## Task 2 - 090512

In a mathematics study group, the following problem from a Soviet textbook was posed:

Wassja bought two stamp albums. Kolja asked him how much he paid for them.

"I used only one type of coin for the payment," Wassja replied, "and for one album exactly 7, for the other exactly 5. I paid a total of 60 kopecks for both albums."

(In the Soviet Union, there are 1-, 2-, 3-, 5-, 10-, 15-, 20-, and 50-kopeck coins, and no other types of kopeck coins.)

How much did one album and how much did the other cost?
Let $a$ be the denomination with which the first album was paid, and $b$ the denomination for the second album. $a$ and $b$ can take values of $1, 2, 3, 5, 10, 15, 20, 50$.

Since 7 and 5 coins were used respectively, we have $7a + 5b = 60$ kopecks.

Since $a$ and $b$ are natural numbers greater than zero, $5b = 60 - 7a$ must be a natural number, meaning $60 - 7a$ is a number divisible by 5.

For the possible values of $a = 1, 2, 3, 5$ (from 10 onwards, $60 - 7a < 0$), we get $60 - 7a = 53, 46, 39, 25$. Only one result, 25, is divisible by 5, and $b = 5$. Thus, $a$ is also 5.

Therefore, the first album cost 35 kopecks and the second album 25 kopecks. The first album was paid with $7 \times$ 5-kopeck coins, and the second album with $5 \times 5$ kopeck coins.