## Task B-1.5.

Determine the last two digits of the number whose square ends with 44.
## Solution.

Let $x, y$ be the last two digits of the natural number $n$ with the given property. The number $n$ can be written as

$$
n=100 a+10 x+y \quad \text { where } a \in \mathbb{N}, \quad x, y \in\{0,1,2 \ldots, 9\}
$$

Then we have

$$
n^{2}=(100 a+10 x+y)^{2}=(100 a)^{2}+100 x^{2}+y^{2}+2000 a x+200 a y+20 x y
$$

or in the form $n^{2}=100 b+y^{2}+20 x y, b \in \mathbb{N}$, and since the last two digits of the number $n^{2}$ are 4, the expression $y^{2}+20 x y=\ldots 44$ ends with 44.

Clearly, $y^{2}$ has the last digit 4, so the only possibilities are $y=2, y=8$.

If $y=2$, then $4+40 x=\ldots 44$ and by direct verification $(x \in\{0,1,2 \ldots, 9\})$ we can conclude that $x=1, x=6$.

If $y=8$, then $64+160 x=\ldots 44$ and by direct verification $(x \in\{0,1,2 \ldots, 9\})$ we can conclude that $x=3, x=8$.

Therefore, the last two digits of the number $n$ can be 12, 62, 38, 88.

## NATIONAL MATHEMATICS COMPETITION

## 2nd grade - high school - B variant

Šibenik, April 29, 2010.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.