SOLUTION: I must find, by a method other than induction, that 1+2^(6K+1)is divisible by 3.

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Question 699138: I must find, by a method other than induction, that 1+2^(6K+1)is divisible by 3.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I can prove that even more similarly defined numbers are divisible by 3.
If you (or the teacher do not like a generalization, you can always adapt the proof to just the subset of 1%2B2%5E%286K%2B1%29 numbers.

All numbers of the form 1%2B2%5E%282n%2B1%29 are divisible by 3.
Some of them (but not all) are of the form 1%2B2%5E%286K%2B1%29 , if n=3K,
as for 1%2B2%5E%282%2A3%2B1%29=1%2B2%5E7=1%2B2%5E%286%2A1%2B1%29 (with n=3, or K=1)
1%2B2%5E7=1%2B128=129=3%2A43 .

2%5E2=4=3%2B1 so .
+ ... + %28n%28n-1%29%2F2%29%2A3%5E2%2Bn%2A3%2B1%5En=3[3%5E%28n-1%29%2Bn%2A3%5E%28n-2%29%2B%28n%28n-1%29%2F2%29%2A3%5E%28n-3%29 + ... + %28n%28n-1%29%2F2%29%2A3%2Bn]+1=3M%2B1 , calling the messy bracket M.
The conclusion so far is (in words) the even powers of 2, when divided by 3, have 1 as the remainder.
In formulas: 2%5E%282n%29=3%2AM%2B1 , where Mis a whole number.
2%5E%282n%2B1%29=+2%5E%282n%29%2A2%5E1=2%5E%282n%29%2A2=%283M%2B1%29%2A2=6M%2B2
1%2B2%5E%282n%2B1%29=1%2B6M%2B2=6M%2B3=3%282M%2B1%29
Since 1%2B2%5E%282n%2B1%29 is 3 times the whole number %282M%2B1%29,
1%2B2%5E%282n%2B1%29 is divisible by 3.