\section*{Exercise 1 - 031221}

Give, without using a table of cube numbers, all two-digit numbers whose third powers begin with the digits of the original number in the same order!
}

If there is a number \(x\) that satisfies the conditions of the problem, then \(x^{3}\) is either four-digit (Case 1), five-digit (Case 2), or six-digit (Case 3), since \(x\) is a two-digit number and \(10^{3}=1000\) and \(99^{3}=970299\).

1. Since \(x^{3}\) begins with the same digits in the same order as \(x\), \(x^{3}\) can be written in the following form: \(x^{3}=100 x+y\) with \(0 \leq y \leq 99\), i.e., it holds that: \(100 x \leq x^{3}<100 x+100,100 x \leq x^{3}<\) \(100(x+1)\) and consequently

\[
100 \leq x^{2}<100\left(1+\frac{1}{x}\right) \leq 100\left(1+\frac{1}{10}\right)=110
\]

due to \(x \geq 10\). Thus, \(100 \leq x^{2}<110\). This relationship is only satisfied for \(x=10\).
2. Since \(x^{3}\) begins with the same digits in the same order as \(x\), \(x^{3}\) can be written in the following form: \(x^{3}=1000 x+y\) with \(0 \leq y \leq 999\), i.e., it holds that:

\[
1000 x \leq x^{3}<1000 x+1000 \quad ; \quad 1000 x \leq x^{3}<1000(x+1)
\]

and consequently \(1000 \leq x^{2}<1000\left(1+\frac{1}{x}\right) \leq 1100\).

This relationship is only satisfied for \(x=32\).

3. Since \(x^{3}\) begins with the same digits in the same order as \(x\), \(x^{3}\) can be written in the following form: \(x^{3}=10000 x+y\) with \(0 \leq y \leq 9999\), i.e., it holds that:

\[
10000 x \leq x^{3}<10000 x+10000 \quad ; \quad 10000 x \leq x^{3}<10000(x+1)
\]

and consequently \(10000<x^{2}\). This is not possible, since \(x\) is two-digit and therefore \(x^{2}\) can be at most four-digit.

Therefore, due to \(10^{3}=1000\) and \(32^{3}=32768\), the numbers 10 and 32 are the only ones that satisfy the condition of the problem.

Adapted from \([2]\)