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Jonatan
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#21
Report 16 years ago
#21
(Original post by JamesF)
If you dont know about modulo arithmetic then you could do it like this instead, the same, but not as neat.
4 = 3+1
4^n = 3^n + ....(nCr)3^r.......+ 1, so each term apart from the last is a multiple of 3. Take the 1 away and your left with a multiple of 3.
Why not generalise it then:

(n^m - 1 ) mod (n-1) = 0
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zaman002
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#22
Report 15 years ago
#22
(Original post by figgetyfig)
lol, I feel a little stupid looking at all this, because I immediately thought of induction, and not any of the other methods. Anyway, I know this has been done, but just to prove that I can...

assume true for n=k
4^k -1 = 3M (where M is an integer)
therefore 4^k = 3M+1

for n=k+1,
4^(k+1) = 4x4^k
therefore 4^(k+1) -1 = 4(3M+1) -1
= 12M+3
= 3(4M + 1)
which is divisible by 3.

If true for n=k, then true for n=k+1
when n=1, 4^n -1 = 1-1 = 0, which is divisible by 3 (even though its not a multiple)
when n=2, 4^n -1 = 16-1 = 15, which is divisible by 3.
therefore true for all positive integer values of n.
Yeh same here.............induction came straight to my head too, i didn't dare think of the way these other peeps did it!!!
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