Question #c9d66

1 Answer
Jan 6, 2017

Answer:

#11#

Explanation:

We need the maximum #n# such that

#(n+1)!-1 le 10^9# or

#(n+1)! le 10^9+1#. We will try to obtain a reasonable firts approximation for #n#.

Applying #log# to both sides

#log((n+1)!) le log(10^9+1)# or

#sum_(k=0)^nlog(k+1) le log(10^9+1)# or

#int_0^n log(xi+1)d xi = (n+1)log(n+1)-n le log(10^9+1)#

Now using an iterative procedure like Newton

#x_(k+1)=x_k-f_k//((df)/(dx))_k# with

#f(x)= (x+1)log(x+1)-x - log(10^9+1)#

we obtain in few iterations the value for #n#

#n = floor11.75 = 11# so with #n=11# we have

#(11+1)!-1 = 479001599 lt 10^9# and

#(12+1)!-1=6227020799 gt 10^9#