Math problem

[Quang Le Minh <n1465236@student.fit.qut.edu.au>]

Hi there !

I am new here, so you guys might not know me. By the way, I saw a message
about  a math problem "Find the smallest number N.....". I have a solution
here for you guy to consider:

Notes: The solution is very brief, so if you want to know more details,
I'll send you a complete solution.

* Suppose N(k) is the number of 7 btw 1 and 7^(k+1) (not include 7^(k+1))
then we have:
             		N(0) = 0, N(1) = 6
			N(k) = 7*N(k-1) + 6*k	(1)
(details of the proof is not shown here)

* From (1) we get N(2) = 54 (number of 7s btw 1 and 7^3). So we need 45 7s
more. But N(3) = 54*7 + 18 = 369 >> 100 we have to split the interval btw
7^3 and 7^4 then.

* From 7^3 to 2*7^3 there are 6 intervals of equal length of 7^2
numbers each of interval contain N(1) 7s (Not include the end limit of the
intervals). Thus from 1 to 2*7^3 there are N(2) + 6*N(1) + 3 +3 = 96
number 7s. (the first '3' comes from 7^3, and the other one comes from
2*7^3)

* The numbers 2*7^3 + 7, 2*7^3 + 14, 2*7^3+21, contains only one 7.
Therefore we have 100 7s from 1 to 2*7^3+21 (=707)


====> So the solution is N = 707 <======

My value might is not correct, because I have not got enough time to work
it out carefully. But I think my algorithm is right. I have to go to the
lecture now. See you guys later !