An algebra problem by j r

$\begin{aligned} X & = \frac 12 \text{ of } \frac 23 \text{ of } \frac 34 \text{ of } \frac 45 \text{ of } \frac 56 \text{ of } \frac 67 \text{ of } \frac 78 \text{ of } \frac 89 \text{ of } \frac 9{10} \text{ of } 1000 \\ & = \frac 12 \times \frac 23 \times \frac 34 \times \frac 45 \times \frac 56 \times \frac 67 \times \frac 78 \times \frac 89 \times \frac 9{10} \times 1000 \\ & = \frac 1{\cancel{2}} \times \frac {\cancel{2}}{\cancel{3}} \times \frac {\cancel{3}}{\cancel{4}} \times \frac {\cancel{4}}{\cancel{5}} \times \frac {\cancel{5}}{\cancel{6}} \times \frac {\cancel{6}}{\cancel{7}} \times \frac {\cancel{7}}{\cancel{8}} \times \frac {\cancel{8}}{\cancel{9}} \times \frac {\cancel{9}}{10} \times 1000 \\ & = \boxed{100} \end{aligned}$

calculate from backwards which is from 1000

u will get 200 times $\frac{1}{2}$ and equals to 100