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