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You are not logged in. #1 20071209 17:16:30
Proofs required!I urgently require two proofs. Character is who you are when no one is looking. #2 20071209 21:02:44
Re: Proofs required!2^n follows a pattern in the last digit of (for natural numbers): (starting at 2^1) 2,4,8,6 Last edited by lucadeltodesco (20071209 21:07:32) The Beginning Of All Things To End. The End Of All Things To Come. #3 20071209 22:17:17
Re: Proofs required!@Ganesh: Why do you need m*n? Just one variable will do! n, where n is a nonnegative integer. etc. Last edited by JaneFairfax (20071209 22:21:12) #4 20071210 01:05:54
Re: Proofs required!m is a constant that determines how many last digits there are (there are m+1), and n is a variable. If you wanted to get even more general, you could even put a "+c" after each n in that array, but that's a simple corollary of ganesh's proposition anyway, so we don't need to prove that bit. Edit: This topic is related at one point, although the subject keeps changing. Edit2: Just realised I should clarify, adding the "+c" will mean that the numbers won't necessarily have the last digits quoted above, but they will be the same for any n. Why did the vector cross the road? It wanted to be normal. #5 20071210 16:56:14
Re: Proofs required!Thanks lucadeltodesco, JaneFairfax and mathsyperson! Character is who you are when no one is looking. 