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02 Feb 2024, 02:46
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If \(x\), \(y\), and \(z\) are integers, \(x < y < z\) and \(xyz = 30\), what is the smallest possible value of y?
A. -6
B. -5
C. -2
D. 2
E. 5
A. -6
B. -5
C. -2
D. 2
E. 5
02 Feb 2024, 02:46
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Official Solution:
If \(x\), \(y\), and \(z\) are integers, \(x < y < z\) and \(xyz = 30\), what is the smallest possible value of y?
A. -6
B. -5
C. -2
D. 2
E. 5
The divisors of 30 are 1, 2, 3, 5, 6, 10, 15, 30, and their negative counterparts.
All \(x\), \(y\), and \(z\) cannot be negative because their product would then be negative. Hence, to obtain the least value of \(y\), we must consider the case when \(x < y < 0 < z\). To minimize \(y\), or equivalently, to maximize the absolute value of \(y\) (\(|y|\)), we need to minimize \(z\). The smallest positive value of \(z\) is thus 1, which leads to \(xy = 30\). Since \(x\) and \(y\) must both be negative, the smallest value of \(y\) is -5 when \(x = -6\).
Answer: B
If \(x\), \(y\), and \(z\) are integers, \(x < y < z\) and \(xyz = 30\), what is the smallest possible value of y?
A. -6
B. -5
C. -2
D. 2
E. 5
The divisors of 30 are 1, 2, 3, 5, 6, 10, 15, 30, and their negative counterparts.
All \(x\), \(y\), and \(z\) cannot be negative because their product would then be negative. Hence, to obtain the least value of \(y\), we must consider the case when \(x < y < 0 < z\). To minimize \(y\), or equivalently, to maximize the absolute value of \(y\) (\(|y|\)), we need to minimize \(z\). The smallest positive value of \(z\) is thus 1, which leads to \(xy = 30\). Since \(x\) and \(y\) must both be negative, the smallest value of \(y\) is -5 when \(x = -6\).
Answer: B
04 May 2024, 07:08
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I think this is a high-quality question and I don't agree with the explanation. Why cannot we consider the combination of -10, -3 and 1?
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