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

(5 points)

Deadline expired on 10 May 2011.

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

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 on problem C. 1075.
 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

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