### 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?

### Reverse to Order

Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?

### Card Trick 2

Can you explain how this card trick works?

# Two Much

##### Stage: 3 Challenge Level:

Rosie Johns, Y8, Davison High School, Worthing and Tufan KÃ½zÃ½ lÃ½rmak, IRMAK Primary and Secondary School, Ãstanbul, Turkey sent really good solutions to this problem. Rosie explained that the terms are all of the form 13N + 1, where N is a whole number, and that the terms of the sequence containing all 2's are based on 222,222,000 + 222. Rosie found you get numbers in the sequence with the digit 2 repeated 3 times, 9 times, 15 times, 21 times and so on.

Explaining a slightly different way of writing down the same method, Tufan says:

"I am a member of the Math Club of my school. Now I am sending the answer to Two Much (The February Six Problems). I noticed the difference of the terms of the sequence is 13. Each term is equivalent to 1 (modulo 13) , so that

1=1 (mod 13)

14 = 13 + 1 = 1 (mod 13)

27 = 2*13 + 1 = 1 (mod13)

40 = 3*13 + 1 = 1 (mod 13) and so on.

The other terms must be equivalent to 1 modulo 13. Now let's search the terms which contain twos; for example when 222 is divided by 13 the quotient is 17 and the remainder is 1. I made a table according to modulo 13:

 Terms Quotient Remainder 222 17 1 2222 170 12 22222 1709 5 222222 17094 0 2222222 170940 2 22222222 1709401 9 222222222 17094017 1

When the last term of the quotient is 7 then the remainder is always 1. The number of the terms etween two sevens is 6. It means the number of the terms which contains twos will increase six by six. The terms which contain twos appear increasing six digits, so that the terms of the sequence which contain twos must be

222,
222 222 222,
222 222 222 222 222,
222 222 222 222 222 222 222,..........going on."

This happens because, as Rosie and Tufan have shown, 222 = 13*17 + 1 so it belongs to the sequence. We use the fact that 222 222 is a multiple of 13 and add 222 222 times a thousand to 222 to get another 'all twos' number in the sequence. To get further 'all twos' numbers we just keep multiplying 222 222 by another million and adding it to the last 'all twos' number so six more twos appear in the successive terms.