Problem 10 · The 60 replaced coins
Classifying each position by the last replacement that touches it.
Integer answer, at most 4 digitsWe place 60 twenty-cent coins in a row. We take the coins in positions $2, 4, 6, \dots, 60$ and replace them with ten-cent coins. Next we take the coins now in positions $3, 6, 9, \dots, 60$ and replace them with five-cent coins. Finally we take the coins now in positions $4, 8, 12, \dots, 60$ and replace them with two-cent coins. How many cents do the 60 remaining coins add up to?
Reasoned solution
Key idea: the final value of a position is decided by the last replacement that touches it: first check multiples of 4, then of 3, then of 2.
Multiples of $4$ (end with $2$¢): $15$ positions $\to 30$¢. Multiples of $3$ but not of $4$ (end with $5$¢): $20 - 5 = 15 \to 75$¢. Even numbers that are multiples of neither $4$ nor $3$ (keep $10$¢): $30 - 15 - 5 = 10 \to 100$¢. The rest keep their $20$¢: $60 - 15 - 15 - 10 = 20 \to 400$¢.