How many positive, relatively prime number pairs (unordered) are there, where the sum of the numbers is $285$?
We are looking for the positive integer solutions of the equation $285=a+(285-a)$, for which $a$ and $285-a$ are coprime. This precisely means that $a$ and 285 are coprime. Indeed, if an integer $d$ is a divisor of both $a$ and 285, then it is also a divisor of $(285-a)$ (and $a$); conversely, if an integer $d$ is a divisor of both $a$ and $(285-a)$, then it is also a divisor of ( $a$ and) $a+(285-a)=285$. Therefore, the greatest common divisor of $a$ and 285 is equal to the greatest common divisor of $a$ and $(285-a)$.

Thus, we need to find the numbers less than 285 that are coprime with 285.

It is known that if the prime factorization of $n$ is

$$
n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}
$$

then the number of integers coprime with $n$ is given by the formula

$$
\varphi(n)=n \cdot\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \ldots\left(1-\frac{1}{p_{k}}\right)
$$

In our case, $n=285=3 \cdot 5 \cdot 19$, and thus

$$
\varphi(285)=285\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{19}\right)=285 \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{18}{19}=144
$$

is the number of coprime numbers. As we have seen, these can be grouped into pairs such that the sum of the pairs is 285. Thus, there are a total of 72 suitable pairs of numbers.

Springer Éva (Budapest, Szent István Gymn., 9th grade)

Remark. The problem can also be solved without knowledge of the $\varphi$ function by counting how many numbers up to 285 are divisible by 3, 5, 19, $3 \cdot 5=15$, $3 \cdot 19=57$, $5 \cdot 19=95$, and $3 \cdot 5 \cdot 19=285$:

3 is a divisor of 95 numbers, 5 is a divisor of 57 numbers, 19 is a divisor of 15 numbers, from the sum of these, we need to subtract the number of numbers divisible by 15, 57, 95, and $3 \cdot 5 \cdot 19=285$, i.e., $95+57+15-18-5-3+1=141$, because no coprime pairs can be formed from these. The remainder $285-141=144$ can indeed form such pairs.

Bohák András (Budapest, Evangelical Gymn., 10th grade)