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Difficulty: MediumOdd and Even Integers (Parity)

For how many positive integers nn less than or equal to 120120 is the expression n3−n+3nn^3 - n + 3^n an even integer?

Answer: 0

Answer

0
Factoring n3−nn^3 - n into (n−1)n(n+1)(n-1)n(n+1) shows that this component is the product of three consecutive integers and is therefore always even for any integer nn. Since 3 is odd, 3n3^n is always odd for any positive integer exponent nn. The sum of an even integer and an odd integer is always odd. Consequently, n3−n+3nn^3 - n + 3^n is odd for all positive integers nn, meaning 0 values of n≤120n \le 120 yield an even integer.

Step-by-Step Solution

1
Analyze the algebraic expression by factoring n3−nn^3 - n.
n3−n=n(n2−1)=(n−1)n(n+1)n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1), which represents the product of three consecutive integers.
The product of consecutive integers contains at least one even number, making n3−nn^3 - n even for all integers nn.
2
Analyze the exponential term 3n3^n.
Since 3 is odd, 3n3^n is odd for every positive integer nn.
An odd integer raised to any positive integer exponent yields an odd integer.
3
Determine the overall parity of the expression n3−n+3nn^3 - n + 3^n.
even+odd=odd\text{even} + \text{odd} = \text{odd}. The expression is odd for all positive integers nn.
Adding an odd integer to an even integer results in an odd integer.
4
Count the number of positive integers n≤120n \le 120 for which the expression is even.
There are 0 such integers.
Because the expression is odd for every positive integer nn, it is never even.

Key Concept

Parity rules of consecutive integer products, odd exponents, and addition of even/odd integers.
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Topics

Odd and Even Integers (Parity)

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