Question #c9d66

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Answer 1 · 2017-01-06T17:05:38

$11$ $11$

We need the maximum $n$ $n$ such that

$(n + 1)! - 1 \leq 10^{9}$ $(n+1)!-1 le 10^9$ or

$(n + 1)! \leq 10^{9} + 1$ $(n+1)! le 10^9+1$ . We will try to obtain a reasonable firts approximation for $n$ $n$ .

Applying $log$ $log$ to both sides

$log ((n + 1)!) \leq log (10^{9} + 1)$ $log((n+1)!) le log(10^9+1)$ or

$n \sum k = 0 log (k + 1) \leq log (10^{9} + 1)$ $sum_(k=0)^nlog(k+1) le log(10^9+1)$ or

$\int_{0}^{n} log (ξ + 1) d ξ = (n + 1) log (n + 1) - n \leq log (10^{9} + 1)$ $int_0^n log(xi+1)d xi = (n+1)log(n+1)-n le log(10^9+1)$

Now using an iterative procedure like Newton

$x_{k + 1} = x_{k} - f_{k} / {(\frac{d f}{d x})}_{k}$ $x_(k+1)=x_k-f_k//((df)/(dx))_k$ with

$f (x) = (x + 1) log (x + 1) - x - log (10^{9} + 1)$ $f(x)= (x+1)log(x+1)-x - log(10^9+1)$

we obtain in few iterations the value for $n$ $n$

$n = ⌊ 11.75 ⌋ = 11$ $n = floor11.75 = 11$ so with $n = 11$ $n=11$ we have

$(11 + 1)! - 1 = 479001599 < 10^{9}$ $(11+1)!-1 = 479001599 lt 10^9$ and

$(12 + 1)! - 1 = 6227020799 > 10^{9}$ $(12+1)!-1=6227020799 gt 10^9$