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This activity produced some good solutions.

Pooja at at the International School in the Seychelles sent this message:

To get my answer I used the two times table. When there were two kids my answer was three. Then I used the five times table when three more kids came along. I said that if each child gets one and two leftover it must mean that the answer must be seven.

Adelle and Amanie, also at the International School in the Seychelles wrote:

With two people it must be odd.

With five people, we know 7 works.

We tried adding 5 but it came to an even number. Then we added 10, it always works.

These numbers work with 2 people: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61

These numbers work with 5 people and 2 people: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97, 107, 117, 127, 137, 147, 157, 167, 177, 187, 197, 207, 217

Children from Culgaith School in the UK sent in the following:

Nathan said that the answer has to be an odd number as it is shared between two children with 1 left over. We found the first solution was 7 - Emma said this. We then tried to find out if there were any other answers. We tried 8 - but then realised that this was an even number so we looked at 9 - too many left over. We looked at 11 - this time too few left over.

Alex suggested 17 - why did he choose this? He said the numbers were "basically the same". Mr Dodd said this wasn't quite true as he would rather have 17 chocolate bars than 7!

Millie said that the 10 lollipops different between 7 and 17 would give each of the two children two lollipops each and so 17 would work. We checked this out and it did! We then used the number square to show us all the solutions up to 100 - they all end in "7". We worked out that this was because the tens digit would alway share out evenly between 5 children and the 7 would give each child 1 lollipop and have two left over.

Deidre at Calgary Science School in Canada had a really lovely suggestion. Her Grade 4 class approached a similar problem they called The Candy Problem. They posted a video of how students arrived at the solutions. Thank you so much for that it's certainly worth watching!

Stella from Chigwell School in England wrote a very thorough account of the challenge:

Any number ending in 7 works as if you divide it by 2 you get 1 left over and if you divide it by 5 you get 2 left over which fits in with the problem.

Because they start off with 2 children and they share them out equally you have to divide by 2 but so that it does't go in exactly you have 1 left over which means that it has to be an odd number. But when they have just finished sharing them out properly 3 more children come along making the total of children go up to 5, so now you have to divide by 5 so it can't be any thing ending in 9, 3, 5 or 1 so that only leaves 7.

Felix from Henleaze Junior wrote:

First I worked out that there were two children and one lolly left over. So I tried 3. But then I worked out that there were then three more children which made 5 so it couldn't be 3. Then I worked out that there were two remaining when 3 more people came and I added the total number of people to the two remaining lollies and got 7. Then I went back to the two people and checked if it would work with the two people and it did, so one possible answer is 7.

But after that I figured out that if you add 10 it would mean that there were 17 lollies and it will still work because 2 people would then have 8 each and one remaining and 5 people would have 3 each and two remaining.

Thank you for those answers which where all thought about in a slightly different way. There usually is a variety of possible ways of approaching problems.