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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]  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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[Reveal] Spoiler: OA
Director
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink]  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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Director
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink]  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.

Director
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Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink]  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.

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]  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]  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^^

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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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
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5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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