Difference between revisions of "2017 AMC 8 Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
| − | 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. Because 0 is a multiple of 11, any number ABCABC is a multiple of 11. | + | 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. Because 0 is a multiple of 11, any number ABCABC is a multiple of 11. |
==See Also== | ==See Also== | ||
Revision as of 13:59, 15 January 2018
Contents
Problem 7
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 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. Because 0 is a multiple of 11, any number ABCABC is a multiple of 11.
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 | ||
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