Solution correction

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

Hi there !

My value obviously is not right (it must be less than 700 right?)
So I correct it as follow:

* Suppose N(k) is the number of 7 btw 1 and 7^(k+1)-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 7^3-1). So we need 45
7s more. 

* From 7^3+1 to 2*7^3-1 there are also N(2)  7s (not include
2*7^3). Thus we already have 54*2  = 108  (7s) include 7^3 which is 111
(7s). From (7^3+5*7^2)+1 to (7^3+6*7^3)-1 there are N(1) number 7.
      From (7^3+6*7^2)+1 to 2*7^3-1 there are N(1) number 7.
Thus,
      From (7^3+5*7^2)+1 to 2*7^3-1 there are 2*N(1) + 2 = 14 (7s)
                                 (number '2' comes from 7^3+ 6*7^2)

We only need throw away 11 7s, so we keep three numbers:
    7^3+5*7^2 + 7  
    7^3+5*7^2 + 14
    7^3+5*7^2 + 21 (609)

========> The solution is 609 <==============

Hope this time it is correct. 
Quang.