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Problem B. 4353. (April 2011)

B. 4353. Let A be a positive integer, and let B denote the number obtained by writing the digits of A in reverse order. Show that at least one of the numbers A+B and A-B is divisible by 11.

(Suggested by J. Mészáros, Jóka)

(3 pont)

Deadline expired on May 10, 2011.

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

Megoldás. Legyenek \(\displaystyle A\) számjegyei \(\displaystyle a_n, a_{n-1},\ldots, a_0\); ekkor \(\displaystyle A=10^na_n+10^{n-1}a_{n-1}+\ldots+a_0\) és \(\displaystyle B=10^na_0+\ldots+10a_{n-1}+a_n\). Ha \(\displaystyle n\) páratlan, akkor az

\(\displaystyle A+B=\sum_{i=0}^n(10^i+10^{n-i})a_i\)

összeg minden tagja osztható \(\displaystyle 11\)-gyel, hiszen \(\displaystyle i<n/2\) esetén

\(\displaystyle 10^i+10^{n-i}=10^{i}(10^{n-2i}+1),\)

ahol \(\displaystyle n-2i\) páratlan lévén \(\displaystyle 10+1\mid 10^{n-2i}+1\), \(\displaystyle i>n/2\) esetén pedig

\(\displaystyle 10^i+10^{n-i}=10^{n-i}(10^{2i-n}+1),\)

ahol \(\displaystyle 10+1\mid 10^{2i-n}+1\), hiszen \(\displaystyle 2i-n\) páratlan. Ezért ilyenkor \(\displaystyle A+B\) osztható 11-gyel.

Hasonlóképpen, ha \(\displaystyle n\) páros, akkor \(\displaystyle n-2i\) és \(\displaystyle 2i-n\) is páros, így az

\(\displaystyle A-B=\sum_{i=0}^n(10^i-10^{n-i})a_i\)

szám 11-gyel való osztatósága leolvasható a

\(\displaystyle 10^i-10^{n-i}=-10^{i}(10^{n-2i}-1)\qquad (i< n/2),\)

illetve a

\(\displaystyle 10^i-10^{n-i}=10^{n-i}(10^{2i-n}-1)\qquad (i\ge n/2)\)



118 students sent a solution.
3 points:87 students.
2 points:16 students.
1 point:7 students.
0 point:7 students.
Unfair, not evaluated:1 solutions.

Problems in Mathematics of KöMaL, April 2011