The difference between the cube and the square of a number is in the form $a b c a b c$ (in the decimal system). Which number is this?
If the sought number is $x$, then according to the problem

$$
\begin{gathered}
x^{3}-x^{2}=100000 a+10000 b+1000 c+100 a+10 b+c100^{3}=10^{6}
$$

If $x=100$, then $x^{2}(x-1)=990,000$, which does not meet the requirement (because it is not in the form $\overline{a b c a b c}$)! Therefore, $x<100$, and $x>\sqrt[3]{100100}$

Since

$$
46<100$; this case must be excluded.
2. $x$ can only be divisible by 11. Then $x-1$ must be divisible by $7 \cdot 13=91$; since $x<100$, $x-1=91$ and $x=92$. However, 92 is not a multiple of 11 and thus this case is not valid!
3. $x$ can only be divisible by 13; then $x-1$ must be divisible by $7 \cdot 11=77$. Since $x<100$, only $x-1=77$ is possible, and thus $x=78$. This indeed fits, because $x=78$ is a multiple of 13 and

$$
78^{3}-78^{2}=468468
$$

which is a number in the form $a b c a b c$.
4. $x$ is divisible by the product $7 \cdot 11$. Then $x=77, x-1=76$. The latter is not a multiple of 13. It does not fit!
5. $x$ is divisible by the product $7 \cdot 13$. Then $x=91, x-1=90$. This also does not fit, because 90 is not a multiple of 13.
6. $x$ cannot be $11 \cdot 13$ and thus cannot be a multiple of $7 \cdot 11 \cdot 13$ either, because $x<100$.

We have exhausted all possible cases and as we can see, the problem has only one solution: $x=78$.

Gyula Ozoróczy (Verbőczy g. VI. o. Bp. I.)[^0]


[^0]:    ${ }^{1} 78^{2} \cdot 77=7 \cdot 11 \cdot 13 \cdot y ; \quad$ from this $\quad 13 y=78^{2}=6^{2} \cdot 13^{2}, \quad y=6^{2} \cdot 13=468$.