Question

Which choice is the best proof of Asha's conjecture that the perimeter of a square whose sides are odd integers is always even?

A. Every time you add four odd numbers, the sum is an even number.

B. Let m and n both represent odd numbers that are the side lengths of a square. And let m + n be an even number. Therefore, m + n = n + m, which shows that the sum of two odd numbers is even. And if two odd numbers are even, then four odd numbers are also even.

C. Let n be an odd number. Because 4n is a multiple of 2, it is an even number.

D. Look at these different examples: 5 + 5 = 10, 9 + 9 + 9 + 9 = 36. So if the side of a square is odd, the perimeter must be even.

Answer 1

The best proof of Asha's conjecture is choice C, which states that the perimeter of a square (4n) with odd integer sides is even because multiplying by four (an even number) results in an even number.

Step-by-step explanation:

The perimeter of a square is calculated by adding all four sides together. Since a square has equal sides, if one side is an odd integer, the perimeter is four times this odd integer. The proof of Asha's conjecture that the perimeter of a square with odd integer sides is always even should fundamentally focus on the properties of adding odd numbers and the result of multiplying an odd number by an even one (in this case, four).

Choice C provides the best proof of Asha's conjecture: Let n be an odd number. Because 4n is a multiple of 2, it is an even number. This statement directly applies the mathematical rules that multiplying an odd integer by an even integer (four, in this case) results in an even product, thereby resulting in an even perimeter.