Recently, the following question was set in a GCSE maths exam:

*Liz buys packets of coloured buttons.*

*There are 8 red buttons in each packet of red buttons.*

*There are 6 silver buttons in each packet of silver buttons.*

*There are 5 gold buttons in each packet of gold buttons.*

*Liz buys equal numbers of red buttons, silver buttons and gold buttons.*

*How many packets of each colour of buttons did Liz buy?*

A certain student answered by calculating 8 × 6 × 5 = 240, concluding that this was the number of beads of each colour and then deducing that Liz bought 30 red, 40 silver and 48 gold buttons.

The student lost marks, apparently because the instructions given to the examiners stated that the answer, that had to be arrived at for the number of buttons, of each colour was 120.

Yet it is the question that is at fault, because although 120 is the least common multiple of 8, 6 and 5, the wording does not include any information pointing to the fact that it is this, which is being sought. In fact any common multiple will do, hence the easy choice of 240.

This is quite unfair on the student, but the damage most probably cannot be undone.

There are better ways of phrasing the question and I have thought of one, which in my view also serves to instruct and teach, besides stating the question for the purposes of the exam:

*Liz buys packets of coloured buttons.*

*There are 8 red buttons in each packet of red buttons.*

*There are 6 silver buttons in each packet of silver buttons.*

*There are 5 gold buttons in each packet of gold buttons.*

*Liz wishes to buy the same number of buttons of each of these three colours and, in such a way, so as to obtain the least possible number of buttons.*

*Calculate how many packets of each colour of buttons Liz should buy, the number of buttons of each colour thus bought and the total number of buttons.*

I believe this way of phrasing the question has some benefits:

- It illustrates a more natural way in which the problem could occur in real life (Why miss an opportunity to bring Maths a little closer to our daily lives for the sake of students?)
- The last sentence helps the student make a distinction between the number of packets and the number of buttons. This is important, because one quite easily could mistake packets for buttons.

And here is a solution to the problem:

*Whatever the number of packets of red buttons Liz chooses, she will always end up with a multiple of 8. Similarly, she will end up with a multiple of 6 for the silver buttons and a multiple of 5 for the gold buttons. *

*Therefore, if Liz wants the same number of buttons of each colour, this common number must be a common multiple of 8, 6 and 5. Since she wants this number to be the least possible, it follows that it must the least common multiple (LCM) of these numbers.*

*To calculate it, let’s decompose these numbers into prime factors:*

*8 = 2³*

*6 = 2 × 3*

*5 is prime*

*Hence the LCM of these three will be 2³ × 3 × 5 = 24 × 5 = 120. *

*[Digression: If one is not happy with this calculation, then one might consider that LCM(8, 6, 5) = LCM(LCM(8, 6), 5). *

*For the LCM of 8 and 6, we list the multiples of these two:*

*for 8: 8, 16, 24, …*

*for 6: 6, 12, 18, 24, …*

*from whence it follows that LCM(8, 6) = 24. The multiples of 24 and 5, now, are:*

*24: 24, 48, 72, 96, 120, …*

*5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125…*

*and the result follows.]*

*So, Liz must buy 120 buttons of each colour. To achieve this, she will have to buy:*

*120 ÷ 8 = 15 packets of red buttons*

*120 ÷ 6 = 20 packets of silver buttons*

*120 ÷ 5 = 24 packets of gold buttons*

*The total number of buttons is 3 × 120 = 360.*