### 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?

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!

### Rotating Triangle

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

# What Numbers Can We Make Now?

##### Stage: 3 and 4 Challenge Level:
Henry, from St. Hugh's, answered our question at the end:

If the bags contained 3s, 7s, 11s and 15s, can you describe a quick way to check whether it is possible to choose 30 numbers that will add up to 412?

He said the following:

It is impossible to make 412.
The starting number is 3 and the difference between the numbers is 4.
If I choose 30 numbers and add them all up, I will get a number that is 30x3=90 more than a multiple of 4.
But, 90÷4=22 remainder 2, so I will get a number that is 2 more than a multiple of 4.
But since 412 is a multiple of 4 (not 2 more than a multiple of 4), it won't work.

Well spotted! Luke, from Cottenham Village College, said this in a more algebraic way, using a tool called 'modular arithmetic':

All of our numbers are of the form 4x-1, for x=1, 2, 3 or 4. Therefore the sum of 30 of them comes to $4(x_1+x_2+\dots +x_{30})-30$, which is 2 mod 4 (i.e. 2 more than a multiple of 4). But 412 is 0 mod 4 (i.e. 0 more than a multiple of 4), so this sum cannot be equal to 412.

If you are unfamiliar with Modular Arithmetic, you might like to take a look at this introductory article.

Luke also gave his thoughts on the interactivity:

The numbers in the bag always form part of a linear arithmetic sequence, and so the number in the x-th bag is mx+c. Consecutive numbers are always a fixed distance m apart. This means that we can read off the value of m easily, and then find the value of c. We can then conclude that, if you choose z numbers from the bags, their sum will be of the form $m(x_1+x_2+\dots +x_z) + cz$, as all the numbers are of the form mx+c, for different values of x. This is obviously cz more than a multiple of m.

To illustrate this method with an example, take the numbers in the bag to be 1, 4, 7 and 10, as in the original example, with z=4.
Their differences are all a multiple of 3, so by analysing them mod 3 (i.e. by looking at their remainders when they are divided by 3) we find that they are all of the form 3x+1.
As z=4 and c=1, they must be cz=4 more than a multiple of 3.
Since 4 is 1 mod 3 (i.e. dividing 4 by 3 gives remainder 1), 4 numbers selected from the bags must add to give 1 more than a multiple of 3.

Nice!