Difference between revisions of "1992 OIM Problems/Problem 1"
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then, <math>S_{20k+p}=k\sum_{i=1}^{20}a_i+\sum_{i=1}^{p}a_i</math> | then, <math>S_{20k+p}=k\sum_{i=1}^{20}a_i+\sum_{i=1}^{p}a_i</math> | ||
− | Now we calculate <math>a_1</math> through <math>a_{20}: | + | Now we calculate <math>a_1</math> through <math>a_{20}</math>: |
− | < | + | <math>\begin{cases} |
a_{1}=\frac{(1)(2)}{2}\text{ mod }10=1\text{ mod }10=1\\ | a_{1}=\frac{(1)(2)}{2}\text{ mod }10=1\text{ mod }10=1\\ | ||
a_{2}=\frac{(2)(3)}{2}\text{ mod }10=3\text{ mod }10=3\\ | a_{2}=\frac{(2)(3)}{2}\text{ mod }10=3\text{ mod }10=3\\ | ||
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a_{19}=\frac{(19)(20)}{2}\text{ mod }10=190\text{ mod }10=0\\ | a_{19}=\frac{(19)(20)}{2}\text{ mod }10=190\text{ mod }10=0\\ | ||
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} | + | \end{cases}</math> |
Revision as of 00:41, 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
:
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.