# Hubble, bubble

Winifred Wytsh bought a box each of jelly babies, milk jelly bears,
yellow jelly bees and jelly belly beans. In how many different ways
could she make a jolly jelly feast with 32 legs?

## Problem

Winifred Wytsh bought a box each of :

- jelly babies.
- milk jelly bears.
- yellow jelly bees.
- and jelly belly beans.

Image

Each box contained 20 jellies.

Winifred wanted to make a jolly jelly feast that had 32 legs.

How many different ways could Winifred make her jelly feast?

## Student Solutions

Henry and Amy of St Mary's Primary School explained how they tackled this problem:

First we counted how many legs babies, bears, bees and beans have. Babies have two legs. Bears have four legs. Bees have six legs. Beans have no legs.

Then we tried to make 32 by adding 2, 4, 6 and 0.

2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 =32.

4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 =32.

6 + 6 + 6 + 6 + 6 + 2 =32.

Each feast we make we can make 21 different ways with the jellies, because we can add 0, 1, 2, 3 all the way up to 20 jelly beans without changing the number of legs, because 32 + 0 + 0 + 0 = 32.

Then we counted how many ways we could make the feast using at least four bees.6 + 6 + 6 + 6= 24, so we wanted to make 32 - 24 = 8 legs from the jellies. We could do this with one bee and one baby, two bears, one bear and two babies or four babies. This makes four different ways not counting the beans, so that makes 21 x 4 = 84 different ways if you can add as many beans as you like.

Then we did the same with exactly three bees. 6 + 6 + 6 = 18, so we wanted 32 - 18 = 14 legs. We did not want to count feasts twice, so we wanted to make the 14 legs without using any more bees. We could do this with three bears and a baby, two bears and three babies, one bear and five babies or seven babies. This made 21 x 4 = 84 different ways again.

We noticed that if we took away a bear we had to add two babies to make the same total, so for each bear we used there was another feast we could make using two babies instead, because bears have four legs and so do two babies. When we used three bears, there were three more feasts.

Then we tried usingexactly two bees. 6 + 6 = 12, 32 - 12 = 20. We can make twenty legs with five bears, so there werefive more feasts we could make, replacing first one bear with two babies, then another bear with another two babies, and so on. That made six feasts not counting the beans, so that made 21 x 6 = 126 feasts altogether.

Then we tried using exactly one bee. 32 - 6 = 26. We can make twenty-six legs with six bears and a baby, so there were six more feasts we could make. That made seven feasts not counting beans, so there were 21 x 7 = 147 feasts.

Then we used no bees. We can make a feast using eight bears, so there were nine feasts we could make using bears and babies so there were 21 x 8 = 168 feasts we could make using no bees at all.

That made 84 + 84 + 126 + 147 + 168 = 609 feasts we could make, which we did on a calculator.

That's a lot of different feasts! Thank you very much, Henry and Amy!

## Teachers' Resources

### Why do this problem?

This
activity is closely linked to The
Pied Piper activity in that it is a problem-solving situation
that requires some knowledge of numbers and their properties.

### Possible approach

It would be good to withhold information that is obvious to
the teacher and just present the facts as on the
problem page.

### Key questions

How are you doing this activity?

### Possible extension

Give pupils the opportunity to invent other challenges of
their own having worked out the answers.

### Possible support

Calculators might be helpful for some children.