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niteshwaghray
GMAT 1: 700 Q45 V40
Posts: 59
23 May 2017, 00:34
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D
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Difficulty:
95%
(hard)
Question Stats:
32% (02:22) correct
68%
(03:00)
wrong
based on 129
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\(Z=2^a×5^b×7^c\)
A positive integer Z can be expressed in terms of its prime factors as above, where a, b and c are positive integers. Is \(3^{|a−b|}<9\)?
(1) Z is divisible by 40 but not by 50
(2) \(Z^{a−b}=2^6×5^2×7^4\)
A positive integer Z can be expressed in terms of its prime factors as above, where a, b and c are positive integers. Is \(3^{|a−b|}<9\)?
(1) Z is divisible by 40 but not by 50
(2) \(Z^{a−b}=2^6×5^2×7^4\)
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:
Data Sufficiency Forum
Data Sufficiency Forum
23 May 2017, 00:50
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\(Z=2^a×5^b×7^c\)
A positive integer Z can be expressed in terms of its prime factors as above, where a, b and c are positive integers. Is \(3^{|a−b|}<9\)?
Is \(3^{|a−b|}<9\)?
Is \(3^{|a−b|}<3^2\)?
Is \(|a−b|<2\)?
(1) Z is divisible by 40 but not by 50
\(40 = 2^3*5\) is a factor of Z while \(50 = 2*5^2\) is not. This implies that a (the power of 2) is at least 3 and b (the power of 5) is exactly 1: \(a > 3\) and \(b = 1\). Thus, \(|a−b|\) will be greater than 2. Sufficient.
(2) \(Z^{a−b}=2^6×5^2×7^4\).
\(Z^{a−b}=2^6×5^2×7^4=(2^a×5^b×7^c)^{a-b}\).
\(2^6×5^2×7^4=(2^a×5^b×7^c)^{a-b}\).
\(2^6×5^2×7^4=2^{a(a-b)}×5^{b(a-b)}×7^{c(a-b)}\).
\(6 = a(a-b)\) and \(2 = b(a-b)\)
Divide one by another \(3 = \frac{a}{b}\) --> \(a = 3b\).
\(|a−b|=|2b|\). Since b is a positive integer, then |2b| is at least 2. Thus, \(|a−b|=|2b|\) will be greater than 2. Sufficient.
Answer: D.
Hope it'clear.
A positive integer Z can be expressed in terms of its prime factors as above, where a, b and c are positive integers. Is \(3^{|a−b|}<9\)?
Is \(3^{|a−b|}<9\)?
Is \(3^{|a−b|}<3^2\)?
Is \(|a−b|<2\)?
(1) Z is divisible by 40 but not by 50
\(40 = 2^3*5\) is a factor of Z while \(50 = 2*5^2\) is not. This implies that a (the power of 2) is at least 3 and b (the power of 5) is exactly 1: \(a > 3\) and \(b = 1\). Thus, \(|a−b|\) will be greater than 2. Sufficient.
(2) \(Z^{a−b}=2^6×5^2×7^4\).
\(Z^{a−b}=2^6×5^2×7^4=(2^a×5^b×7^c)^{a-b}\).
\(2^6×5^2×7^4=(2^a×5^b×7^c)^{a-b}\).
\(2^6×5^2×7^4=2^{a(a-b)}×5^{b(a-b)}×7^{c(a-b)}\).
\(6 = a(a-b)\) and \(2 = b(a-b)\)
Divide one by another \(3 = \frac{a}{b}\) --> \(a = 3b\).
\(|a−b|=|2b|\). Since b is a positive integer, then |2b| is at least 2. Thus, \(|a−b|=|2b|\) will be greater than 2. Sufficient.
Answer: D.
Hope it'clear.
23 May 2017, 15:59
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Bunuel, 40 = 2^3*5 and 50 = 2*5^2.
I agree that a is atleast 3.. but how is b EXACTLY 1.. I can do Z= 200, which is 2^3*5^2*7^0 .. so b can be 2 as well right..
am I missing something?
I agree that a is atleast 3.. but how is b EXACTLY 1.. I can do Z= 200, which is 2^3*5^2*7^0 .. so b can be 2 as well right..
am I missing something?
