Difference between revisions of "2017 AIME I Problems/Problem 3"
(→Solution) |
|||
| (7 intermediate revisions by the same user not shown) | |||
| Line 6: | Line 6: | ||
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. | 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 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 | + | Adding up the first <math>17</math> of the cycle of <math>20</math>, we can see that the answer is <math>\boxed{069}</math>. |
| + | ~ Maths_Is_Hard | ||
==Video Solution== | ==Video Solution== | ||
| − | https://youtu.be/BiiKzctXDJg ~ | + | https://youtu.be/BiiKzctXDJg ~ Shrea 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 22:37, 6 February 2024
Contents
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 
and the cycles are ![\[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\]](http://latex.artofproblemsolving.com/7/8/b/78b3ef83b37bec9ad8bb199f6c02360123c5e907.png)
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 can see that the answer is 
.
~ Maths_Is_Hard
Video Solution
https://youtu.be/BiiKzctXDJg ~ Shrea 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. 
