Arbitrary digit of an exponential number

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SUMMARY

This discussion focuses on finding an arbitrary digit of the exponential number 23^234, which contains 219 digits. The method involves understanding the repetition of the last N digits in the sequence of powers of 23, where the length between repeats is determined by the formula 4*5^(N-1). To find the 187th digit, one must generate a list of 4*5^(219-187-1) numbers and apply modulo arithmetic to locate the desired digit. The conversation emphasizes the use of number theory rather than complex mathematics.

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cap.r
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I am looking to find an arbitrary digit of let's say 23^234. This one has 219 digits. so let's 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.
 
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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)
 
interesting... How do I get the formula for finding repeating length in an arbitrary exponential X^N.
 

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