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

m (Solution 2)
(One intermediate revision by one other user not shown)
Line 10: Line 10:
  
 
==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>.
+
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>.
  
 
==Video Solution==
 
==Video Solution==

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png