Difference between revisions of "2017 AIME I Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | We see that <math> | + | We see that <math>d_n</math> appears in cycles of <math>20</math> and the cycles are <cmath>1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,</cmath> adding a total of <math>70</math> each cycle. |
− | 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 | + | 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 and <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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/BiiKzctXDJg ~Shreyas S | ||
==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}} |
Latest revision as of 19:00, 17 June 2020
Contents
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 and the cycles are adding a total of each cycle. Since , we know that by , there have been cycles and 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 .
Video Solution
https://youtu.be/BiiKzctXDJg ~Shreyas S
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 |
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