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.
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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.
 
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 $n$, let $d_n$ be the units digit of $1 + 2 + \dots + n$. Find the remainder when \[\sum_{n=1}^{2017} d_n\]is divided by $1000$.

Solution

We see that $d(n)$ appears in cycles of $20$, adding a total of $70$ each cycle. Since $\left\lfloor\frac{2017}{20}\right\rfloor=100$, we know that by $2017$, there have been $100$ cycles, or $7000$ has been added. This can be discarded, as we're just looking for the last three digits. Adding up the first $17$ of the cycle of $20$, we get that the answer is $\boxed{069}$.

See Also

2017 AIME I (ProblemsAnswer KeyResources)
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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