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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For some integer q, q^2 - 5
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gaurav_gaur
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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
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
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Anju@Gurome
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For this type of problem apparently our only option is to check all the options one by one. Even if we can prove that (q² - 5) is not divisible by 3 and hence not by 30, someone may ask why not we check for other ones also?guerrero wrote:For some integer q, q² - 5 is divisible by all of the following EXCEPT
The hint is : whenever there is (square of some number - some constant) form in divisibility problems, always try to proceed as follows...
q² - 5 = (q² - 1) - 4 = (q - 1)(q + 1) - 4
Now, (q - 1), q, and (q + 1) are three consecutive integer.
Hence, at least one among them is multiple of 3.
Now,
- # If q is multiple of 3, q² will be a multiple of 3.
So, (q² - 5) cannot be a multiple of 3
# If either (q - 1) or (q + 1) is multiple of 3, (q - 1)(q + 1) will be a multiple of 3.
So, (q - 1)(q + 1) - 4, i.e. (q² - 5) cannot be a multiple of 3
So, (q² - 5) cannot be a multiple of 30.
The correct answer is B.
Anju Agarwal
Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
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Quant Expert, Gurome
Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.
§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §
