40(1282). A schoolboy had several coins worth 15 k. and 20 k., with more 20 k. coins than 15 k. coins. The schoolboy spent one-fifth of all his money, paying two coins for movie tickets. He then spent half of the remaining money on lunch, paying with three coins. How many coins of each denomination did the schoolboy have initially?
Solution. The problem reduces to solving the equation $20 x+15 y=n$ in natural numbers, where $x$ is the number of 20-kopeck coins, $y$ is the number of 15-kopeck coins, and $x>y$; the natural number $n$ is determined from the condition of the problem.

Since one fifth of all the money was paid with two coins, one fifth of the money is either 40 k. (two 20 k. coins), 30 k. (two 15 k. coins), or 35 k. (20 k. + 15 k.). Therefore, the total amount of money the schoolboy had was either 200 k., 150 k., or 175 k. Half of the remaining money after the cinema is either 80 k. $\left(\frac{200-40}{2}\right)$, 60 k. $\left(\frac{150-30}{2}\right)$, or 70 k. $\left(\frac{175-35}{2}\right)$.

It is easy to establish that half of the remaining money is 60 k. (70 k. and 80 k. cannot be paid with three coins of 15 k. and 20 k.), and the total amount of money is 150 k.

Solving the equation $20 x+15 y=150$ for $x>y$ in natural numbers, we find $x=6, y=2$.

Answer: The schoolboy had 6 twenty-kopeck coins and 2 fifteen-kopeck coins.