Three consecutive integers a, b, and c satisfy a < b < c. If a + b + c = -9 and a \cdot b … — GRE General Test

Three consecutive integers $a$ , $b$ , and $c$ satisfy $a < b < c$ . If $a + b + c = -9$ and $a \cdot b \cdot c < 0$ , what is the value of $(-1)^a + (-1)^b + (-1)^c$ ?

Answer: 1

Answer

The value of the expression is 1.

The sum of three consecutive integers

a + b + c = 3b = -9

determines

b = -3

, making

a = -4

and

c = -2

. The product

(-4)(-3)(-2) = -24

is negative, confirming the given condition. Applying exponent sign rules, raising

-1

to an even integer power yields

1

, while raising

-1

to an odd integer power yields

-1

. Thus,

(-1)^{-4} = 1

(-1)^{-3} = -1

, and

(-1)^{-2} = 1

. Summing these three terms gives

1 + (-1) + 1 = 1

Step-by-Step Solution

Find the values of integers

a

b

, and

c

a = -4

b = -3

c = -2

Three consecutive integers centered at

b

sum to

3b = -9

, so

b = -3

Check the sign condition of the product

a \cdot b \cdot c

(-4)(-3)(-2) = -24 < 0

The product of three negative numbers is negative.

Evaluate

(-1)^n

for each integer power.

(-1)^{-4} = 1

(-1)^{-3} = -1

(-1)^{-2} = 1

Negative one raised to an even integer power is 1; raised to an odd integer power is -1.

Sum the three evaluated terms.

1 + (-1) + 1 = 1

Addition of the resulting values.

Key Concept

Even-odd exponent rules for negative bases and sign rules for product of signed integers.

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