## Task $32 / 64$

Which cube numbers under $10^{9}$ end with their last three digits being a cube number again?
The basis of the sought cubic number has the form $10 x+y$. The cases $x=0$ and $y=0$ are trivial and can therefore be disregarded. Thus, it holds that $0<x \leq 99$ and $0<y \leq 9, x$ and $y$ are integers. The requirement of the problem states that

$$
(10 x+y)^{3}=10^{3} k+z^{2}=10^{3} k+y^{3}
$$

First, we prove that $z=y$. If

$$
(10 x+y)^{3}=10^{3} k+z^{2} \quad \text { with } z \neq y \text { and } 0<z \leq 9
$$

it would follow that

$$
\begin{array}{r}
10^{3} x^{3}+3 \cdot 10^{2} x^{2} y+3 \cdot 10 x y^{2}+y^{3}=10^{3} k+z^{3} \\
10\left(10^{2} x^{3}-10^{2} k+30 x^{2} y+3 x y^{2}\right)=z^{3}-y^{3}
\end{array}
$$

i.e., $z^{3}-y^{3}$ would have to be divisible by 10. This would only be possible if two of the first nine cubic numbers had the same units digit, but they are all different: $1 ; 8 ; 7 ; 4 ; 5 ; 6 ; 3 ; 2 ; 9$. A rearrangement of (1) yields

$$
\begin{array}{r}
10^{3} x^{3}+3 \cdot 10^{2} x^{2} y+3 \cdot 10 x y^{2}+y^{3}=10^{3} k+z^{3} \\
10 \cdot 3 x y(10 x+y)=10^{3}\left(k-x^{3}\right) \\
3 x y(10 x+y)=100\left(k-x^{3}\right)
\end{array}
$$

Thus, the product on the left must be divisible by 100. Since 3 is coprime to 100, it follows that $x y(10 x+y)$ must be divisible by 100. We now distinguish three cases:

1. For $y=1 ; 3 ; 7 ; 9$, this means: $x$ must be divisible by 100. This contradicts the assumption $0<x \leq 99$, and thus these values for $y$ are excluded.
2. For $y=5$, $(10 x+y)$ is not divisible by 4; therefore, $x$ must be divisible by 4: $x=4 n$ with $0<n \leq 24$, $n$ an integer.

This yields the 24 values $x=4 ; 8 ; 12 ; \ldots ; 92 ; 96$.

3. For $y=2 ; 4 ; 6 ; 8$, it follows that, since neither $y$ nor $(10 x+y)$ is divisible by 5, $x$ must be divisible by 25. This yields the values $x=25 ; 50 ; 75$.

Thus, we obtain the following result:

Of the 999 cubic numbers below $10^{9}$, 36 have the desired property, namely the following:

| $45^{3}$ | $85^{3}$ | $125^{3}$ | $165^{3}$ | $205^{3}$ | $245^{3}$ | $285^{3}$ | $325^{3}$ | $365^{3}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $405^{3}$ | $445^{3}$ | $485^{3}$ | $525^{3}$ | $565^{3}$ | $605^{3}$ | $645^{3}$ | $685^{3}$ | $725^{3}$ |
| $765^{3}$ | $805^{3}$ | $845^{3}$ | $885^{3}$ | $925^{3}$ | $965^{3}$ | $252^{3}$ | $254^{3}$ | $256^{3}$ |
| $258^{3}$ | $502^{3}$ | $504^{3}$ | $506^{3}$ | $508^{3}$ | $752^{3}$ | $754^{3}$ | $756^{3}$ | $758^{3}$ |