Exponent Math PuzzleDate: 07/09/2006 at 03:23:35 From: Gaurav Subject: Mathematics Puzzle - with Power and Remainders What would be the remainder when 3^4^5^6^7 ... so on til infinity is divided by 17? I'm not sure how to even start the question. Will power cycles and the remainder theorm work in this? Date: 07/09/2006 at 05:01:23 From: Doctor Ricky Subject: Re: Mathematics Puzzle - with Power and Remainders Hey Gaurav, Thanks for writing Dr. Math! The easiest way to attack this problem is to understand what we are dealing with. We have the number: 3^4^5^6^... Obviously, to find out what the remainder is when divided by 17, we need to know what the exponent will be. For us to find THAT out, we need to find out what the exponent of THAT will be. We can actually stop there because we can make some simple deductions that will provide us with our answer. First, let us figure out the exponent of the exponent, i.e. 5^6^7^... We notice something immediately: 5^n for any number 'n' ends in a 5. That means that 5^n must be odd for any number 'n'. Now let's deal with the original exponent: 4^5^6^7^... We notice that since 5^n is odd and a power of 5, we can write this as: 5^n = 2k+1, for some integer k>=0. This means we are looking at: 4^(2k+1) Let's look at some examples to see what this means: 4^1 = 4 4^3 = 64 4^5 = 1024 ... We see that 4 to an odd power ends in a 4 and will obviously be divisible by 4, which is even. Numbers like this are 4, 64, etc... Now we use this to look at our original question, 3^4^5^6^7^... Hopefully this has helped you enough that you can solve this problem, but if you have any more questions, please let me know! - Doctor Ricky, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/