Difference between revisions of "2017 AIME I Problems/Problem 3"
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==Solution== | ==Solution== | ||
We see that <math>d(n)</math> appears in cycles of <math>20</math>, adding a total of <math>70</math> each cycle. | We see that <math>d(n)</math> appears in cycles of <math>20</math>, adding a total of <math>70</math> each cycle. | ||
− | Since <math>\lfloor\frac{2017}{20}\rfloor=100</math>, we know that by <math>2017</math>, there have been <math>100</math> cycles, or <math>7000</math> has been added. This can be discarded, as we're just looking for the last three digits. | + | Since <math>\left\lfloor\frac{2017}{20}\right\rfloor=100</math>, we know that by <math>2017</math>, there have been <math>100</math> cycles, or <math>7000</math> has been added. This can be discarded, as we're just looking for the last three digits. |
Adding up the first <math>17</math> of the cycle of <math>20</math>, we get that the answer is <math>\boxed{069}</math>. | Adding up the first <math>17</math> of the cycle of <math>20</math>, we get that the answer is <math>\boxed{069}</math>. | ||
Revision as of 22:58, 8 March 2017
Problem 3
For a positive integer , let be the units digit of . Find the remainder when is divided by .
Solution
We see that appears in cycles of , adding a total of each cycle. Since , we know that by , there have been cycles, or has been added. This can be discarded, as we're just looking for the last three digits. Adding up the first of the cycle of , we get that the answer is .
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.