Difference between revisions of "2017 AMC 8 Problems/Problem 7"
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==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 | + | 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 16:19, 13 February 2021
Problem
Let 
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 ?
Solution 1
Let 
Clearly, is divisible by 
.
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 .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
