Let n be an integer such that -15 \le n \le 15. How many integer values of n satisfy both … — GRE General Test

Let $n$ be an integer such that $-15 \le n \le 15$ . How many integer values of $n$ satisfy both of the following conditions?

1. $(-1)^{n^2 + n + 1} < 0$
2. $(n - 3)(n + 4)^2 < 0$

Answer: 17

Condition 1 is satisfied by every integer because

n^2 + n = n(n+1)

is always even (as the product of two consecutive integers), which makes

n^2 + n + 1

always odd and

(-1)^{\text{odd}} = -1 < 0

. Condition 2 requires

(n - 3)(n + 4)^2 < 0

. Since

(n+4)^2 > 0

for all

n \neq -4

, this reduces to

n < 3

while excluding

n = -4

. Within the range

-15 \le n \le 15

, there are 18 integers strictly less than 3, and removing

n = -4

leaves 17 valid values.

Determine the parity of the exponent

n^2 + n + 1

n^2 + n + 1

is always odd for any integer

n

, making

(-1)^{n^2 + n + 1} = -1 < 0

unconditionally true.

The product of consecutive integers

n(n+1)

is always even, so adding 1 results in an odd number.

Solve the sign inequality

(n - 3)(n + 4)^2 < 0

n < 3

with

n \neq -4

A squared expression is strictly positive except when its base is zero. At

n = -4

, the product becomes 0, violating the strict inequality.

Count the integer solutions in the range

-15 \le n \le 15

17 integers satisfy both conditions.

There are 18 integers less than 3 in the interval

[-15, 15]

, and excluding

n = -4

gives

18 - 1 = 17

Parity of consecutive integer products and sign rules for squared terms in inequalities

Estimated Time:2m 0s

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