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Difficulty: Very hardOdd and Even Integers (Parity)

Let f(n)=n5+4n3+3n+1f(n) = n^5 + 4n^3 + 3n + 1 for any positive integer nn. For how many integer values of nn in the range 1≤n≤1501 \le n \le 150 is the expression E(n)=(f(n))n+nf(n)E(n) = (f(n))^n + n^{f(n)} an even integer?

Answer: 75

Answer

The total number of integer values of nn in the given range for which the expression is even is 75.
Analyzing f(n)(mod2)f(n) \pmod 2 reveals that n5≡n(mod2)n^5 \equiv n \pmod 2 and 3n≡n(mod2)3n \equiv n \pmod 2, making f(n)≡n+0+n+1≡2n+1≡1(mod2)f(n) \equiv n + 0 + n + 1 \equiv 2n + 1 \equiv 1 \pmod 2. Since f(n)f(n) is unconditionally odd, (f(n))n(f(n))^n is an odd number raised to a positive integer power, which remains odd. The second term nf(n)n^{f(n)} has positive exponent f(n)≥1f(n) \ge 1, so its parity matches the base nn. Therefore, E(n)=Odd+nE(n) = \text{Odd} + n is even precisely when nn is odd. In the range 1≤n≤1501 \le n \le 150, exactly half of the 150 integers are odd, yielding 75.

Step-by-Step Solution

1
Evaluate the parity of f(n)f(n)
f(n)=n5+4n3+3n+1≡n+0+n+1≡2n+1≡1(mod2)f(n) = n^5 + 4n^3 + 3n + 1 \equiv n + 0 + n + 1 \equiv 2n + 1 \equiv 1 \pmod 2, so f(n)f(n) is always odd.
Powers of an integer retain the base parity (n5≡n(mod2)n^5 \equiv n \pmod 2), and coefficients modulo 2 simplify terms.
2
Evaluate the parity of (f(n))n(f(n))^n and nf(n)n^{f(n)}
(f(n))n(f(n))^n is always odd because an odd base raised to any positive integer exponent is odd. nf(n)n^{f(n)} has the same parity as nn because f(n)≥1f(n) \ge 1.
Exponent parity rules state that oddk=odd\text{odd}^k = \text{odd} and evenk=even\text{even}^k = \text{even} for positive integer exponents kk.
3
Formulate the condition for E(n)E(n) to be even
E(n)=Odd+nE(n) = \text{Odd} + n. Thus, E(n)E(n) is even if and only if nn is odd.
The sum of two integers is even if and only if both integers have the same parity (Odd + Odd = Even).
4
Count odd integers in the specified interval
There are 75 odd integers in {1,2,3,…,150}\{1, 2, 3, \dots, 150\}.
Half of the integers in any consecutive sequence of even length starting at 1 are odd.

Key Concept

Odd and Even Integers (Parity)
Estimated Time:2m 0s
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Topics

Odd and Even Integers (Parity)

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