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21 Jun 2024, 01:31
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
81% (00:59) correct
19%
(01:34)
wrong
based on 162
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If \(n\) is a positive integer, how many values of \(n\) are there such that \(n^5 = n!\)?
A. 0
B. 1
C. 2
D. 3
E. Infinitely many
A. 0
B. 1
C. 2
D. 3
E. Infinitely many
06 Jul 2024, 04:30
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Official Solution:
If \(n\) is a positive integer, how many values of \(n\) are there such that \(n^5 = n!\)?
A. 0
B. 1
C. 2
D. 3
E. Infinitely many
It's clear that \(n=1\) satisfies the given equation. The real question now is, is it the only solution? First of all, note that if \(n\) is greater than 1, \(n!\) becomes even, therefore, if there is another solution to \(n^5 = n!\), then \(n\) must be even for \(n^5\) to also be even. Let's test some even values:
If \(n=2\), \(n^5 = 2^5 = 32\), while \(n! = 2\). The values do not match: \(n^5 > n!\).
If \(n=4\), \(n^5 = 4^5 = 2^{10} = 1024\), while \(4! = 24\). The values do not match: \(n^5 > n!\).
If \(n=6\), \(n^5 = 6^5 = 6*6*6*6*6\), while \(6! = 2*3*4*5*6\). The values do not match: \(n^5 > n!\).
If \(n=8\), \(n^5 = 8^5 = 8*8*8*8*8\), while \(8! = 2*3*4*5*6*7*8\). The values do not match. Notice that in this case, it becomes \(n^5 < n!\) instead.
For \(n\geq 8\), observe that the factorial function \(n! = 1*2*3*...*n\) grows much faster than the exponential function \(n^5 = n*n*n*n*n\). This is because when \(n \geq 8\), there are additional factors greater than or equal to \(n\) in the factorial, and their number is more than 5. These factors cause the factorial to grow much more rapidly compared to the exponential function. For example, if \(n=10\), \(n^5 = 10*10*10*10*10\), while, \(n! = 1*2*3*4*5*6*7*8*9*10=(2*5)*(3*4)*(6*7)*(8*9)*10\). Consequently, we can conclude that there is no solution for \(n \geq 8\).
Based on the above analysis, the only positive integer solution for the given equation is \(n=1\). Therefore, \(n^5 = n!\) has only one solution.
Answer: B
If \(n\) is a positive integer, how many values of \(n\) are there such that \(n^5 = n!\)?
A. 0
B. 1
C. 2
D. 3
E. Infinitely many
It's clear that \(n=1\) satisfies the given equation. The real question now is, is it the only solution? First of all, note that if \(n\) is greater than 1, \(n!\) becomes even, therefore, if there is another solution to \(n^5 = n!\), then \(n\) must be even for \(n^5\) to also be even. Let's test some even values:
If \(n=2\), \(n^5 = 2^5 = 32\), while \(n! = 2\). The values do not match: \(n^5 > n!\).
If \(n=4\), \(n^5 = 4^5 = 2^{10} = 1024\), while \(4! = 24\). The values do not match: \(n^5 > n!\).
If \(n=6\), \(n^5 = 6^5 = 6*6*6*6*6\), while \(6! = 2*3*4*5*6\). The values do not match: \(n^5 > n!\).
If \(n=8\), \(n^5 = 8^5 = 8*8*8*8*8\), while \(8! = 2*3*4*5*6*7*8\). The values do not match. Notice that in this case, it becomes \(n^5 < n!\) instead.
For \(n\geq 8\), observe that the factorial function \(n! = 1*2*3*...*n\) grows much faster than the exponential function \(n^5 = n*n*n*n*n\). This is because when \(n \geq 8\), there are additional factors greater than or equal to \(n\) in the factorial, and their number is more than 5. These factors cause the factorial to grow much more rapidly compared to the exponential function. For example, if \(n=10\), \(n^5 = 10*10*10*10*10\), while, \(n! = 1*2*3*4*5*6*7*8*9*10=(2*5)*(3*4)*(6*7)*(8*9)*10\). Consequently, we can conclude that there is no solution for \(n \geq 8\).
Based on the above analysis, the only positive integer solution for the given equation is \(n=1\). Therefore, \(n^5 = n!\) has only one solution.
Answer: B
General Discussion
21 Jun 2024, 01:59
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Bunuel
\(n^5 = n!\)
This is only applicable when \(n = 1\)
Option B
