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26 May 2017, 04:47
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If \(n!g(n) = (n+3)!\) and n are all integers from 1 to 30, inclusive, what is the probability that g(n) is divisible by 6?
A. 1/6
B. 1/5
C. 1/3
D. 1/2
E. 1
A. 1/6
B. 1/5
C. 1/3
D. 1/2
E. 1
26 May 2017, 04:48
Kudos
Bookmarks
Official Solution:
If \(n!g(n) = (n+3)!\) and n are all integers from 1 to 30, inclusive, what is the probability that g(n) is divisible by 6?
A. 1/6
B. 1/5
C. 1/3
D. 1/2
E. 1
In general, \(n! = n(n-1)(n-2)\ldots .(2)(1)\), then \((n-1)! = (n-1)(n-2)\ldots .(2)(1)\), and then \((n-2)!=(n-2)(n-3)\ldots ..(2)(1)\), so you get \(n! = n(n-1)!=n(n-1)(n-2)!\). Then, it becomes \((n+3)! = (n+3)(n+2)(n+1)n!\). Therefore,
you get \(n!g(n) = (n+3)! \rightarrow n!g(n)=(n+3)(n+2)(n+1)n!\), and if you divide both sides by \(n!\), you get \(g(n) = (n+3)(n+2)(n+1)\) since \((n+3)(n+2)(n+1)\) is always the multiple of 3 consecutive integers. Thus, they always become multiples of 6. Therefore, \(g(n)\) is always a multiple of 6, so the probability of dividing \(g(n)\) by 6 without a remainder is \(\frac{30C1}{30C1} = \frac{30}{30} = 1\). Therefore, E is the answer.
Answer: E
If \(n!g(n) = (n+3)!\) and n are all integers from 1 to 30, inclusive, what is the probability that g(n) is divisible by 6?
A. 1/6
B. 1/5
C. 1/3
D. 1/2
E. 1
In general, \(n! = n(n-1)(n-2)\ldots .(2)(1)\), then \((n-1)! = (n-1)(n-2)\ldots .(2)(1)\), and then \((n-2)!=(n-2)(n-3)\ldots ..(2)(1)\), so you get \(n! = n(n-1)!=n(n-1)(n-2)!\). Then, it becomes \((n+3)! = (n+3)(n+2)(n+1)n!\). Therefore,
you get \(n!g(n) = (n+3)! \rightarrow n!g(n)=(n+3)(n+2)(n+1)n!\), and if you divide both sides by \(n!\), you get \(g(n) = (n+3)(n+2)(n+1)\) since \((n+3)(n+2)(n+1)\) is always the multiple of 3 consecutive integers. Thus, they always become multiples of 6. Therefore, \(g(n)\) is always a multiple of 6, so the probability of dividing \(g(n)\) by 6 without a remainder is \(\frac{30C1}{30C1} = \frac{30}{30} = 1\). Therefore, E is the answer.
Answer: E
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