You may also like

problem icon

Calendar Capers

Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat this for a number of your choice from the second row. You should now have just one number left on the bottom row, circle it. Find the total for the three numbers circled. Compare this total with the number in the centre of the square. What do you find? Can you explain why this happens?

problem icon

Adding All Nine

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

problem icon

Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

Take Three from Five

Stage: 3 and 4 Challenge Level: Challenge Level:2 Challenge Level:2
The group at Henry Cavendish Primary School thought they found 5 numbers that worked - 6, 17, 5, 41, 96 - and asked for some feedback! This was a really good try, but unfortunately there is a multiple of three in there: 17 + 5 + 41 is equal to 3 times 21. Never mind!

The answer is that it's not possible. Lots of people got this, but David, from Sha Tin College, gave a particularly well written answer:


Firstly, each one of the five integers can be expressed in the form 3x + 0, 3x + 1 or 3x + 2, where x is an integer.

When we take three integers (say 3x + a, 3y + b, 3z + c) from a group of 5 and add them together, we will always get an integer in the following form:

3x + 3y + 3z + k, which equals 3(x + y +z) + k (where k is the sum of a, b and c)

It is obvious that the term 3(x + y + z) must be a multiple of 3. Therefore, we only need to consider k in order to see whether 3(x + y + z) + k is a multiple of 3. For 3(x + y + z) + k to be a multiple of 3, k must therefore also be a multiple of 3.

As a, b and c are each equal to 0, 1 or 2, there are 4 different ways in which k can be a multiple of 3:

1. 0 + 0 + 0 (i.e. a=0, b=0, c=0)
2. 1 + 1 + 1
3. 2 + 2 + 2
4. 0 + 1 + 2

Out of a group of 5 integers, it is always possible to find an a, b and c that add up to a multiple of 3. This is because in a group of 5 zeros, ones or twos, there is either at least one of each number (0,1,2), or at least 3 of just one of the numbers (0,0,0 or 1,1,1 or 2,2,2).

Therefore it is always possible to choose three numbers that will add up to a multiple of 3 from any group of 5 numbers.

The Year 5 group at St. John's C of E Primary remarked that it is possible to make a set of 5 numbers like this if you don't insist that they have to be whole numbers: they came up with 0.2, 0.2, 0, 0, 1. That's definitely thinking outside the box!

Thanks to everyone - lots of clever answers here!