35 61. Let $x=[x]+\alpha, 0 \leqslant \alpha<1$. When $0 \leqslant \alpha<\frac{1}{4}$, $f(x)=10[x]$; when $\frac{1}{4} \leqslant$ $\alpha<\frac{1}{3}$, $f(x)=10[x]+1$; when $\frac{1}{3} \leqslant \alpha<\frac{1}{2}$, $f(x)=10[x]+2$; when $\frac{1}{2} \leqslant$ $\alpha<\frac{2}{3}$, $f(x)=10[x]+4$; when $\frac{2}{3} \leqslant \alpha<\frac{3}{4}$, $f(x)=10[x]+5$; when $\frac{3}{4} \leqslant$ $\alpha<1$, $f(x)=10[x]+6$. Therefore, when $0 \leqslant x \leqslant 10$, if $[x]=0,1,2, \cdots, 9$, it is known from the above that $f(x)$ can take $6 \times 10=60$ different values. If $[x]=10$, then $f(x)$ only takes 1 value, making a total of 61 different values.
35 Let $[x]$ denote the greatest integer not exceeding $x$. Then, when $0 \leqslant x \leqslant 10$, the number of all different integers represented by $f(x)=[x]+[2 x]+[3 x]+[4 x]$ is $\qquad$