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q, q^2 – 5

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q, q^2 – 5

by sanju09 » Sat Feb 20, 2010 12:15 am
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
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by harsh.champ » Sat Feb 20, 2010 12:26 am
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???
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by vijay_venky » Sat Feb 20, 2010 12:34 am
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
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by shashank.ism » Sat Feb 20, 2010 12:44 am
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.....
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by ajith » Sat Feb 20, 2010 2:12 am
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]
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by sanju09 » Sat Feb 20, 2010 2:19 am
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?

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by ajith » Sat Feb 20, 2010 2:22 am
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
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by shashank.ism » Sat Feb 20, 2010 8:03 am
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....
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