Non-routine question

[alchemy]

Need help doing the following question:
Show that 43^n + 83 x 92^(3n-1) is divisible by 7 for all positive integers of n.
Thanks.

[alchemy]

Bump...need help ASAP.

[keltingmeith]

This is weird and I've never seen anything like this before... But, here's a yahoo answers thingo on it using the binomial theorem.

[nerdgasm]

My suggestion would be to write 43 as 6(7)+1, 83 as 12(7)-1, and 92 as 13(7)+1. This is because we are interested in checking divisibility by 7 (hence all the 7s in the expression above).

Then you have, [ 6(7) + 1 ]^n + [ 12(7)-1] [ 13(7) + 1 ]^(3n-1).

Now, by the binomial theorem, it follows that every term in the binomial expansion of [ 6(7) + 1 ]^n is divisible by 7, except for the last one, which is 1^n = 1. In other words, the 'remainder' you get when you divide [ 6(7) + 1 ]^n by 7, is 1.

One of the very cool things about divisibility problems is that generally, numbers with the same 'remainder' behave the same. In other words, 5 behaves in the same way as 19 (both have remainder 4) , and -34 behaves in the same way as 15 (both have remainder 1). If you know some modular arithmetic or other number theory, you're probably aware of this - but I'm avoiding going into the formalism since I think the main thing is to communicate the idea across.

So, we've seen that the remainder of [ 6(7) + 1 ]^n when divided by 7 is 1. So in terms of checking for divisibility by 7, it behaves just like a 1. Similarly,  [12(7)-1] just behaves like a -1, and [ 13(7) + 1 ]^(3n-1) just behaves like 1^(3n-1), which is 1.

So, informally, [ 6(7) + 1 ]^n + [ 12(7)-1] [ 13(7) + 1 ]^(3n-1) behaves just like 1 + (-1)(1), which is of course 0, which is divisible by 7.

More formally, we would write,

 





for all positive integers n. Here, the symbol for the three horizontal lines means "is congruent to", which basically means "has the same remainder as", and the "mod 7" means we are talking about remainders when we divide by 7. So, because the value of the expression always has a remainder of 0 when you divide it by 7, you can conclude that it is divisible by 7.

This sort of stuff tends to crop up more in high-school competition maths - and in university, you might be able to take a unit involving number theory, where you'll learn more about this kind of stuff too.

I realise I haven't explained things very in-depth, but if you would like to discuss this a bit more, feel free to ask!