dimitri92 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

The way I would approach this question:

So q^2 - 5 is divisible by all of the following except:

29, 31, 41 - big prime numbers, don't know any divisibility rules for these, forget them for the time being.. 38 = 19*2. (q^2 - 5) can be divisible by 2 (e.g. when q^2 ends with a 5, q^2 - 5 ends with a 0). As for 19, again a big prime number. Leave it for the time being.

(If the question is anywhere close to an actual GMAT question, they will not expect you to do many calculations with 29, 31, 41 etc. I see these big prime numbers and am quite convinced that they are just a smokescreen.Try and focus on what they could ask you like divisibility by 2, 3 etc. )

As for 30, q^2 - 5 is divisible by 10 (using the logic shown above). What about 3?

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

In any 3 consecutive numbers, (e.g. \((q - 1), q, (q + 1)\)), one and only one number will be divisible by 3.

If either (q - 1) or (q + 1) is divisible by 3, (q - 1)(q + 1) is divisible by 3, which means \((q - 1)(q + 1) - 4\) cannot be divisible by 3. If q is divisible by 3, then q^2 will be divisible by 3 and q^2 - 5[/m] will not be divisible by 3.

_________________

Karishma

Veritas Prep | GMAT Instructor

My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews