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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 points)

This problem is for grade 9 students only.

Deadline expired on 12 April 2010.

Google Translation (Sorry, the solution is published in Hungarian only.)

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 on problem K. 252.
 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

•  Támogatóink: Morgan Stanley