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Re: n is an integer, and n^2 < 39 [#permalink]
Wow you've made the problem look so easy.

sandy wrote:
Explanation

We know that n is an integer.

And \(n^2 < 39\). The largest perfect square less than 39 is 36

So \(n^2 = 36\). Solving for n we get

\(n = +6\) and \(-6\).

The greatest possible value of n minus the least possible value of \(n = 6 - (-6)\).

Hence both quantities are equal. Hence option C is correct.
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Re: n is an integer, and n^2 < 39 [#permalink]
1
we know that n is integer and n*n < 39
so the range of possible values of n is {-6, -5, -4,-3,-2,-1,0,1,2,3,4,5,6}
and n*n = {36,25,16,9,4,1,0}
A. 6-(-6) = 12
So the answer is C
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Re: n is an integer, and n^2 < 39 [#permalink]
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If n is an integer and n*n<39,
then n is in the set { -6, -5, ..., 0, 1, ..., 6}

Quantity A = 6 - (-6) = 12
Quantity B = 12

The two quantities are always equal

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Re: n is an integer, and n^2 < 39 [#permalink]
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