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

(Created page with "== Problem == For each positive integer 4n<math>, let </math>a_n<math> be the last digit of the number. </math>1+2+3+\cdots +n<math>. Calculate </math>a_1 + a_2 + a_3 + \cdots...")
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
For each positive integer 4n<math>, let </math>a_n<math> be the last digit of the number. </math>1+2+3+\cdots +n<math>. Calculate </math>a_1 + a_2 + a_3 + \cdots + a_{1992}$.
+
For each positive integer <math>n</math>, let <math>a_n</math> be the last digit of the number. <math>1+2+3+\cdots +n</math>. Calculate <math>a_1 + a_2 + a_3 + \cdots + a_{1992}</math>.
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Revision as of 12:53, 13 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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

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