If n is an integer such that 1/9

Question

If n is an integer such that \frac{1}{9} < \frac{1}{(n^2-1)} < \frac{1}{2} , what is the value of n? (1) 1/3 > 1/(1 - n) > 1/7 (2) n is not an even integer.

Kinshook · Answer

Asked: If n is an integer such that \frac{1}{9} < \frac{1}{(n^2-1)} < \frac{1}{2} , what is the value of n? 1/(n^2-1) > 1/9 1/(n^2 - 1) - 1/9 > 0 {9 - (n^2 - 1)}/9(n^2 - 1) > 0 (10 - n^2)/(n^2 - 1) > 0 \frac{(n + \sqrt{10})(n-\sqrt{10})}{(n-1)(n+1)} < 0 -4 < -\sqrt{10} < n < -1 or 1 < n < \sqrt{10} < 4 n = {-3,-2,2,3} 1/(n^2 - 1) < 1/2 1/(n^2 - 1) - 1/2 < 0 {2 - (n^2 - 1)}/2(n^2 - 1) < 0 (3 - n ^2)/(n^2 - 1) < 0 (n^2 - 3)/(n^2 - 1) > 0 n < - \sqrt{3} or -1 < n < 1 or n > \sqrt{3} n = {-3-2,2,3} (1) 1/3 > 1/(1 - n) > 1/7 -7 < n - 1 < -3 -6 < n < -2 n = -3 SUFFICIENT (2) n is not an even integer. n = -3 or 3 NOT SUFFICIENT IMO A

bellbell · Answer

Please tell me if I'm wrong.
n is an integer
1/9<1/(n^2-1)<1/2 => 9>(n^2-1)>2
(a) 9>(n^2-1) => 10>n^2
(b) (n^2-1)>2 => n^2>3
=> 10>n^2>3
=> n^2 can only be 4 or 9 => n=-2, 2, -3, 3
Let's go to the options:
(1)1/3 > 1/(1 - n) > 1/7 => 3<1-n<7 => -2<n<-6
=> n=-3 => sufficient
(2) => not sufficient because we already solved that n can be integer from the beginning

Bunuel · Answer

Official Solution: If n is an integer such that \frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2} , what is the value of n ? Given that n is an integer, for the inequality \frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2} to hold true, \frac{1}{n^2-1} must be \frac{1}{8} , \frac{1}{7} , \frac{1}{6} , \frac{1}{5} , \frac{1}{4} , or \frac{1}{3} . Hence, n^2-1 must be 8, 7, 6, 5, 4, or 3. Consequently, n^2 must be 9, 8, 7, 6, 5, or 4. However, n^2 cannot be 8, 7, 6, or 5, because in these cases n won't be an integer. Therefore, n^2 is either 9 or 4, making n equal to -3, 3, -2, or 2. The question essentially asks: if n is -3, 3, -2, or 2, what is the value of n ? (1) \frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7} . Given that n is an integer, for the inequality \frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7} to hold true, \frac{1}{1 - n} must be \frac{1}{4} , \frac{1}{5} , or \frac{1}{6} . Hence, n is -3, -4, or -5. Since from the stem n is -3, 3, 2, or -2, then n must be -3. Sufficient. (2) n is not an even integer. This statement rules out n being 2 and -2. However, n can still be 3 or -3. Not sufficient. Answer: A

Bunuel · Answer

Practice slightly easier of this question  HERE .

If n is an integer such that 1/9 < 1/(n^2-1) < 1/2, what is the value

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