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Salty_Titanium wrote: Since we have $\sum x_i \geq n$, we have $\sum x_i = pn$ where $p \mid 100$ Sadly, this is not true. Let $(x_1, x_2, x_3) = (2, 3, 5)$. Then $\sum x_i = 10$, but $3 \nmid 10$.

If $n\ge 1$, then all $x_i\ge 1$. Number of members $i$, suth that $x_i>1$ less than 7 ($2^7=128>100$). Therefore we can take $x_7=x_8=...=x_n=1$ and solve $$x_1x_2x_3x_4x_5x_6(x_1+x_2+x_3+x_4+x_5+x_6+n-6)=100n.$$One solution is $x_1=50, n=49,x_2=x_3=...=x_{49}=1.$ Let $x_1\ge x_2\ge ...\ge x_k>1, x_{k+1}=...=x_n=1, k\le 6$. Let $P=\prod_{i=1}^k x_i, S=\sum_{i=1}^k x_i$. Then $P(S-k+n)=100n$ or $$n=\frac{P(S-k)}{100-P}.$$Maximum value of n, when $P=99, k=1$, then $S-k=98, n=99*98.$

Let $x_{k+1}=...=x_n=1, x_i>1 \ \forall \ 1\leq i \leq k$. Note that $x_1=x_2=...=x_n=1$ is impossible (else $100n=n$) So $\sum_{i=1}^{n} x_i > n \Rightarrow \prod_{i=1}^{n} x_i \leq 99$ and $k \geq 1$ Now note that $\prod_{i=1}^{k} x_i \geq \sum_{i=1}^{k} x_i$ Hence $100n = \prod_{i=1}^{k} x_i \cdot (\sum_{i=1}^{k} x_i + n-k) \leq \prod_{i=1}^{k} x_i \cdot (\prod_{i=1}^{k} x_i + n-k) \leq 99 \cdot (99+n-1)$ So $n \leq 99 \cdot 98,$ and for this value of $n$ take $x_1=99, x_2=...=x_n=1$

Let $x_1 \ge x_2 \ge \cdots \ge x_k >1 $ and $x_{k+1} = \cdots =x_n = 1$. Computation gives $n=\frac{x_1x_2 \cdots x_k (x_1 + \cdots x_k - k)}{100-x_1x_2 \cdots x_k}$. Fix the value of $S=x_1x_2 \cdots x_k$. It is easy to show, since the only remaining term is $x_1+ \cdots x_k - k$, that $n$ is the largest when $k=1$ and $x_1=S$. Therefore, $n=\frac{S(S-1)}{100-S}$, which is largest when $S=99$, so $n=99 \cdot 98$.

To rkm0959: why don't you use Bernoulli's Inequality? it's easy to show that x1+x2+...+xk-k=(x1-1)+(x2-1)+...+(xk-1)<=x1x2...xn-1<=98 so n<=99*98/(100-98)=9702