Marcab wrote:
Another approach to tackle this problem:
The given relation is:
2^98=256L+N
This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder.
Here X=2^98, Quotient=256 and Remainder=0.
We just have to check whether 256 OR 2^8 divides 2^ 98. If yes, then the value of N is straightaway 0. No need to check choices even.
The given relation is:
2^98=256L+N
This is actually written in "remainder format" i.e. X=Quotient*some integer + Remainder.
Here X=2^98, Quotient=256 and Remainder=0.
We just have to check whether 256 OR 2^8 divides 2^ 98. If yes, then the value of N is straightaway 0. No need to check choices even.
Could you tell me if I'm understanding it correctly?
What if we had 12/5 = 2 remainder 2? Then in remainder format we would have 2 * 5 + 2 = 12.
So according to what you're saying, if we just have to check if 2 can divide 12? In this case it does, but the remainder is 2 not 0.
Please, help me in understanding where I'm going wrong.