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  MTGAP
  Daedalian Member
  
   Posted: Sun Oct 12, 2008 11:19 pm    Post subject: 1 I was thinking the other day, "What is the last digit of [url="http://everything2.com/e2node/Graham%2527s%2520Number"]Graham's number?[/url]" In base three, it would obviously be zero. But it's trickier in base ten. It turns out that 3^3^...^3^3 any number of times always ends in a seven. I also noticed that each digit has a cycle. 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729. The ones digit repeats. It goes 1, 3, 9, 7. You can see why. 1x3 is 3, 3x3 is 9, 9x3 ends in 7, and 7x3 ends in 1. This type of pattern works for every place value. The tens place has a 40-digit cycle, the hundreds place has a 100-digit cycle, and the thousands place has a 500-digit cycle. I was unable to calculate any cycle above that. So what I'm wondering is, is there a predictable pattern to the increasing values of these cycles? Here's a number that's simpler than three: 5. The ones digit is always 5, the tens is always 2. The hundreds cycles through 1 and 6. The thousands cycles through 3, 5, 8, and 0. The cycles get much longer after that. So what's the pattern?_________________This statement is false.
  ralphmerridew
  Daedalian Member
  
   Posted: Mon Oct 13, 2008 12:10 am    Post subject: 2 First, the tens place has a 20-digit cycle. So for the special case of powers of 3, modulo 10^i, the cycle length is a simple GP. More generally, define phi(n) as the number of numbers 1 <= i <= n such that gcf(i,n) = 1. If p_1, p_2, ... p_j are distinct primes, and n = p_1^e_1 * p_2^e_2 * ... * p_j^e_j, then phi(n) can be calculated as phi(n) = (p_1-1)*p_1^(e_1-1) * (p_2-1)*p_2^(e_2-1) * ... * (p_j-1)*p_j^(e_j-1) = n * (1-1/p_1) * (1-1/p_2) * ... * (1-1/p_j). IIRC, the period of any base mod n will be a factor of phi(n)
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