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I have figured out that:

Given $N = 10^K$, $D = K + 1$.

What is the formula for a natural number $D = {???}(N)$

How to solve this in Mathematica/WolframAlpha ?

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3 Answers 3

You have it (presuming that you see 000 as having 2 leading zeros and make a special case for it-the log won't work well). Just round Z up to the next $10^k-1$, or use the ceiling function: $\lceil log_{10}(N+1)\rceil$

With the change to the question, see the Hristo's comment below

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I'm sorry, I figured out that the question in it's original form is incorrect, so I changed it. Thanks for the tip! –  Hristo Hristov Feb 9 '11 at 14:19
2  
Maybe it is Floor[Log10[N]] + 1 ? –  Hristo Hristov Feb 9 '11 at 14:27
    
@Hristo: I agree –  Ross Millikan Feb 9 '11 at 14:44
up vote 1 down vote accepted

I found out that the answer is

D = Floor[Log10[N]] + 1

http://www.wolframalpha.com/input/?i=Floor[Log10[2011]]+%2B+1

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IntegerLength[n]

gives the number of digits in the base 10 representation of the integer n.

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