A right-angled triangle has side lengths that are integers. What could be the last digit of the area's measure, if the length of the hypotenuse is not divisible by 5?
Let the lengths of the legs be $a$ and $b$, and the length of the hypotenuse be $c$. According to the Pythagorean theorem, we then have

$$
a^{2}+b^{2}=c^{2}
$$

A square number can give a remainder of 0, 1, or 4 when divided by 5. If neither $a$ nor $b$ is divisible by 5, then the left side of (1) can give a remainder of 2 (in the form $1+1$), 0 (in the form $1+4$), or 3 (in the form $4+4$) when divided by 5. Since 2 and 3 cannot be the remainder of a square number $-c^{2}$ when divided by 5, and 0 is excluded by the condition, if the length of the hypotenuse is not divisible by 5, then the length of one of the legs will be divisible by 5.

If both legs are even, then their product is divisible by 4. We claim that this is also true if there is an odd leg.

First, note that both legs cannot be odd. In this case, in (1), $a^{2}+b^{2}$ would be even, and thus $c^{2}$, as an even square number, would be divisible by 4, while $a^{2}$ and $b^{2}$, when divided by 4, would each give a remainder of 1, making their sum 2. In this case, however, equality could not hold in (1).

If now one of the legs, say $a$, is odd, then $b$ is even, and $c$ is odd. From (1),

$$
b^{2}=(c+a)(c-a)
$$

Here, both factors on the right side are even. If neither is divisible by 4, then they each give a remainder of 2 when divided by 4, so their sum, $2c$, is divisible by 4. This would, however, make $c$ even, which is not true.

Thus, in (2), the right side is also divisible by $2^{3}=8$. We know that a square number, in this case $b^{2}$, has every prime factor in its prime factorization with an even exponent. This means that $b^{2}$ is divisible by $2^{4}$, and thus $b$ is divisible by 4.

This proves that if (1) holds, then $ab$ is divisible by 4. The area of the triangle, $ab/2$, is thus an even number. Since there is a leg divisible by 5, the area is also divisible by $2 \cdot 5=10$, so the last digit of its measure is 0.

Remark. The statement - that on the one hand, in every Pythagorean triple there is a number divisible by 5, and on the other hand, the product of the legs is divisible by 4 - can also be easily shown based on the known characterization of Pythagorean triples. According to this characterization, Pythagorean triples can be written in the form $a=k\left(u^{2}-v^{2}\right)$, $b=2 k \cdot u v$, and $c=k\left(u^{2}+v^{2}\right)$, where $k$ is any natural number, and $u$ and $v$ are relatively prime numbers of different parities, with $u>v$. See, for example, Rademacher and Toeplitz's book *Numbers and Figures*.