I  have got AIR 1100 and state (tamilnadu) 28 in aieee 2008 and can get both cse or ece in NIT-Trichy.......

Over the past years, ece has been preferred more by ppl, who have the option to take it in NIT-T. I heared that u need to memorise a lot in ece, while in cse, u need to be great in math and logic. I like taking cse. But i heared from any ppl, as it came in the newspapers that comp science is a declining course and not as assuring as other engg. groups.

I want to take cse, but every1 around me is suggesting to go for ece. What r the advntgs n disadvantages ? what should i do ? i m worried, as i feel i might regret my choice......

Plz help.

This is my mehod to find GCD of 2 big numbers A and B and expressing that gcd of the form mA + nB

First, let us find the gcd. Let us call it as d

1) Divide the bigger number by the smaller one.

2) Divide the smaller number by the remainder u get in step 1.

3) Divide the step 1 remainder by step 2, then each remainder by the next  remainder and so on.

4) Reach the step when u get no remainder at all. The divisor will be the LCM.

Let me explain it with an example.

Q. Find the gcd of 858 and 325 and express it in the form of m858 + n325.

Solution :

858 = 325*2 + 208

325 = 208*1 + 117

208 = 117*1 + 91

117 = 91*1 + 26

91 = 26*3 + 13

26 = 13.2

thr4, gcd 0f 858 and 325 is d = 13

d = 13 = 91 - 26*3

= 91 -3(117 - 91*1)

= 91*4 - 117*3

=4(208-117) - 117*3

= 4*208 - 7*117

= 4*208 - 7(325-208)

=11*208 -7*325

=11*(858-325*2) - 7*325

=11*858 - 29*325

thr4, d = 13 = m*858 + n*325 where m =11 and n = -29

As u can see, just alternate between the 2 values and write them as differences of bigger values from the initial steps whr we found the gcd.

Hope this article is worth the effort i put in, thinking n typing it, to help u .