2. Petya has several 5-ruble coins and several 2-ruble coins. Vanya has as many 5-ruble coins as Petya has 2-ruble coins, and as many 2-ruble coins as Petya has 5-ruble coins. Petya has 60 rubles more than Vanya. Which coins does Petya have more of - 5-ruble or 2-ruble? By how many?
Solution. Answer: Petya has 20 more 5-ruble coins.

Suppose Petya has more 2-ruble coins than 5-ruble coins. Let Petya remove one 2-ruble coin, and Vanya remove one 5-ruble coin. Then the difference in the sums of Petya's and Vanya's coins will increase by 3. If they repeat this operation until Petya has an equal number of 5-ruble and 2-ruble coins, the sums of Petya's and Vanya's coins should become equal (since the composition of the coins will be the same), but their difference will be no less than 60. This is a contradiction. Therefore, Petya has more 5-ruble coins.

Let Petya then remove one 5-ruble coin, and Vanya remove one 2-ruble coin. Then the difference in the sums of Petya's and Vanya's coins will decrease by 3. If they repeat this operation until Petya has an equal number of 5-ruble and 2-ruble coins, the sums of Petya's and Vanya's coins will become equal (since the composition of the coins will be the same). This means they performed this operation $60: 3=20$ times. Therefore, initially, Petya had 20 more 5-ruble coins than 2-ruble coins.

If briefly,

$$
60=5 n+2 k-(5 k+2 n)=3(n-k)
$$

Therefore, $n-k=60: 3=20$.