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If x, y, and z are positive integers such that x < y < z and xyz = 24,
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17 Dec 2021, 06:52
17 Dec 2021, 06:52
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Question Stats:
55% (01:00) correct
44% (00:42) wrong based on 9 sessions
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If x, y, and z are positive integers such that \(x < y < z\) and \(xyz = 24\), then the least possible value of z is
(A) 2
(B) 4
(C) 6
(D) 8
(E) 12
(A) 2
(B) 4
(C) 6
(D) 8
(E) 12
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Re: If x, y, and z are positive integers such that x < y < z and xyz = 24,
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17 Dec 2021, 08:57
17 Dec 2021, 08:57
1
Start by factorizing 24:
2^3 * 3 = 2 * 2 * 2 * 3
There are a number of different ways that we can make this work, but we just have to rearrange the numbers in this multiplication for two conditions:
1) It must be a group of three numbers (that is, x, y, and z)
2) x < y < z
We want to make the value of z the smallest possible, of course. The lowest that x could be is 2. This is a place to start.
x = 2
We can't have y be the same value (that is, y = 2 for 2*2*6=24). Rather, y must be larger.
That means that y is 3 or 4. Yet if we choose 4, we have no place to put the three, because y < z as well.
So we're stuck with y = 3.
This means that the z = 4 and that is the lowest case where all three numbers match x < y < z.
It's just a bit of testing holding the two above conditions together. Answer B.
2^3 * 3 = 2 * 2 * 2 * 3
There are a number of different ways that we can make this work, but we just have to rearrange the numbers in this multiplication for two conditions:
1) It must be a group of three numbers (that is, x, y, and z)
2) x < y < z
We want to make the value of z the smallest possible, of course. The lowest that x could be is 2. This is a place to start.
x = 2
We can't have y be the same value (that is, y = 2 for 2*2*6=24). Rather, y must be larger.
That means that y is 3 or 4. Yet if we choose 4, we have no place to put the three, because y < z as well.
So we're stuck with y = 3.
This means that the z = 4 and that is the lowest case where all three numbers match x < y < z.
It's just a bit of testing holding the two above conditions together. Answer B.
General Discussion
Re: If x, y, and z are positive integers such that x < y < z and xyz = 24,
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17 Dec 2021, 08:05
17 Dec 2021, 08:05
1
\(x, y,\) and \(z\) = positive integers & \(x < y < z\)
xyz = 24
We need the least value of z so \(z = 4\) \(x = 2\) , \(y = 3\)
Answer B
xyz = 24
We need the least value of z so \(z = 4\) \(x = 2\) , \(y = 3\)
Answer B
