So many remainders

Suppose that $R_{100} = m$ , where $0 \le m \le 99$ . It is a simple induction (working down from $100$ ) to show that $R_j \; = \; \left\{ \begin{array}{ll} m & j \ge m+1 \\ 0 & j \le m \end{array} \right.$ (so long as we are dividing by a number greater than $m$ , $m$ yields remainder $m$ . Dividing $m$ by $m$ gives remainder $0$ , and dividing $0$ by anything gives remainder $0$ ). Thus $H+1 \; = \; \prod_{j=2}^{100} (1+R_j) \; = \; (m+1)^{100-m}$ which is maximized when $m=23$ . The smallest positive integer value of $n$ for which $R_{100} = 23$ is, of course, $\boxed{23}$ itself.

To see why $m=23$ gives the maximum, the function $f(x) \; = \; \ln\big(x^{101-x}\big) \; = \; (101-x)\ln x \hspace{2cm} x > 0$ has derivative $f'(x) \; =\; \frac{101}{x} - 1 - \ln x$ which is a strictly decreasing function, with $f'(24) > 0 > f'(25)$ . Thus $f(x)$ has a single maximum at a point $\alpha$ , where $24 < \alpha < 25$ , and since $f(24) > f(25)$ we see that $f$ is maximised over the integers at $x=24$ . This gives the answer of $m=23$