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Difficulty: EasyOdd and Even Integers (Parity)

If nn is an integer such that 1≤n≤501 \le n \le 50, how many values of nn satisfy the condition that n2+3nn^2 + 3n is an odd integer?

Answer: 0

Answer

0
For any integer nn, the expression n2+3nn^2 + 3n can be rewritten as n(n+3)n(n + 3). Since nn and n+3n + 3 differ by 3 (an odd number), one factor must be even and the other must be odd. The product of an even integer and an odd integer is always even. Therefore, n2+3nn^2 + 3n is even for all integer values of nn, meaning there are exactly 0 values of nn in the given range for which the expression is odd.

Step-by-Step Solution

1
Analyze the algebraic expression n2+3nn^2 + 3n by testing parity cases for any integer nn.
The parity of the expression depends on whether nn is even or odd.
Every integer is either even or odd.
2
Evaluate n2+3nn^2 + 3n assuming nn is even.
n2n^2 is even and 3n3n is even; their sum is even.
Even ×\times Even = Even, and Even ++ Even = Even.
3
Evaluate n2+3nn^2 + 3n assuming nn is odd.
n2n^2 is odd and 3n3n is odd; their sum is even.
Odd ×\times Odd = Odd, and Odd ++ Odd = Even.
4
Count the number of integers nn in 1≤n≤501 \le n \le 50 for which n2+3nn^2 + 3n is odd.
Because n2+3nn^2 + 3n is even for all integer values of nn, zero values of nn make the expression odd.
No integer nn satisfies the condition.

Key Concept

Odd and Even Integers (Parity)
Estimated Time:1m 0s
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Topics

Odd and Even Integers (Parity)

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