Difference between revisions of "1992 OIM Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | {{ | + | |
| + | <math>a_n=\frac{n(n+1)}{2}\text{ mod } 10</math> | ||
| + | |||
| + | Let <math>S_n=a_1 + a_2 + a_3 + \cdots + a_n</math> | ||
| + | |||
| + | Let <math>k</math> be an integer and <math>p</math> be | ||
| + | |||
| + | {{alternate solutions}} | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe7.htm | https://www.oma.org.ar/enunciados/ibe7.htm | ||
Revision as of 23:23, 13 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 
Let 
be an integer and 
be
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
