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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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