Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem K. 252. (March 2010)

K. 252. The sum of six consecutive integers is multiplied by the sum of the next six integers. Prove that the product obtained in this way will always leave the same remainder when divided by 36.

(6 pont)

Deadline expired on April 12, 2010.

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

Megoldás. Legyen az első szám $\displaystyle a$. Ekkor a szorzat $\displaystyle (6a+15)(6a+51)=36a^2+396a +765=36(a^2+11a+21)+9$. A keresett osztási maradék 9.

### Statistics:

 116 students sent a solution. 6 points: 77 students. 5 points: 15 students. 4 points: 1 student. 3 points: 4 students. 2 points: 5 students. 1 point: 2 students. 0 point: 7 students. Unfair, not evaluated: 5 solutions.

Problems in Mathematics of KöMaL, March 2010