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18 Nov 2021, 03:28
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If \(n\) is a positive integer and \(n^2 + 19n - n! = 0\), then \(n\) can take how many different values ?
A. \(0\)
B. \(1\)
C. \(2\)
D. \(4\)
E. \(5\)
A. \(0\)
B. \(1\)
C. \(2\)
D. \(4\)
E. \(5\)
19 Nov 2024, 03:37
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Bunuel
\(n^2 + 19n - n! = 0\)
We have n! as a term. The only thing we can do is cancel out the common n which is present in all three terms.
\(n^2 + 19n - n! = n(n + 19 - (n-1)!) = 0\)
n cannot be 0 because it is a positive integer. So
\((n + 19 - (n-1)!) = 0\)
\(n+ 19 = (n-1)!\)
The factorial term is greater than 19. We know that the first factorial greater than 19 is 4! = 24
If n = 5, we get 5 + 19 = (5-1)! which is valid. Hence n can be 5.
As n will increase, (n-1)! will increase a whole lot and cannot be equal to n + 19. e.g. when n = 6, we get 6 + 19 = 5! = 120 Not valid.
Hence n can take only 1 value (which is 5)
Answer (B)
General Discussion
18 Nov 2021, 03:28
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Official Solution:
If \(n\) is a positive integer and \(n^2 + 19n - n! = 0\), then \(n\) can take how many different values ?
A. \(0\)
B. \(1\)
C. \(2\)
D. \(4\)
E. \(5\)
\(n^2 + 19n - n! = 0\);
Re-arrange: \(n^2 + 19n = n!\);
Factor out \(n\): \(n(n + 19) = n!\);
Reduce by \(n\): \(n + 19 = (n-1)!\).
\(n + 19 \geq 20\), so start with \(n=5\): \(n + 19 = 24\) and \((n-1)!=4!=24\).
Notice that when \(n > 5\), then \((n-1)!\) will for sure be greater than \(n+19\), so no larger values of \(n\) are possible.
\(n\) can take only one value: 5.
Answer: B
If \(n\) is a positive integer and \(n^2 + 19n - n! = 0\), then \(n\) can take how many different values ?
A. \(0\)
B. \(1\)
C. \(2\)
D. \(4\)
E. \(5\)
\(n^2 + 19n - n! = 0\);
Re-arrange: \(n^2 + 19n = n!\);
Factor out \(n\): \(n(n + 19) = n!\);
Reduce by \(n\): \(n + 19 = (n-1)!\).
\(n + 19 \geq 20\), so start with \(n=5\): \(n + 19 = 24\) and \((n-1)!=4!=24\).
Notice that when \(n > 5\), then \((n-1)!\) will for sure be greater than \(n+19\), so no larger values of \(n\) are possible.
\(n\) can take only one value: 5.
Answer: B
