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Maths help

Prove that the mean of any 4 consecutive even integers is an integer.
Original post by JasmineJohal14
Prove that the mean of any 4 consecutive even integers is an integer.


Do you know how to represent an even integer in terms of a general integer nn ?
Original post by RDKGames
Do you know how to represent an even integer in terms of a general integer nn ?

It says even consecutive so would it be...
n, n+2, n+4, n+6
Original post by JasmineJohal14
It says even consecutive so would it be...
n, n+2, n+4, n+6


Not quite. I said that nn is any general integer. If nn is odd, then n, n+2, n+4 and n+6 are all odd numbers. Not what we want.

Don't jump to the consecutive integers part yet, let's start simple. What's the form of a general *even* number ?
Original post by RDKGames
Not quite. I said that nn is any general integer. If nn is odd, then n, n+2, n+4 and n+6 are all odd numbers. Not what we want.

Don't jump to the consecutive integers part yet, let's start simple. What's the form of a general *even* number ?

2n
Original post by JasmineJohal14
2n


And in relation to this 2n, what is the next even number along?
Consider 2n to be an even integer.

If 2n is an even integer, its consecutive even integers would be 2n+2, 2n+4 and 2n+6. The 4 consecutive even integers to use can be 2n, 2n+2, 2n+4 and 2n+6. To calculate the mean, we add the values and divide by how many there are. In this case, we do (2n + 2n + 2 + 2n + 4 + 2n + 6)/4 = 2n + 3.

In the beginning, we established that 2n is an integer. 2n+3 must be an integer, as an integer plus an integer makes an integer.

Hope this helped.
Original post by RDKGames
And in relation to this 2n, what is the next even number along?

2n, 2n+2, 2n+4, 2n+6
Original post by minyeonkim
Consider 2n to be an even integer.

If 2n is an even integer, its consecutive even integers would be 2n+2, 2n+4 and 2n+6. The 4 consecutive even integers to use can be 2n, 2n+2, 2n+4 and 2n+6. To calculate the mean, we add the values and divide by how many there are. In this case, we do (2n + 2n + 2 + 2n + 4 + 2n + 6)/4 = 2n + 3.

In the beginning, we established that 2n is an integer. 2n+3 must be an integer, as an integer plus an integer makes an integer.

Hope this helped.

Thank youu!
Original post by minyeonkim
Consider 2n to be an even integer.

If 2n is an even integer, its consecutive even integers would be 2n+2, 2n+4 and 2n+6. The 4 consecutive even integers to use can be 2n, 2n+2, 2n+4 and 2n+6. To calculate the mean, we add the values and divide by how many there are. In this case, we do (2n + 2n + 2 + 2n + 4 + 2n + 6)/4 = 2n + 3.

In the beginning, we established that 2n is an integer. 2n+3 must be an integer, as an integer plus an integer makes an integer.

Hope this helped.


Again, full solutions are not allowed. Just give hints. OP would've very well got to the answer by themselves with a little push in the right direction.

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