Difference between revisions of "2017 AMC 10A Problems/Problem 20"
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We can find out that the least number of digits the number <math>N</math> is <math>142</math>, with <math>141</math> <math>9</math>'s and one <math>5</math>. | We can find out that the least number of digits the number <math>N</math> is <math>142</math>, with <math>141</math> <math>9</math>'s and one <math>5</math>. | ||
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 <math>1</math> 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. | + | Knowing that this number is ONLY divisible by <math>9</math> when <math>6</math> is subtracted, we can subtract <math>6</math> 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 <math>6</math> from every number, we can conclude that <math>1233</math> (originally <math>1239</math>) is the only number divisible by <math>9</math>. |
So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>. | So our answer is <math>\boxed{\textbf{(D)}\ 1239}</math>. | ||
Revision as of 12:39, 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 to this number, we get: ....... Knowing that this number is ONLY divisible by when is subtracted, we can subtract from every available choice, and see if the number is divisible by afterwards. After subtracting from every number, we can conclude that (originally ) is the only number divisible by . 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 |
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