Difference between revisions of "2017 AMC 10A Problems/Problem 20"
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By randomly mixing the digits up, we are likely to get: <math>9999</math>...<math>9995999</math>...<math>9999</math>. | By randomly mixing the digits up, we are likely to get: <math>9999</math>...<math>9995999</math>...<math>9999</math>. | ||
By adding 1 to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>. | By adding 1 to this number, we get: <math>9999</math>...<math>9996000</math>...<math>0000</math>. | ||
− | + | Knowing that this number is ONLY divisible by <math>9</math> when <math>6</math> is subtracted, we can subtract 6 from every available choice, and see if the number is divisible by <math>9</math> afterwards. | |
After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9. | After subtracting 6 from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by 9. | ||
So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>. | So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>. |
Revision as of 12:38, 10 February 2017
Contents
Problem
Let equal the sum of the digits of positive integer . For example, . For a particular positive integer , . Which of the following could be the value of ?
Solution 1
Note that . This can be seen from the fact that . Thus, if , then , and thus . The only answer choice that is is .
Solution 2
We can find out that the least number of digits the number is , with 's and one . By randomly mixing the digits up, we are likely to get: ....... By adding 1 to this number, we get: ....... Knowing that this number is ONLY divisible by when is subtracted, we can subtract 6 from every available choice, and see if the number is divisible by afterwards. After subtracting 6 from every number, we can conclude that (originally ) is the only number divisible by 9. So our answer is .
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AMC 10 Problems and Solutions |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.