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I understand more or less how to do this problem, however, I am having trouble actually showing that n can be divisible by 16.

Here's what I have done so far

If n is an odd integer, then n = 2k + 1, where k is any integer.

2k + 1 * (2k + 3) * (2k + 5) must be an odd number (due to multiply odd numbers)

If n is an even integer, then n = 2k, where k is any integer.

2k * (2k + 2) * (2k + 4)

How do I show that 2k * (2k + 2) * (2k + 4) is a multiple of 16?

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You're doing fine. For your question, note that you can factor out 2 all the way across:

$$ 2k \cdot (2k+2) \cdot (2k+4) = 2\cdot 2 \cdot 2 \cdot \underbrace{k(k+1)(k+2)}_{\text{focus here}}. $$ Now, notice that the final three terms are three consecutive integers, so one of them must be even. So, factor the two out of that one, and get $2^4=16$ out front.

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  • $\begingroup$ This makes sense thankyou. Would I just simply state that because there are 3 consecutive integers, that one must even and hence as an even number would be equal to 2k (or 2l whatever for this example), by factoring out the 2 we get 2^4? $\endgroup$ – Akaion Mar 8 '19 at 3:12
  • $\begingroup$ Yes, you'll have to figure out how to write it up. The terms either go odd-even-odd, in which case $k+1=2m$ or they go even-odd-even in which case $k=2m$. This allows you to move the $2$ correctly in the proof. $\endgroup$ – Randall Mar 8 '19 at 3:13
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Your approach is fine and, as noted by Randall, will lead to a solution. The other way to handle the problem is to note that if $n$ is even, either $n \equiv 2 \pmod{4}$ or $n \equiv 0 \pmod {4}$, and in either case the result will follow.

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For the divisible by 16 case, note that n,n+2,n+4, contains at least 1 multiple of 4, being consecutive even numbers.

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