Mathematical and Physical Journal
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Problem C. 1184. (October 2013)

C. 1184. Prove that 52013.21008+31008.22013 is divisible by 19.

(5 pont)

Deadline expired on November 11, 2013.


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

Megoldás. Egy lehetséges átalakítás:

\(\displaystyle 5^{2013}\cdot2^{1008}+3^{1008}\cdot2^{2013}=5\cdot2^{2}\cdot5^{2012}\cdot2^{1006}+3^{2}\cdot2\cdot3^{1006}\cdot2^{2012}=\)

\(\displaystyle =20\cdot50^{1006}+18\cdot12^{1006}=19\cdot50^{1006}+19\cdot12^{1006}+50^{1006}-12^{1006}.\)

Az összeg első két tagja osztható 19-cel. Tudjuk, hogy \(\displaystyle 50^{1006}-12^{1006}\) osztható \(\displaystyle (50-12)\)-vel, azaz \(\displaystyle 2\cdot19\)-cel. Ezzel az állítást igazoltuk.


Statistics:

216 students sent a solution.
5 points:147 students.
4 points:19 students.
3 points:19 students.
2 points:4 students.
1 point:6 students.
0 point:16 students.
Unfair, not evaluated:5 solutionss.

Problems in Mathematics of KöMaL, October 2013