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Joined: 19 Nov 2020
Posts: 326
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GRE 1: Q160 V152
x, y and z are positive integers such that 1<x<y<z
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23 Sep 2021, 04:16
23 Sep 2021, 04:16
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\(x\), \(y\) and \(z\) are positive integers such that \(1 < x < y < z\). If the product of \(x\), \(y\) and \(z\) is 399, what is \(z-y\) ?
Source: Manhattan Review, GMAT Question Bank
Show: ::
12
Source: Manhattan Review, GMAT Question Bank
Re: x, y and z are positive integers such that 1<x<y<z
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23 Sep 2021, 06:20
23 Sep 2021, 06:20
1
Since there are 3 integers that are needed to get the product of 399, divide 399 by 3 to get 133.
From here, factor 133 to find 19x7.
Since x<y<z, the numbers are x=3, y=7 and z=19.
Z-Y or 19-7=12.
The answer is 12.
From here, factor 133 to find 19x7.
Since x<y<z, the numbers are x=3, y=7 and z=19.
Z-Y or 19-7=12.
The answer is 12.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
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Re: x, y and z are positive integers such that 1<x<y<z
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10 Oct 2021, 06:26
10 Oct 2021, 06:26
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motion2020 wrote:
\(x\), \(y\) and \(z\) are positive integers such that \(1 < x < y < z\). If the product of \(x\), \(y\) and \(z\) is 399, what is \(z-y\) ?
Source: Manhattan Review, GMAT Question Bank
Show: ::
12
Source: Manhattan Review, GMAT Question Bank
Given: xyz = 399
This suggests that we should find the prime factorization of 399.
When we do this we get: 399 = (3)(7)(19)
Since x < y < z, we can conclude that x = 3, y = 7 and z = 19
So, z - y = 19 - 7 = 12
Answer: 12
