### 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?

# Tis Unique

##### Stage: 3 Challenge Level:

Congratulations to all the following for their correct solutions:

Andrei Lazanu Lazanu (School number 205, Bucharest),
Fiona Watson (Stamford High School),
Musab Khawaja,
Chong Ching Tong, Chen Wei Jian and Teo Seow Tian (River Valley High School, Singapore),
Clement Goh Tian (River Valley High School, Singapore),
Emma Holmes

It is good to see you explain how you came to your answer. Some of you said you solved the problem by "trial and error" but I am sure you made some decisions along the way. Sometimes it is necessary to try several possible numbers before finding the one that works for other constraints. Next time try to explain your thinking. I have included Andrei's solution as one of those that gave a very complete explanation of how they thought the problem through.

First I observed that the first digit of the result must be 1:

 * * 4 + 2 8 * _ _ _ _ 1 * * *

I used for the other letters the following notation:

 a b 4 + 2 8 c _ _ _ _ 1 d e f

2 + a + n = , where n can be 0 or 1.
a + n = 8 + d, where a and d are smaller than 10.

 1.1. n = 0

a = 8 + d.
a can be 8 or 9 and d 0 and 1 respectively, but in the first case 8 was used, and in the second case 0 was used.

 1.2. n = 1

a + 1 = 8 + d
a can be 7, 8 and 9 and d can be 0, 1 and 2. In the first situation it works because the digits weren't used another time, in the second combination it doesn't work, and the last situation doesn't work. The result is:

 7 b 4 + 2 8 c _ _ _ _ 1 0 e f

4 + c =

c can be : 3, 5, 6, 9. Then is: 7, 9, 10, 13. The possibilities are:

c = 5; f = 9

c = 9; f = 3

 2.1. c = 5; f = 9

 7 b 4 + 2 8 5 _ _ _ _ 1 0 e 9

b + 8 =

Using only 3 and 6 there isn't any possibility.

 2.2. c = 9; f = 3

 7 b 4 + 2 8 9 _ _ _ _ 1 0 e 3

b + 8 + 1 =

b + 9 =

In this situation only b = 6 and e = 5 satisfies the condition.

This is the only solution for the problem, because using a step-by-step method, I obtained only one solution.