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Re: DS-Remainder of Quad Equation [#permalink]
02 Oct 2008, 09:10

vr4indian wrote:

can anyone explain this?

IMO A 1) t divided by 7 leaves remainder 6. t^2 divided by 7 leaves remainder same as 36/7 & i.e. 1 5t divided by 7 leaves remainder as 5*6/7 i.e. 2 6 divided by 7 leaves remainder as 6. All 3 plus leaves remainder same as (1+2+6)/7 or 9/7 & i.e. 2

2) With this we cant calculate remainder for 5t. Hence insufficient.

Logic used: If X divided by A leaves remainder as x & Y divided by A leaves remainder as y, then X+Y when divided by A leaves the same remainder as (x+y)/7 would. Also, XY when divided by A leaves the same remainder as (xy)/7 would.

Re: DS-Remainder of Quad Equation [#permalink]
02 Oct 2008, 21:42

Thats a new approach. I wasnt aware of it.

jatinrai wrote:

vr4indian wrote:

can anyone explain this?

IMO A 1) t divided by 7 leaves remainder 6. t^2 divided by 7 leaves remainder same as 36/7 & i.e. 1 5t divided by 7 leaves remainder as 5*6/7 i.e. 2 6 divided by 7 leaves remainder as 6. All 3 plus leaves remainder same as (1+2+6)/7 or 9/7 & i.e. 2

2) With this we cant calculate remainder for 5t. Hence insufficient.

Logic used: If X divided by A leaves remainder as x & Y divided by A leaves remainder as y, then X+Y when divided by A leaves the same remainder as (x+y)/7 would. Also, XY when divided by A leaves the same remainder as (xy)/7 would.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...