Difference between revisions of "1992 OIM Problems/Problem 1"
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a_{20}=\frac{(20)(21)}{2}\text{ mod }10=210\text{ mod }10=0 | a_{20}=\frac{(20)(21)}{2}\text{ mod }10=210\text{ mod }10=0 | ||
\end{cases}</math> | \end{cases}</math> | ||
+ | |||
+ | Using the table above we calculate: <math>\sum_{i=1}^{20}a_i=70</math>, | ||
+ | |||
+ | Hence, <math>S_{20k+p}=70k+\sum_{i=1}^{p}a_i</math> | ||
+ | |||
Revision as of 00:43, 14 December 2023
Problem
For each positive integer , let
be the last digit of the number.
. Calculate
.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Let and
be integers with
, and
Since , then
Let
Since
then,
Now we calculate through
:
Using the table above we calculate: ,
Hence,
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.