Arbitrary digit of an exponential number

  • Thread starter cap.r
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  • #1
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Main Question or Discussion Point

I am looking to find an arbitrary digit of lets say 23^234. This one has 219 digits. so lets find the 187th digit....

This isn't homework or anything, I just think it would be interesting and I couldn't find it on google. I would rather use simple number theory that can be taught to a college student. but if this requires more math, go ahead and use whatever you need.
 

Answers and Replies

  • #2
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There's no easy way to do it. However, note that the last N digits in the sequence 23, 23^2, 23^3, 23^4, ... must eventually repeat. For 23, the length between repeats is 4*5^(N-1). So all you need to do to find the 187th digit in 23^234 is to print out that list of 4*5^(219-187-1) numbers and use modulo arithmetic to figure out where in that list your desired digit is. ;o)
 
  • #3
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interesting... How do I get the formula for finding repeating length in an arbitrary exponential X^N.
 

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