## Task 31/63

Given a four-digit number $Z$ with the following property: If you remove the first two digits, you get the square root of $Z$. What is the number $Z$?
Let $Z=100 x+y$, where $1 \leq y \leq 99$ and $10 \leq x \leq 99$, with $x$ and $y$ being integers. Then, due to the properties of $Z: 100 x+y=y^{2}$ or $100 x=y(y-1)$.

It is therefore necessary to find the product of two consecutive integers that is an integer multiple of 100. If we set $x=a b$, it follows that either

1.) $y=4 a ; y-1=25 b$ or 2.) $y=25 a ; y-1=2 b$

It is easy to see that further factorizations

3.) $y=2 a ; y-1=50 b$ or 4.) $y=50 a ; y-1=2 b$ or 5.) $y=10 a ; y-1=10 b$

are not useful.

From this, it follows: $25 b=4 a-1$ or $25 a=4 b+1$. The left sides of the equations are divisible by 25, so the right sides must also be. Therefore, only values that end in

75 or 25 are possible, so $b=3 ; a=19$ or $a=1 ; b=6$.

Since $a b=x \geq 10$, the values $a=1 ; b=6$ are not feasible, and since $a b=x \leq 99$, the values $b=7 ; a=44$ or $a=5 ; b=31$ and all larger values are also not feasible.

Thus, the problem is solved. It follows that

$$
x=a b=57 \quad ; \quad y=4 a=25 b+1=76
$$

so $Z=100 x+y=5776=76^{2}$. As the solution process shows, this is also the only solution.