$1+3+5+...+x=1600$
⇒$(x+1).[(\frac{x-1}{2}+1):2]=1600$
⇒$(x+1).\frac{x-1+2}{2}.$ $\frac{1}{2}=1600$
⇒$\frac{(x+1)(x+1)}{4}=1600$
⇒$(x+1)^2=6400$
⇒\(\left[ \begin{array}{l}x+1=80\\x+1=-80\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=79(tm)\\x=-81(loại)\end{array} \right.\)
Vậy $x=79$