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Two Much

Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Rosie Johns, Y8, Davison High School, Worthing and Tufan Kýzý lýrmak, IRMAK Primary and Secondary School, Ýstanbul, 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:

























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,..........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.