\textbf{(A)}\ 11}$ ~PEKKA

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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