Given the decimal integers N and B. How can the number of digits in N! in base B be calculated without calculating the value of N!?  Example: If N = 5 and B = 2, then N! = 5! = 120. Representing the value in base B, we get 1111000 which has 7 digits.  But I'm trying to get the result without calculating the value of N!.
I am interested in good integer approximation from below and from above for binary Log(N!).  and  provides only a general idea but not exact values. In other words I need integers A and B so that  A &lt;= Log(N!) &lt;= B
This was a question my friend found in a book.  A block attached to a spring pulled by a constant horizontal force is kept on a smooth horizontal surface. Initially the spring is in its natural state. Then the maximum work that the applied force F can do is:  (a) $F^2/k$ (b) $2F^2/k$ (c) $4F^2/k$ (d) $F^2/2k$  My friend thought that the answer is (a) since the maximum displacement is $x=F/k$ and $W=Fx=F^2/k$. But he saw that the book's answer is different and he asked this question to me. I at first thought that the answer will be (a), but then I realised that this situation is somewhat similar to a charging capacitor. The final energy of the spring is actually $\frac12kx^2=F^2/2k$. So probably the answer is (d)? But the work done is still (a)? But the book says neither. It's (b). 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