The number 76 has an interesting property: the last two digits of the number $76^{2}=5776$ are 76 again.

a) Are there any other two-digit numbers with this property?

b) Find all such three-digit numbers $A$ such that the last three digits of the number $A^{2}$ form the number $A$.

c) Does there exist an infinite sequence of digits $a_{1}, a_{2}, a_{3}, \ldots$, such that for any natural number $n$, the square of the number $\overline{a_{n}} \overline{a_{n-1}} \bar{\ldots} \bar{\ldots} \overline{\overline{2}} \overline{a_{1}^{-}}$ ends with these same $n$ digits? The obvious answer $a_{1}=1$ and $0=a_{2}=a_{3}=\ldots$ is excluded.
a) If $A^{2}$ ends with the same two digits as $A$, then $A^{2}-A=A(A-1)$ is divisible by $100=25 \cdot 4$. Since the numbers $A$ and $A-1$ are coprime, one of them must be divisible by 25, and the other by 4. Let's try whether each of the numbers 25, 50, and 75 can serve as $A$ or $A-1$ (both these numbers are two-digit). For this, we only need to check which of the numbers adjacent to them are divisible by 4. These will be only 76 and 24, so $A$ can only be 25 and 76.

![](https://cdn.mathpix.com/cropped/2024_05_06_49e49f22191b45b40a57g-02.jpg?height=74&width=1987&top_left_y=2098&top_left_x=1)
the digit $a_{n}$ can be uniquely chosen so that the square of the number $A=\overline{a_{n}^{-}} \overline{a_{n-1}} \ldots a_{1}^{-}$ ends with $\bar{a}_{n}^{-} \bar{a}_{n-1} \ldots \bar{a}_{1}^{-}$. Proof. Let $B^{2}=\ldots \overline{b_{n}^{-}} \overline{a_{n-1}} \overline{-\ldots} \bar{a}_{1}^{-}$. Then $(n \geq 2 \Rightarrow 2 n>n+1) A^{2}=\left(10^{n} a_{n}+B\right)^{2}=10^{2 n} a_{n}+2 \cdot 10^{n} a_{n} B+$ $B^{2}=\ldots \bar{\ldots} c_{n}^{-}-\overline{a_{n-1}} \ldots a_{1}^{-}$, where $c_{n}$ is the last digit of the number $2 a_{n} a_{1}+b_{n}$. We need to choose $a_{n}$ so that $c_{n}$ equals $a_{n}$, that is, so that $2 a_{n} a_{1}+b_{n}-a_{n}=\left(2 a_{1}-1\right) a_{n}+b_{n}$ is divisible by 10. If $a_{1}=5$, then $a_{n}=b_{n}$, and if $a_{1}=6$, then $a_{n}=$ $10-b_{n}$ for $b_{n}>0$ and $a_{n}=0$ for $b_{n}=0$.

Therefore, in b) we get the numbers 625 and $376\left(76^{2}=5766\right)$, and in c) the sequences ... 8212890625 and 1787109376

## Answer

a) There are;
b) 376 and 625;
c) exists.

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