### 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?

# Weekly Problem 26 - 2013

##### Stage: 3 Challenge Level:
It is not possible. In proving this, we call the sum of the two numbers at the ends of an edge the 'weight' of that edge. Note that three of the nodes are connected to exactly two nodes, while the other three are connected to exactly four nodes. Thus the total of the nine weights must always be an even number, whichever numbers are placed at the nodes.

Now the smallest possible weight is $1+2=3$ while the largest possible weight is $5+6=11$. As there are exactly nine edges, we deduce that for the weights for each edge to be different they must take the values $3, 4, 5, 6, 7, 8, 9, 10$ and $11$. However, the total of these is $63$, an odd number, so the task is impossible.

This problem is taken from the UKMT Mathematical Challenges.

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