Currency Exchange
Dan and Ann have 9 and 8 coins respectively. What is the smallest number of coins they must swap so they end up with equal amounts of money.
Age
11 to 14
Challenge level
Dan has nine 20p coins and his sister Ann has eight 50p coins. What is the smallest number of coins that must change hands so that Dan and Ann end up with equal amounts of money?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
Answer: 5 coins
The difference between the amounts
Dan: 9$\times$20p = £1.80
Ann: 8$\times$50p = £4.00
Difference: £2.20
Half of £2.20 = £1.10 needs to change hands (from Ann to Dan)
£1.10 = 50p + 3$\times$30p $\rightarrow$ not OK because Ann has no 20p s to give to Dan
£1.10 = 3$\times$50p $-$ 2$\times$20p $\rightarrow$ 5 coins exchanged
Finding the total amount of money
In total, Ann and Dan have 180p + 400p = 580p, so they need to end up with 290p each. Therefore Dan needs to receive at least three 50p coins from Ann to have enough money. Then Ann has 250p, so she needs at least two 20p coins from Dan, at which point they have 290p each. Therefore, at least 5 coins must change hands.
Algebra
Suppose Dan gives Ann $x$ $20$p coins and Ann gives Dan $y$ $50$p coins. Then Dan has $(180-20x+50y)$p and Ann has $(400-50y+20x)$p.
We need $180-20x+50y=400-50y+20x$, which is the same as $10y-4x=22$, that is $5y-2x=11$. The solution to this equation which minimises $x+y$ is $x=2$, $y=3$, so the smallest number of coins that must change hands is $5$.