be $a_1,..a_r = 1$ and $a_{r+1},a_{r+2},...a_{100} \ge 2$ $P_n=\prod_{i=n}^{100}a_i$ und $S_n=\sum_{i=n}^{100}a_i$ $P_1 = P_{r+1}=S_1=r+S_{r+1}$ $P_{r+1}-S_{r+1}=r$ lhs is monotonously increasing in all a_i if r should be a minimimum all a_i have to be 2 hence: $r = P_{r+1}-S_{r+1} \ge 2^{100-r}-2(100-r)$ Trial gives $r \ge 93$ hence there at least 93 1's in the sequence

Actually for $r=93$ that gives $93 < 2^7 - 2\cdot 7$, so we need $r\geq 94$ to have for $r=94$ that $94 > 2^6 - 2\cdot 6$. However, it seems that there is no actual solution (equality) for $r=94$, so we need increase to $r=95$, for which, denoting the $100$ integers by $1=a_1\leq a_2 \leq \cdots \leq a_{100}$, there exists the solution $a_k=1$ for $1\leq k \leq 95$ and $(a_{96},a_{97},a_{98}, a_{99},a_{100}) = (2,2,3,3,3)$