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!
What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
Well done to Maulik aged 11 who sent in some nice work on this problem. Neil's solution is given below.
For a 2 by 2 square with column headings of x and x+1, and row headings of y and y+1, Neil says that:
So the diagonal from top right to bottom left is:
$(x+1)y+x(y+1) = xy+y+xy+x = 2xy+x+y$
Lets call that Z.
The diagonal from top left to bottom right is:
$xy+(x+1)(y+1) = xy+xy+x+y+1 = 2xy+x+y+1$
So the first diagonal is Z and the second Z+1 so the diagonal from top left to bottom right is always 1 more than the diagonal from top right to bottom left.
For a 3 by 3 square with column headings of x, x+1 and x+2, and row headings of y, y+1 and y+2, Neil says that:
The 3 by 3 square looks like this:
The diagonal from top right to bottom left is:
Let's say that $3xy+3x+3y = W$
The diagonal from top right to bottom left is W+1.