341
Hari Shankar
·2009-09-07 18:40:26
Look for pairs (m,n) such that
(1) m and n are both odd and 4|m-n
(2) m and n are both even and |m-n| = 2
62
Lokesh Verma
·2009-09-08 01:51:49
A small addition to what prophet sir said.. You just have to see the last digit...
62
Lokesh Verma
·2009-09-15 01:26:46
If a+b is a multiple of 10, then last digit of a is 10-(last digit of b)
that is what i meant
341
Hari Shankar
·2009-09-15 01:35:11
The conditions I mentioned are sufficient for this. However the theory behind getting at those conditions comes from using congruecnes and Euler Totient theorem.
11
Devil
·2009-09-15 01:43:08
This can be sorted out by congruencies as well....
3^2\equiv -1(mod10)\\ \ 7^4\equiv 1(mod10)
So setting setting n=4, we have
m\epsilon (6,10,14,18)
If u ask how I generated this series, I shall ask u to keep raising both sides of the congruence 3^2\equiv -1(mod10) to odd powers on both sides.
Simialrly apply the logic to
3^4\equiv 1(mod10)\\ \ 7^2\equiv -1(mod10)
So m=4, and thus a similar set for n's can be generated!
11
Devil
·2009-09-15 01:43:59
Prophet sir's post came while I was typing mine.