DS-Remainder of Quad Equation

Question

DS Question.

gmatnub · Answer

A

if you just post the picture instead of a word doc, you would get more answers, not to mention, it would be easier too.

domleon · Answer

to our experts - please help us with this out!!

scthakur · Answer

From stmt1, t = 7n + 6 for n = 0,1,2,.....

Hence, t^2 + 5t + 6 = 49n^2 + 119n + 72
When you divide this expression by 7, the remainder will be 2. Hence, sufficient.

For stmt2, t^2 = 7n + 1 for n = 0,1,2,3....
Hence, t = sqrt(7n+1) and since t is an integer, hence, it can only be 1,5,9,....

For t = 1, the given expression will have remainder as 5.
For t = 5, the given expression will have remainder as 0.

Hence ,stmt2 is not sufficient.

jatinrai · Answer

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.

ritula · Answer

Thats a new approach. I wasnt aware of it.

lylya4 · Answer

It's a easiest way to solve 1, but not 2

In stmt2, t^2 + 7 divisible by 7

leaves 5t - 1

t = 2 => remainder 2
t = 3 => remainder 0

DS-Remainder of Quad Equation

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