Determine the value of a smallest positive integer N greater than 1 such that the base ten expansion of N^1/N contains precisely six zeroes immediately following the decimal point. How about seven zeroes?

Extra Challenge: A non computer program solution.

In general, the mth qualifying number (counting 38 as the zeroeth such number) of the form  (n-1)^(1/(n-1)) has the form (as string):

"1.(m zeroes)1(m+2 zeroes)1"; since the last part is so small it can be considered as 1

1.1001, 1.010001, 1.00100001, 1.0001000001, etc.

This is the form 1+1/(10^(m+1))+1/(10^(2m+4)) 

So the function (1+1/(10^(m+1))+1/(10^(2m+4)))^(N-1)+1 = N returns n as the integer part of N.

When m=5, N=16626518.569717461698..., so n = 16626518.

When m=6, N=190660035.08011929935..., so n = 190660035.

When m=7, N= 2148818406.4441059949..., so n= 2148818406.

And so on.

 

Edited on October 2, 2014, 11:07 am

Posted by broll on 2014-10-02 10:59:28