Question

In a particular country, the coins come in the denominations 1, 15 and 50 schillings. Whilst visiting this country, Peter bought a book for his wife, Rachel, paying with several (more than 2) coins. He paid with the minimum number of coins needed to make that amount.

His change contained one more coin than the handful of coins with which he paid, but again it was the minimum number of coins needed to make that amount.

What are the eight lowest possible prices for the book?

What is the lowest price of the book where the coins Peter pays with is a sensible set of coins to have handed over?

Answer 1

Explanation:

Depending on the tools available to your 11y.o., it's a typical question for linear inequalities.

Where a,b,c are the # of 1,15,50 coins given by the buyer, and d,e,f the # of coins given back by the seller.

1×a + 15×b + 50×c > 1×d + 15×e + 50×f

a+b+c+1>=d+e+f

a+b+c >=2

Now, there are other clues in the text :

If the seller gives back money, then you didn't pay with 1's, otherwise, it means you overpay and you could have given fewer coins. So a=0.

Similarly, the seller cannot give you back a coin of 50, otherwise, you could have given fewer coins as the buyer. So f=0.

So that gives :

15×b + 50×c > 1×d + 15×e

b+c+1>=d+e

b+c >=2

From that point on, there is only a handful of scenarios to test.