sufal Says
Thanks, Rohit Bhai. I have not given any mock apart from AIMCATS. Any other than that works fine for me. Btw can you please explain that remainder question again in detail? I didn't get it. I always loose out on such questions.
Sufal bhai you need to get well versed with chinese remainder theorem and Eulers number concept and you can solve these questions easily.
something similar has been discussed at the starting of this thread. go thru that...
In very short i will try to explain you:
Euler number is given after u have spplit you number into prime factors.
like 164 = 41*2^2
now E(164) = 164(1-1/2)*(1-1/41) = 80
Now any number to the power of 80 when divided by 164 will give 1 as remainder.
so 2^164/164 = (2^80)^2 * 2^4/164 = 2^4/164 = 16
This is one way of solving. This concept is extremely important as it reduces big powers to small values.
now to make it little complex lets find the remainder for 2^164^164/164
Rem(2^164^164/164) = Rem(2^4^164/164) - this is using E(164) = 80 as above.
Ab aar dhang se dekhein toh hum log power ka remainder nikal rahe hain when it is divided by E(164)= 80
toh lets find that, i.e. Rem(4^164/80) = Rem(4^162/5) = 1
this reduces ou problem from 2^164^164/164 to 2/164 = 2
Lemme know if somethin confuses you here...
there is another way of solving it, CRT... read that and solve these questions again.
Some exercise question for all of us:
1. What are the last two digits in 1372^482
2. remainder when 1234^5678/91
3. Remainder when 12^34 * 56^78/9
questions are self created... might be abrupt. Try kariye sab log and full approach post kariyega...
PS : i got that question wrong in this AIMCAT

silly max mistake.
Rohit Mishra,
Faculty of Management Studies, Delhi
2012-14