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n^^ = n! - n^2, where n is a positive integer. For how many

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n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post 01 Oct 2017, 04:43
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n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?

A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^

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Director
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post 01 Oct 2017, 07:20
Bunuel wrote:
n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?

A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^

Kudos for correct solution.


Not clear with the ques?

It has asked for how many values of n, n^^<0, or should it be; for what value of n, n^^<0. PLz clarify??
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post 01 Oct 2017, 07:59
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I interpreted the first way.

Thus,
1^^ = 1! - 1 = 0
2^^ = 2! - 4 = -2
3^^ = 3! - 9 = -3
4^^ = 4! - 16 = 18

So, I would say there are two cases in which n^^<0. Now, we have to find which one of the choice has value 2.
Given our prior computations, it is easy to notice that 1^^-2^^ = 0 - (-2) = 2.

Thus, the answer is B!
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post 01 Oct 2017, 08:23
IlCreatore wrote:
I interpreted the first way.

Thus,
1^^ = 1! - 1 = 0
2^^ = 2! - 4 = -2
3^^ = 3! - 9 = -3
4^^ = 4! - 16 = 18

So, I would say there are two cases in which n^^<0. Now, we have to find which one of the choice has value 2.
Given our prior computations, it is easy to notice that 1^^-2^^ = 0 - (-2) = 2.

Thus, the answer is B!


you are right, i missed it :(
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post 02 Nov 2018, 23:51
C & E
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post 03 Nov 2018, 04:50
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Bunuel wrote:
n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?

A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^

Kudos for correct solution.



n^^ = \(n! - n^2\).... n^^ < 0 means \(n!-n^2<0...........n^2>n!......n^2>n(n-1)!.......n>(n-1)!\)
so the number must be greater than the product of numbers less than itself.
3>2*1 but 4 is NOT > 3*2 so values are 1, 2 and 3 but 1 will give us a 0, thus 2 values...
1^^ is 0, so cannot be the answer.
3^^ is negative, so cannot be the answer
so let us check 1^^-2^^ = 0-(2!-2^2)=0-(2-4)=2...answer

B
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Re: n^^ = n! - n^2, where n is a positive integer. For how many   [#permalink] 03 Nov 2018, 04:50
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