Prove (w/o induction) that B^n and 0 are congruent mod B. Hint from professor: let n = m + 1
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Originally Posted by jzellt Prove (w/o induction) that B^n and 0 are congruent mod B. Hint from professor: let n = m + 1 Hint isn't really necessary as long as , which i think you are assuming note that this is obviously the case. So begin by saying where can we go with this?
I guess I should've post this in my post...but I actually got that far. Since B^n = B^(m+1) then B|(BB^m) - 0 or B|BB^m. But, Im not sure what to do from here.
Originally Posted by jzellt I guess I should've post this in my post...but I actually got that far. Since B^n = B^(m+1) then B|(BB^m) - 0 or B|BB^m. But, Im not sure what to do from here. since we have that which means?
You did say without induction this time, didn't you? B is congruent to 0, mod n, if and only B is a multiple of n: say B= na where a is some integer. If B is congrunt to 0 mod n, then B= na. is sufficient. And, of course, 0= 0n.
Last edited by Jhevon; Mar 19th 2009 at 06:37 PM. Reason: fixed LaTeX
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