Danielle and Sally noticed that 7 works.
Sally noticed that 27 worked, as well as 107.
So we think that any number that ends
with 7 works. This is because any number that ends in 7 must be
odd, and is two more than any number divisible by 5 (there are 5
friends).
It is very helpful to see how your thinking developed as you tackled this problem - thank you Class 7Q3. Poppy from Steeple Morden Primary wrote:
If the two children end up with one lolly
it must be an odd number of lollies. Then 3 more children come
making the total number of children 5. Say they had 1 lolly each
when they shared them, the number of lollies would be seven because
1 times 5 is 5 add on 2 for the left over ones and it makes
seven.
If we carry this on to 10 lollies each it
shows:
1 lolly each- 7 lollies
2 lollies each- 12 lollies
3 lollies each- 17 lollies
4 lollies each- 22 lollies
5 lollies each- 27 lollies
6 lollies each- 32 lollies
7 lollies each- 37 lollies
8 lollies each- 42 lollies
9 lollies each- 47 lollies
10 lollies each- 52 lollies
You can see here that it goes up in 5s
each time. The answers that are possible are the ones where the
children have an odd number of lollies.
It is most probable that they will have
less than 27 lollies in their bag but this is not certain they
could have 5007 lollies in the bag and 1001 each leaving two left
over!
Thank you Poppy. I like the way you have thought about the practical side of the problem at the end and have decided how many lollies the children are likely to have had. Phoebe and Alice from Cambridge used a 100 square to help solve the problem. Here is Alice's square:
