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I am trying to use algebra to prove that

odd + even + odd = even

so I have three numbers, where n is odd.

I know that odd + odd = even but I can't prove this idea.

$$n, n + 1, n+ 2$$

where $n$ is odd, $n+1$ is even and $n+2$ is odd.

When I add them together, I get

$$3n+3 = 3(n+1)$$

but this can't make sense because when $n = 4$, $3 \times 5 = 15$

Thanks in advance.

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    $\begingroup$ You choose $n=4$ as even, so no wonder your example does not work. $\endgroup$ – M. Winter Oct 7 '17 at 11:17
  • $\begingroup$ We can simply see that we have $2$ odd numbers added together, which is an even number times an odd number, which is even. Then you are just left with even + even, which is even. $\endgroup$ – Sam Anderson Oct 7 '17 at 11:23
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For all an odd number there is an integer $n$, for which this number is equal to $2n-1$.

For all an even number there is an integer $n$, for which this number is equal to $2n$.

Thus, $$odd+even+odd=2n-1+2k+2m-1=2(n+k+m-1)=even.$$

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  • $\begingroup$ best answer, just fabulous. Thanks! $\endgroup$ – vik1245 Oct 7 '17 at 19:31
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Even number: $2k$, $k \in \mathbb{Z};$

Odd number: $2k+1.$

Even + even = even:

$2k +2l =2(k+l);$

Odd +odd = even:

$(2k+1) +(2l+1) = 2[(k+l)+1]$.

Odd +even is odd:

$(2k+1) + 2l = 2(k+l) +1.$

Even + odd ?

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write $$2k_1+1+2k_2+1+2m=2(k_1+k_2+1+m)=2m'$$ and this is even.

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In your example $n,n+1,n+2$ for $n=4$ we have the numbers $4,5,6$, where it is not true, that they are odd, even and odd. Indeed, the result $15$ is not even.

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When $n =4$, you're adding up $4, 5, 6$, and the total is 15, but that's not a problem, because you're adding up even + odd + even, not the other way around. The example you're probably thinking of is $3 + 4 + 5$, which is $n = 3$, and the sum is $3 \times 4$ which is 12 (an even number, as expected)

More generally, if you have

$$ odd + even + odd $$ you can swap the last two to get $$ odd + odd + even $$ and now the first two sum to an even number, giving you $$ even + even $$ which is always even.

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This question requires a different approach.

To ensure an odd number is produced, I use $2n-1$ or $2n+1$

To ensure an even number is produced, I use $2n$

So, now I have

$$2n-1, 2n, 2n+1$$

The sum of these numbers:

$$2n-1 + 2n + 2n+1 = 6n = 2(3n)$$

Hence as any number multiplied by 2 is even, an odd odd even combination will always be even.

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    $\begingroup$ It should be noted that you must use different variables for each of the numbers, as your current notation implies you're using the same exact two odd numbers, with the even number following directly afterwards. $\endgroup$ – AleksandrH Oct 7 '17 at 11:50
  • $\begingroup$ @AleksandrH sorted. $\endgroup$ – vik1245 Oct 7 '17 at 11:53
  • $\begingroup$ Looks the same on my end. You need to change two of the "n"s to different variables. $\endgroup$ – AleksandrH Oct 7 '17 at 12:28
  • $\begingroup$ @AleksandrH I have, haven't I? I now used $2n-1$ and $2n+1$ $\endgroup$ – vik1245 Oct 7 '17 at 19:31
  • $\begingroup$ Nope. $2n-1$ and $2n+1$ are two consecutive odd numbers. For example, if $n = 3$, then $2n-1 = 5$ and $2n+1 = 7$. Furthermore, the even number in your case is sandwiched between the two odd numbers (with that example, $6$). What you're trying to prove here holds for any odd numbers and any even numbers. You need to use three different (and therefore independent) variables—say $x$, $y$, and $z$—to cover all possible cases. $\endgroup$ – AleksandrH Oct 7 '17 at 23:19
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You have supposed that $n$ is odd, $n+1$ is even and $n+2$ is again odd ... but at the end you have taken the $n=4$ which is not odd & it is even. So $3n+3=3(n+1)$ can't make sense.

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