Is this using current British coinage, or pre-decimalisation? There is a quite famous problem by Dudeney about this, but it uses old coinage - farthings and suchlike. The book "Amusements in Mathematics by Henry Ernest Dudeney" is available ibn Project Gutenberg, if you want to look it up.
Still, it's a nice example of proving by exhaustion, using current coinage.
We have: 1p, 2p, 5p, 10p, 20p and 50p
Let's consider the values up to 10, in terms of numbers of coins needed
1 -> 1
2 -> 1
3 -> 2
4 -> 2
5 -> 1
6 -> 2
7 -> 2
8 -> 3
9 -> 3
OK, so second digit can be satisfied with at most 3 coins.
But first digit is same case as second digit, multiplied by 10, since for every unit, there exists an equivalent 10 times bigger. Therefore we have proved that 10p, 20p, ... 90p can be given with three coins.
Therefore every integer 1 through 99 nclusive can be paid with no more than 6 coins.
But do check out Dudeney - there are some fun puzzles there! :-)
which amount up to £1 cannot be paid exactly with fewer than 6 coins?
is this a trick question and can all the amounts be paid?! I am really struggling with this one
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thankyou so much for your help I am vey grateful, I felt you could do every value but wasn't sure if I was missing something.
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