Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem C. 1075. (April 2011)

C. 1075. Find the three-digit number that is twelve times the sum of its digits.

(5 pont)

Deadline expired on May 10, 2011.

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

Megoldás. $\displaystyle N=\overline{abc}=100a+10b+c=12(a+b+c)$ a feladat szerint: $\displaystyle 88a-2b=11c$ szerint $\displaystyle c$ páros, ugyanakkor $\displaystyle 2b=11(8a-c)$ miatt $\displaystyle b$ osztható 11-gyel. Mivel $\displaystyle b\le 9$, ezért $\displaystyle b=0$ és így $\displaystyle 8a=c$, amiből $\displaystyle c\le 9$ figyelembevételével $\displaystyle a=1$, $\displaystyle c=8$: $\displaystyle \mathbf{N=108}$.

### Statistics:

 218 students sent a solution. 5 points: 128 students. 4 points: 39 students. 3 points: 27 students. 2 points: 14 students. 1 point: 7 students. 0 point: 1 student. Unfair, not evaluated: 2 solutions.

Problems in Mathematics of KöMaL, April 2011