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
OAB
(A) 29
(B) 30
(C) 31
(D) 38
(E) 41
OAB
The answer is 'b'.
1) For an integer to be divisible by 30, it should be divisible by 10. Thus it should have units digit as 0.
2) in order to achieve units digit as 0 for q*q should have unit digit as 5.
3) 5 as a unit digit is only possible for the squares which are multiple of 5 for ex: 5*5=25, 15*15=225, 25*25=625 etc. That means q*q should be divisible 5.
4) if a number which is divisible by 5(q*q) and then 5 is subtracted from it(q*q-5). The result will never be divisible by 3.
Alternatively
We can take every option and then check if it's multiple added 5 by forms a square or not.
A) 29*4+5=121=>11*11 No
C) 31*1+5=6=>6*6 No
D) 38*2+5=81=>9*9 No
E) 41*4+5=169=>13*13 No
30 could not find anything. Bingo, this is the answer.
PS: 30 was looking suspicious to me. Everything else is a prime number including 38, which is a multiple of 17. But 30 has factorization as 2*3*5, everything below 10.
Cheers,
GG
