Difference between revisions of "2017 AIME I Problems/Problem 3"
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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. | 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{69}</math>. | Adding up the first <math>17</math> of the cycle of <math>20</math>, we get that the answer is <math>\boxed{69}</math>. | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2017|n=I|num-b=2|num-a=4}} | {{AIME box|year=2017|n=I|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:43, 6 November 2017
Problem 3
For a positive integer 
, let 
be the units digit of 
. Find the remainder when
![\[\sum_{n=1}^{2017} d_n\]](http://latex.artofproblemsolving.com/c/2/5/c25a797fd18625aea3a70ae1b6906027405232ff.png)
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. 
