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n^2 is a positive integer

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n^2 is a positive integer

by sanju09 » Wed Mar 04, 2009 3:54 am
If n is an integer and if n^2 is a positive integer, which of the following must also be a positive integer?

(A) n^2 + n
(B) 2 n^2 – n
(C) n^2 – n^3
(D) n^3 + n
(E) 2 n^3 + n
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by DanaJ » Wed Mar 04, 2009 5:57 am
You get that, since n^2 is a positive integer, n will be any integer except for 0.
Eliminate a few options quickly, particularly D and E, by plugging in any negative number, like for instance n = -2.

In order to asses the other stmts, we need to remember a property of the quadratic equation. Take the general case of a quadratic equation:
ax^2 + bx + c
When a is positive, the above equation will be negative only between the roots, while it will be positive for the rest of the bunch.

Now take A. n^2 + n = n(n + 1) - this is a quadratic equation, which we know is negative only between the roots, 0 and -1. But since there are no integers between -1 and 0, the equation is never negative. However, take n = -1 and you get that n^2 + n = 0, which is of course, not positive.

Time for B. 2n^2 - n = n(2n - 1) - again, negative between the roots, with roots 1/2 and 0. Since there are no integers between 0 and 1/2 and 0 has been eliminated at the beginning, this is what we're looking for. So my answer is B
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Re: n^2 is a positive integer

by El Cucu » Wed Mar 04, 2009 7:22 am
sanju09 wrote:If n is an integer and if n^2 is a positive integer, which of the following must also be a positive integer?

(A) n^2 + n
(B) 2 n^2 – n
(C) n^2 – n^3
(D) n^3 + n
(E) 2 n^3 + n
Plug in -2, -1, 1 & 2.(as n is an integer)

Only B) satisfies the condition.
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Re: n^2 is a positive integer

by marcusking » Wed Mar 04, 2009 7:25 am
sanju09 wrote:If n is an integer and if n^2 is a positive integer, which of the following must also be a positive integer?

(A) n^2 + n
(B) 2 n^2 – n
(C) n^2 – n^3
(D) n^3 + n
(E) 2 n^3 + n
I try to prove the equation wrong to do so I try the smallest negative integer n=-1 to try and make the answers negative
A)-1^2 = 1 + -1 = 0 Not a positive integer
B) 2 * 1 - (-1) = 2 + 1 = 3 (Works)

Now try B with the smallest positive integer +1
2 * 1 - 1 = 1 (Works)

It has satisfied both a positive and negative integer, if I go further in either direction the 2*n^2 will only become a larger positive number

ANS. B.

Please start posting the OA.
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by sanju09 » Thu Mar 05, 2009 2:23 am
If n is an integer and if n^2 is a positive integer, then 2n^2 – n = 2n^2 + (–n) is the sum of two integers, and must therefore be an integer. Since n^2 is a positive integer, it follows that |n| ≥ 1. The integer 2n^2 – n will be positive if 2n^2 > n. Since |n| ≥ 1, it follows that n^2 > n, and so 2n^2 > n. Therefore, if n^2 is a positive integer, 2n^2 – n must also be a positive integer.

If n^2 is a positive integer, then the expressions in the other choices must be integers, but they need not be positive integers:

For n^2 + n, if n = –1, then n^2 + n = (–1)^2 + (–1) = 1 – 1 = 0, which is not positive.

For n^2 – n^3, if n = 2, then n^2 – n^3 = 2^2 – 2^3 = 4 – 8 = –4, which is not positive.

For n^3 + n, if n = –2, then n^3 + n = (–2)^3 + (–2) = –8 – 2 = –10, which is not positive.

For 2n^3 + n, if n = –2, then 2n^3 + n = 2(–2)^3 + (–2) = –16 – 2 = –18, which is not positive.

Therefore, the correct answer is [spoiler](B)[/spoiler].
The mind is everything. What you think you become. -Lord Buddha



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