GMATPREP:DS4

Question

please explain your answers

Maple · Answer

I have difficulty testing cases one and two independently; however, I will show my work nevertheless:

(t+2)*(t+3)

1)
t=13,20,27,34,41,48,55,62,69,76,83,90,97

2)
t^2/7=x+1
t^2=7*x+7
t=sqrt(7*x+7)
too complicated to look for a pattern

1&2) 
I am adding the remainders in this test

t^2/7+5*t/7+6/7

1 t^2 and 5 ts and 6

(1*1+5*6+6)/7

=37/7=5+2/7
remainder is 2

guess  C  ( i don't know about A and B)

rpmodi · Answer

Incorrect

, I am looking for some clear understading on these kind of patterns , like if d is divided by 7 and remainder 6, is there any pattern for d2 and d2+d ...etc

rpmodi · Answer

OKAY , after a lot of thinking I came up with some really weird solution to this problem :-)

) please post your creative solutions if you think of any . A) any number can be represented as quotient*divisor +remainder (qd+r) so we can say t=qd+r condition 1) t when divided by 7 remainder is 6 so t=7q+6 plugging this value in t^2+5^t+6 we get (7q+6)^2+5*(7q+6)+6 = 49q+84q+36+35q+30+6 = now all the terms with q in them will be divisible by 7 . remainder of this equation here is 72 , which will be the case for any value of t . so 72/7 =remainder 2 ALWAYS , hence A is sufficient 2) t^2 divided by 7 remainder is 1 for this to happen 7q+1 must be a perfect square - try q=5 and 9 for both these values t^2+5*t+6 remainder value will be different so not sufficient .....phew ... so answer is A . More solutions people ............

Maple · Answer

How did you come up with 5 and 9 in your 2nd condition solution? I would have some trouble coming up with those values quickly. Your condition 1 solution is the optimal way to solve condition 1.

GMATBLACKBELT · Answer

This one looks harder than it really is.

EDIT: oops I miscalculated on S1.

Test out choices.

1: (t+2)(t+3)/7

t/7 has a remainder of 6. Well if we make t=6 or 13 we have remainder of 6.

Now plug in 6 and 13 in for t. 15*16/7 r=2.

8*9/7 --> r=2.

You can test more but this one seems pretty consistent.

2: t^2/7 has a remainder of 1. 1/7 or 6^2/7 both have a remainder of 1.

Well lucky for us we save time b/c we have already got the remainder for when t=6.

now plug in 1 for t. (3)(4)/7 --> r =5 insuff.

A

rpmodi · Answer

How did you come up with 5 and 9 in your 2nd condition solution? ---this cabe happen in two scenarios , higly lucky to pick random numbers and see the magic happen so I got lucky in this case , or know this rule , that any number represented by n*d+1 will be a perfect square if d is chosen to be (n+2) or (n-2) --- because (n)(n+2) +1 = (n+1)^2 and n(n-2)+1 = (n-1)^2... with this can anyone come up with generic formula which will make n*d+r a perfect square :-)

prasannar · Answer

A for me

I plugged numbers to check the validity

using 1

13,20,27....

169+65+6=240/7 = 2 is the reminder
400+100+6=506 = 2 is the reminder
729+135+6=870=2 is the reminder

thus -1 is sufficient

using 2
36+30+6 =72 => remainder is 2
64+40+6=110 =>reminder is 5 thus 2 is insufficient

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