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 Question from PT, a parent: Given that x is a non-zero integer, how do you show that for all values of x, (x3 + 11x) is divisible by 6? I know it works but how do I answer the "all values of x" part? Thanks in advance!

The standard way to answer questions like this is to use "modular arithmetic". The idea is that every integer is equal to
6n+r where r=0,1,2,3,4, or 5; and that addition, subtraction, and multiplication work on these remainders in the obvious way. For instance, if a = 6n+3 and b = 6m+2, a+b = 6p + 5. If the remainder is too big just take away sixes till it fits: (6n + 5) + (6m + 4) = 6p + 3 (since 9 = 6+3).
Because 6 is not prime, division does not work! It does for prime "bases".

There are shorter notations that we can use here. We can say (eg) 26 is "equivalent to 2 modulo 6", or we can say it belongs to the equivalence class [2].

Now work case by case. For instance,

[2]3 = [2][4] = [2] (the class containing 8)
11*[2] = [5][2] = [4] (the class containing 10)
[2]3 + 11*[2] = [0] (the class containing 6, and all
the numbers divisible by 6.

Do the same for [0],[1],[3],[4],[5] and you're done.

It's hardly worth doing here, but for numbers that aren't prime powers you can factor and check each individually: in other words, to show x3 + 11x is divisible by 6, show it's divisible by 2 and by 3.

Good Hunting!
RD

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