Difference between revisions of "2017 AMC 8 Problems/Problem 7"
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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>. | 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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==See Also== | ==See Also== |
Revision as of 20:15, 20 July 2019
Contents
[hide]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 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 |
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