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 and <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 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}} |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.