### Pebbles

Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?

### It Figures

Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?

### Bracelets

Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads?

# Which Numbers? (1)

##### Age 7 to 11 Challenge Level:

We had lots of correct solutions to this problem and the majority identified the Green and Blue sets as multiples of $5$ and $9$ respectively. A few question marks arose when looking at the Red set.

Max of St Mary's Catholic High School said:

\eqalign{ Red &= 10,19,28,37,46,55,64,73,82\cr Green &= 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\cr Blue &= 9,18,27,36,45,54,63,72,81,90,99}

So the red set is $9$ times table but adding $1$ on every time.
The green set is the $5$ times table.  There are $20$ numbers in this set less than $101$.
The blue set is the $9$ times table. There are $11$ numbers in this set less than $101$.

Max correctly identifes the Green and Blue sets. He also noticed that there was a common difference of $9$ in the red set. However we would need to justify why $91$ and $100$ are not in this set.

Mr Harrison's maths group of Northdown Primary school had a different suggestion for the Red set:

We got a hundred square grid and crossed out all the numbers they gave us at the beginning. We tried to find all the patterns, for example the red set were in a diagonal line.

We looked for gaps in the patterns. We found:
Red set: (digits sum $=10$)
$\{19,28,37,46,55,64,73,82,91\}$

Green set: (Multiples of $5$)
$\{5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\}$

Blue set: (Multiples of $9$)
$\{9,18,27,36,45,54,63,72,81,90,99\}$

The red set has $9$ numbers and there are no other possible numbers less than $101$ which satisfy the condition Mr Harrison's maths group imposed.

There was a big divide between those of you who agreed with Max and those who agreed with Mr Harrison's maths group.

Megan of Crookhill Primary agreed with Mr Harrison's maths group:

First I plotted the hints on a grid and looked at how many there were in each colour group. Then I looked at the pattern on the board to gain more information. To find out the Green set we knew the numbers that they gave us were already in the $5$ times table. On the grid there were two lines of $10$ numbers which made $20$ in that group so we saw that it was Green set.

Then we solved the Red set. The first thing we spotted was the pattern on the page so then we looked really carefully at the numbers then we solved it. The digits all add up to ten; that was the hardest to spot I think The easist to spot was the Blue set. I enjoy doing my $9$ times table that's probably why so I didn't need to work anything out.

Well done Megan, and I enjoy my $9$ times table as well.