### 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:

We had a good number of responses to this challenge. The first ones are from Culgaith School in the UK who 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.

We were really pleased with our thinking and problem solving.  Looking forward to the next problem!

From Deidre at Calgary Science School in Canada we had a really lovely suggestion. Grade 4 class, she says approached a similar problem they called the Candy problem. They posted a Video of how the students arrived at solutions at - http://savouringtheish.blogspot.com. 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:-

They shared them out evenly and had one left over. Just as they had finished sharing them their friends Kishan, Hayley and Paul came along. They wanted some lollies too so the children shared them out again between all of them. This time they had two lollies left over.

How many lollies could there have been in the bag?

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 $7$.

Luke from Locks Heath Junior School in England presented us with the following good account.

There were two people at the start and there was remainder $1$, so it had to be odd  then , $3$ more people came over and there was remainder $2$, so I thought of every odd number up to $9$ so there is $1 3 5 7 9$ I chose every odd . On number $1$ only $1$ person could get a lolly  then I thought of $3$ then, only $3$ people could get a lolly, then I tried $5$ then only everybody could get $1$ each then there was no reminder. Then I skipped $7$ for no reason then I went on to $9$ there would of been to many left. So the answer is $7$ .

Well done all of you. Thank you so much for letting us into your thoughts and ideas while sorting out solutions.

We had a bit of blogging going on in our .  We hope to hear from you all again soon.