v1rok wrote:

Nevermind. I found my mistake.

ST2

Let's represent t=7p+j, then t^2=49i^2+14pj+j^2. We know that t^2 has a remainder 1 when it is divided by 7, therefore j^2=1, and j=1.

Piece in RED is wrong. It does not mean that j^2=1. It means that when j^2 is divided by 7 the remainder will be 1. Two different things!

j can be either 1 or 6. (1^2 and 6^2 both have remainder 1 when dividing by 7.) Hence, not sufficient!!!!

This one trapped me

You know I first answered A and then after seeing your post I changed mine to D. I handled statement 2 differently.

This was my initial analysis:

t^2 = 7y+1 where y is a +ve integer.

Now EQ1 becomes

49y^2 + 14y + 1 + 5 * SQRT(7y+1) + 6

=

49y^2 + 14y + 7 + 5 * SQRT(7y+1)

Bold part is fully divisible so drop that part. Now we have to take care of 5 * SQRT(7y+1)

7y+1 should be a perfect square

when y = 0 -------> 5 * SQRT(7y+1) = 5--------> Remainder = 5

when y = 5 -------> 5 * SQRT(7y+1) = 30--------> Remainder = 2: INSUFF

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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008