1. Let the three consecutive even numbers be 2x, 2y and 2z.
There product: (2x)*(2y)*(2z) = 8 xyz, and thus it is a multiple of 8.
Infact, the numbers need not be even consecutive numbers. They hold true for ANY three even numbers.
For e.g. (2)(4)(8) = 64 = 8*8

2.Let the consecutive numbers 2n - 2, 2n and 2n + 2,
Sum: (2n - 2) + 2n + (2n + 2) = 6n
So, it is always a multiple of 6.
For e.g. 2 + 4 + 6 = 12 = 6*2

hannah_lewis11 wrote:1. Prove that the product of three consecutive even numbers is always a multiple of eight. Your answer must be supported by reasoning and not just specific examples


2. Prove that the sum of three consecutive even numbers is always a multiple of six. Your answer must be supported by reasoning and not just by specific examples.

1. Any even no. has 2 as its factor. So 3 even nos. will have 3 2's. 2*2*2 makes 8. So divisible by 8.

2. The second even no. will be greater than first by 2 and third one will be greater than first by 4, hence collectively we get 6 in total. Say nos. are 2x, 2x+2, 2x+4. Sum is 6x+6 = 6(x+1). It is multiple of 6.