Mathematical and Physical Journal
for High Schools
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Problem K. 391. (November 2013)

K. 391. Add all positive integers for which the remainder on division by 2013 equals the quotient.

(6 pont)

Deadline expired on December 10, 2013.


Sorry, the solution is available only in Hungarian. Google translation

Megoldás. A 2013-mal való osztásnál 2013-féle maradék lehet. A 0 ezek közül most nem felel meg, mert nem pozitív szám. Vagyis megfelelő szám 2012 darab van: \(\displaystyle 1\cdot2013+1\), \(\displaystyle 2\cdot2013+2\), …, \(\displaystyle 2012\cdot2013+2012\). Ezek összege:

\(\displaystyle S=(1\cdot2013+1)+(2\cdot2013+2)+\ldots+(2012\cdot2013+2012)=\)

\(\displaystyle =2013\cdot(1+2+\ldots+2012)+(1+2+\ldots2012)=2014\cdot(1+2+\ldots+2012)=2014\cdot\frac{2013\cdot2012}{2}=4078507092.\)


Statistics:

211 students sent a solution.
6 points:142 students.
5 points:8 students.
4 points:11 students.
3 points:7 students.
2 points:5 students.
1 point:14 students.
0 point:16 students.
Unfair, not evaluated:8 solutions.

Problems in Mathematics of KöMaL, November 2013