Difference between revisions of "2017 AMC 8 Problems/Problem 7"
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<math>\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111</math> | <math>\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111</math> | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/7an5wU9Q5hk?t=647 | ||
==Solution 1== | ==Solution 1== | ||
Revision as of 20:03, 27 October 2020
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 ?
Video Solution
https://youtu.be/7an5wU9Q5hk?t=647
Solution 1
Let 
Clearly, is divisible by 
.
Solution 2
We can see that numbers like 
can be written as 
. We can see that the alternating sum of digits is 
, which is 
. Because is a multiple of 
, any number is a multiple of , so the answer is .
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. 
