\section*{Problem 1 - 031231}

Give all two-digit numbers that have the following property!

If one forms their third power and deletes all digits of this number except the last two, one obtains the original number again.
}

If there is a number \(x\) of the required kind, it can be written in the form

\[
x=10 a+b
\]

with \(a=1,2, \ldots, 9\) and \(b=0,1,2, \ldots, 9\), since \(x\) is a two-digit number.

\(b=0\) is not possible, as then the last three digits of \(x^{3}\) would be zero, which is only possible for \(x=0\). According to the problem statement, all numbers \(x\) are sought for which the following holds:

\[
x^{3}=100 n+10 a+b
\]

where \(n\) is a natural number. From (1) it follows that:

\[
x^{3}=1000 a^{3}+300 a^{2} b+30 a b^{2}+b^{3}
\]

From (2) and (3) it follows that \(b^{3}-b\) is divisible by 10, and can thus be represented in the form \(b^{3}-b=10 m \quad\) (4);

\(m\) a natural number, can be represented. Furthermore, from (2) and (3) it follows that \(30 a b^{2}+b^{3}-b-10 a\) is divisible by 100, and can thus be represented in the form

\[
30 a b^{2}+b^{3}-b-10 a=100 k \quad \text { or } \quad 3 a b^{2}+\frac{1}{10}\left(b^{3}-b\right)-a=10 k
\]

\(k\) a natural number, can be represented.

Due to (4), only the digits 1, 4, 5, 6, and 9 from the digits 1, ..., 9 are possible for \(b\). Substituting these in (5) one after the other, we get:

For \(b=1\), \(a=5\) results, i.e., \(x=51\).

For \(b=4\), \(a=2\) results, i.e., \(x=24\).

For \(b=5\), \(a=2\) or \(a=7\) results, i.e., \(x=25\) or \(x=75\).

For \(b=6\), \(a=7\) results, i.e., \(x=76\).

For \(b=9\), \(a=4\) or \(a=9\) results, i.e., \(x=49\) or \(x=99\).

Thus, for the solution of the problem, only the numbers 24, 25, 49, 51, 75, 76, 99 are possible.

By forming the third powers of these numbers, one confirms that each of them satisfies all the conditions of the problem.

Taken from \([2]\)