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Does a number start with 9?

Posted on: March 30th, 2013 by
4

You know that 9^{100000000000} is a 95424250944 digit number. Consider the set \{9^1,9^2,9^3,\dots 9^{100000000000}\}. How many of these numbers start with a 9? (Example: 91 starts with a 9, but 729 does not start with a 9). (Obviously you are not allowed to use a computer. That's why the number is bloated up by so much that even a computer can't feasibly solve it).

Source: One of the practice problems from the Indian National Mathematics Olympiad training programme.


4 Responses to Does a number start with 9?

  1. Bharatram had this to say about that:

    I have a vague solution which you could refine for me:
    I reduced the problem to solving for positive integers n such that the fractional part of nlog9 is at least log9.
    Now in the additive group [0,1) with addition modulo 1 (i.e.floor of addition), consider the subgroup generated by log9. We would like to show that the elements of this subgroup which are at least log9 is a 1-log9 fraction.
    Here is where I go into the discrete case: Let 0<c<1 be such that c=a/b in lowest form. Then in the group modulo b, c can be identified with a, and as gcd(a,b)=1, the subgroup generated by a is the whole group modulo b. In which case the elements "at least a" in the group are a 1-a/b or a 1-c fraction. We can take limits in the case of c being irrational (can we?)
    Thus the number of powers of 9 with 9 as the first digit is 100000000000(1-log9) which is
    4575749057.
    Is that okay? I'd love to hear a simple way of saying this without resorting to groups(though I guess modulo arithmetic, which is the same thing, can be used without the group-theoretic terminology)

    • Gowtham had this to say about that:

      Whoa! A big solution! You are pretty close! The exact answer is 100000000000-95424250944+1=4575749057. The simple solution I have:
      ...
      ...
      ...
      ...
      ...
      Suppose 9^n starts with a 9. Then 9^{n-1}={9^n\over 9} has the same number of digits as 9^n. Otherwise, 9^{n-1} has 1 digit lesser. Since the number of digits has to reduce from 95424250944 to 1, you have the answer.

      I like your solution though because it helps us estimate the number of elements that start with an 8 for example. Well, I am not sure if the answer will be exact. So the exact answer for this question is 100000000000-\lfloor 100000000000\log 9\rfloor

    • Gowtham had this to say about that:

      By the way, your solution made me realize that this puzzle is associated to Bemford's law: http://en.wikipedia.org/wiki/Benford's_law

      • Bharatram had this to say about that:

        Close? Dude, its the same answer!
        Also, the answer is actually shorter, except for my interpretation of the distribution of n*c(mod 1) in [0,1) for 0<ci>0 is a log(i+1)-log(i) fraction. Holy shit Benford's law! This log(i+1)-log(i) fraction just made me realize it!


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