### Biscuit Decorations

Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?

### Constant Counting

You can make a calculator count for you by any number you choose. You can count by ones to reach 24. You can count by twos to reach 24. What else can you count by to reach 24?

### Skip Counting

Find the squares that Froggie skips onto to get to the pumpkin patch. She starts on 3 and finishes on 30, but she lands only on a square that has a number 3 more than the square she skips from.

# Lots of Lollies

##### Stage: 1 Challenge Level:

This activity produced some good solutions being sent in.

Luke at Mortimer Primary School said:

I think the answer is seven lollies because I think first they had three lollies each with one left over then when the extra three friends came they had one lolly each with two lollies left over.  They had seven lollies in total.

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, sent in a copy of their work but unfortunately it was a little too faint for us to put on the site, except for the little diagram at the end. They 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

We had the following from Emily at King Edward VI Camp Hill School for Girls:

To work this out, I made a list of numbers that could be divided by 5 with 2 left.
This list included 7, 12, 17, 22.
Now it can't be an even number because that would divide by 2 with none left over but it could be either 7 or 17 because they both work.
Therefore I have come to the conclusion that any number ending with 7 could be the solution to this problem.

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.

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.