Difference between revisions of "2017 AMC 10A Problems/Problem 20"
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~ProGameXD | ~ProGameXD | ||
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+ | ==Solution 3== | ||
+ | The number <math>n</math> can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice. | ||
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+ | If <math>\boxed{\textbf{(A)}\ 1}</math> is correct, then <math>n</math> must be some number <math>99999999...9</math>, because when we add one to <math>99999999...9</math> we get <math>10000000...00</math>. Thus, if <math>1</math> is the correct answer, then the equation <math>9x=1274</math> must have an integer solution (i.e. <math>1274</math> must be divisible by <math>9</math>). But since it does not, <math>1</math> is not the correct answer. | ||
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+ | If <math>\boxed{\textbf{(B)}\ 3}</math> is correct, then <math>n</math> must be some number <math>29999999...9</math>, because when we add one to <math>29999999...9</math>, we get <math>30000000...00</math>. Thus, if <math>2</math> is the correct answer, then the equation <math>2+9x=1274</math> must have an integer solution. But since it does not, <math>3</math> is not the correct answer. | ||
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+ | Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if <math>S(n+1)=N</math>, then <math>n</math> must be a number whose initial digits sum to <math>N-1</math>, and whose other, terminating digits, are all <math>9</math>. Thus, we can evaluate the three final possibilities by seeing if the equation <math>(N-1)+9x=1274</math> has an integer solution. | ||
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+ | The equation does not have an integer solution for <math>N=12</math>, so <math>\boxed{\textbf{(C)}\ 12}</math> is not correct. However, the equation does have an integer solution for <math>N=1239</math> (<math>x=4</math>), so <math>\boxed{\textbf{(D)}\ 1239}</math> is the answer. | ||
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==See Also== | ==See Also== |
Revision as of 18:59, 30 April 2017
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 .
~ProGameXD
Solution 3
The number can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.
If is correct, then must be some number , because when we add one to we get . Thus, if is the correct answer, then the equation must have an integer solution (i.e. must be divisible by ). But since it does not, is not the correct answer.
If is correct, then must be some number , because when we add one to , we get . Thus, if is the correct answer, then the equation must have an integer solution. But since it does not, is not the correct answer.
Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if , then must be a number whose initial digits sum to , and whose other, terminating digits, are all . Thus, we can evaluate the three final possibilities by seeing if the equation has an integer solution.
The equation does not have an integer solution for , so is not correct. However, the equation does have an integer solution for (), so is the answer.
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.