Problem 4 · Carme's and Daniel's coins
Equalising money: $5d - 2c = 11$ with the fewest coins.
Integer answer, at most 4 digitsCarme has 9 two-cent coins. Daniel has 8 five-cent coins. What is the minimum number of coins that must change hands so that both have the same amount of money?
Copa Cangur · SCM
Medium
Closed answer
Reasoned solution
Key idea: Carme has $18$ cents and Daniel $40$: $58$ in total, i.e. $29$ each. Carme must gain $11$ cents net.
If Daniel hands over $d$ five-cent coins and Carme hands over $c$ two-cent coins, Carme's net gain is $5d - 2c = 11$. Since $2c$ is even, $5d$ must be odd: $d$ odd.
With $d = 1$: $2c = -6$, impossible. With $d = 3$: $2c = 4 \Rightarrow c = 2$. Total coins changing hands: $3 + 2 = 5$.
Check: Carme $18 + 15 - 4 = 29$ ✓, Daniel $40 - 15 + 4 = 29$ ✓.
Answer: 5