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?
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?
Can you explain how this card trick works?
Rosie Johns, Y8, Davison High School, Worthing and Tufan
KÃ½zÃ½ lÃ½rmak, IRMAK Primary and Secondary
School, Ãstanbul, 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.