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?
How about the combination
1, 2, 3, 4? Of course they have exchanges cards with one another. 1, 5, 6, 7? Yes of course, they have.
1, 8, 9, 10? Sure they did!
1, 11, 12, 13? Yes!
1, 14, 15, 16? Yes!
1, 17, 18, 19? Yes too!
1, 20, 21, 22?
1, 23, 24, 25?
1, 26, 27, 28?
1, 29, 30, 2?
Now comes the conclusion: student number 1 has exchanged cards with everyone else in the class.
How about student number 2? Using the same method, you can find out that he or she has exchanged cards with everyone else as well.
And so did number 3, 4, and everyone else in the class.
The truth is, everyone (30 students) had exchanged cards with everyone else.
Further question: What if, in each subset of 4 students, at least one (but not necessarily all four) had exchanged cards with the other three?