Question

Use the definitions of even, odd, prime, and compositive numbers to justify your answers for (a)-(c). Assume that rands are particular integers.

(a) Is 6rs even?
A. Yes, because 6rs = 2(3rs) + 1 and 3rs is an integer.
B. Yes, because 6rs = 2(3rs) and 3rs is an integer.
C. No, because 6rs = 2(3rs) + 1 and 3rs is an integer.
D. No, because 6rs = 2(3rs) and 3rs is an integer.

Answer 1

The number 6rs is even because it is the result of multiplying 6, which is an even number, by other integers r and s.

Step-by-step explanation:

When considering the expression 6rs with r and s defined as particular integers, we apply the definitions of even, odd, prime, and composite numbers. An even number is an integer that can be divided by 2 without a remainder. Using this definition, we can determine if 6rs is even. The number 6 is already known to be even because it is equal to 2 times 3, hence any integer multiplied by 6 must also be even. Therefore, 6rs is indeed an even number because 6rs = 2(3rs), and since 3rs is an integer, multiplying it by 2 ensures that the result remains even.