sanju09 wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

Putting q=6,we get 31
q=7,we get 44
q=8,we get 59
..........
......
......
......

Like this I tested by plugging numbers but the approach seems very long.

Anyboady has any formal approach???

Had to do a bit of calculation..
here we must take each answer option and check whether any multiple of it+5 is a perfect square.

where(x=1,2,3,4,5........)
[spoiler]29x+5 {34,63,92,121-bingo this is a perfect square}
30x+5
31x+5 {36-This is a perfect square}
38x+5 {43,81-This is a perfect square}
41x+5 {46,87,128,169-This is a perfect square}[/spoiler]

So the only thing that remains is 30x+5 and is my answer

harsh.champ wrote:

sanju09 wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

Putting q=6,we get 31
q=7,we get 44
q=8,we get 59
..........
......
......
......

Like this I tested by plugging numbers but the approach seems very long.

Anyboady has any formal approach???

and in this way also we can't be sure whether we are correct or not.
Well I was going through the same way.....

sanju09 wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

q^2-5=( q^2-1) +4
= ( q+1) (q-1) +4

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3
if p is a multiple of 3 q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3, so it will not be a multiple of 30
[spoiler][Sorry for blatantly copy pasting my own solution from another thread, now do not ask me if this were a GMAT problem and if I did not see that solution would I have done the same]

So, B[/spoiler]

ajith wrote:

sanju09 wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

q^2-5=( q^2-1) +4
= ( q+1) (q-1) +4

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3
if p is a multiple of 3 q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3, so it will not be a multiple of 30
[spoiler][Sorry for blatantly copy pasting my own solution from another thread, now do not ask me if this were a GMAT problem and if I did not see that solution would I have done the same]

So, B[/spoiler]

Isn't that q^2 - 5 = ( q^2 - 1) - 4?

Rest is fabulous!!

sanju09 wrote:
q^2-5=( q^2-1) +4
= ( q+1) (q-1) +4
Isn't that q^2 - 5 = ( q^2 - 1) - 4?

Rest is fabulous!!

Yes it is.... I keep on making silly mistakes.. nevertheless, the approach remains

ajith wrote:

sanju09 wrote:For some integer q, q^2 - 5 is divisible by all of the following EXCEPT
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41

q^2-5=( q^2-1) +4
= ( q+1) (q-1) +4

Now q-1, q, q+1 are three consecutive integers , one of them should be a multiple of 3

If q-1 or q+1 is a multiple of 3; q^2-5 is not because 4 is not a multiple of 3
if p is a multiple of 3 q^2 is a multiple of 3 but q^2 -5 is not a multiple of 3

So q^2 -5 is never a multiple of 3, so it will not be a multiple of 30
[spoiler][Sorry for blatantly copy pasting my own solution from another thread, now do not ask me if this were a GMAT problem and if I did not see that solution would I have done the same]

So, B[/spoiler]

Ajith that was a nice solution...I just want to know how did you approached to the "q-1, q, q+1 are three consecutive integers" thing.. I was helpless when i first saw the question....