Question

Use Theorem 5.2.1 to prove that if m and n are any positive integers and m is odd, then m^21 ⋅k^50 ⋅(n^1k) is divisible by m. Does the conclusion hold if m is even? Justify your answer.

Answer 1

Theorem 5.2.1 states that if a positive integer m is odd, then m^2 is divisible by m. Therefore, m^21 * k^50 * (n^1k) is divisible by m. However, if m is even, Theorem 5.2.1 does not hold.

Step-by-step explanation:

Theorem 5.2.1 states that if a positive integer m is odd, then m^2 is divisible by m. To prove that m^21 * k^50 * (n^1k) is divisible by m, we need to show that m^2 is divisible by m. Since m is odd, we can rewrite m^2 as m^(2*10 + 1) = (m^2)^10 * m. Since m^2 is divisible by m according to Theorem 5.2.1, m^(2*10) * m is divisible by m. Therefore, m^21 * k^50 * (n^1k) is divisible by m.

If m is even, Theorem 5.2.1 does not hold. This is because for even numbers, m^2 is not necessarily divisible by m. For example, if m = 2, then 2^2 = 4, which is not divisible by 2. Therefore, the conclusion does not hold if m is even.