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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.

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]
03 Nov 2018, 04:50

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Expert's post

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^^

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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