Difference between revisions of "1992 OIM Problems/Problem 1"

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== Solution ==
 
== Solution ==
{{solution}}
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<math>a_n=\frac{n(n+1)}{2}\text{ mod } 10</math>
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Let <math>S_n=a_1 + a_2 + a_3 + \cdots + a_n</math>
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Let <math>k</math> be an integer and <math>p</math> be
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{{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 00:23, 14 December 2023

Problem

For each positive integer $n$, let $a_n$ be the last digit of the number. $1+2+3+\cdots +n$. Calculate $a_1 + a_2 + a_3 + \cdots + a_{1992}$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

$a_n=\frac{n(n+1)}{2}\text{ mod } 10$

Let $S_n=a_1 + a_2 + a_3 + \cdots + a_n$

Let $k$ be an integer and $p$ be

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

https://www.oma.org.ar/enunciados/ibe7.htm