1992 OIM Problems/Problem 1
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,
Using the table above we calculate:
Therefore,
Hence,
- Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. Although I did get the correct answer, I was only given 8 out of 10 points because I didn't prove that the pattern repeats every 20 numbers. I simply listed the calculations for the first 23 to 26
's and realized that the pattern repeated. Without proof I proceeded to calculate 20 times 99 plus the summation from
through
and despite getting the correct numerical answer, I was not awarded the full points for not sufficient proof. I submitted such proof on this page.
~Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.