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14 Apr 2019, 10:35
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E
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Difficulty:
75%
(hard)
Question Stats:
36% (01:33) correct
64%
(01:34)
wrong
based on 107
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If p and q are composite integers, is p/q an integer?
(1) Each prime factor of q is also a factor of p.
(2) The total number of factors of q is 12, and that of p is 24
(1) Each prime factor of q is also a factor of p.
(2) The total number of factors of q is 12, and that of p is 24
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
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14 Apr 2019, 17:30
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if P=3 and Q=9, every prime factor of q is also a factor of p, but p/q is not an integer; if we invert p and q, p/q is. Therefore 1 is not sufficient.
If every factor of q is a factor of p, p/q is an integer; however, if they differ by a single prime, it is not. therefore 2 is not sufficient
If every factor of q is a factor of p, p/q is an integer; however, if they differ by a single prime, it is not. therefore 2 is not sufficient
15 Apr 2019, 09:18
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composite numbers are non-prime numbers
(1) Each prime factor of q is also a factor of p.
let a and b prime factors for q and p.
if \(p = a^5b^3\) and \(q = a^3b^2\) then \(\frac{p}{q}\) is integer
if \(p = a^3b^2\) and \(q = a^5b^3\) then \(\frac{p}{q}\) is non-integer
insufficient ---> the trick here is that the statement does't say that "each factor of q is a factor of p"
(2) The total number of factors of q is 12, and that of p is 24
let a,b and c prime factors
if \(p = a^5b^3\) [no. of factors = (5+1)(3+1)=24] and \(q = a^3b^2\) [no. of factors =(3+1)(2+1)=12] then \(\frac{p}{q}\) is integer
if \(p = a^2b^3c^1\) [no. of factors = (2+1)(3+1)(1+1)=24] and \(q = a^3b^2\) [no. of factors = (3+1)(2+1)=12] then \(\frac{p}{q}\) is non-integer
insufficient
combining 1 & 2:
the same assumptions stated to test statement (2) shows that combining is still insufficient. E
(1) Each prime factor of q is also a factor of p.
let a and b prime factors for q and p.
if \(p = a^5b^3\) and \(q = a^3b^2\) then \(\frac{p}{q}\) is integer
if \(p = a^3b^2\) and \(q = a^5b^3\) then \(\frac{p}{q}\) is non-integer
insufficient ---> the trick here is that the statement does't say that "each factor of q is a factor of p"
(2) The total number of factors of q is 12, and that of p is 24
let a,b and c prime factors
if \(p = a^5b^3\) [no. of factors = (5+1)(3+1)=24] and \(q = a^3b^2\) [no. of factors =(3+1)(2+1)=12] then \(\frac{p}{q}\) is integer
if \(p = a^2b^3c^1\) [no. of factors = (2+1)(3+1)(1+1)=24] and \(q = a^3b^2\) [no. of factors = (3+1)(2+1)=12] then \(\frac{p}{q}\) is non-integer
insufficient
combining 1 & 2:
the same assumptions stated to test statement (2) shows that combining is still insufficient. E
