Question

find the number of possible values of x where 0<x<1000 and
[x/2]+[x/3]+[x/3]=(31/30)x​

Answer 1

Answer:

$\frac{x}{2} + \frac{x}{3} + \frac{x}{3} = ( \frac{31}{30} )x \\ \frac{3x + x + x}{6} = ( \frac{31}{30} )x \\ \frac{5x}{6} = \frac{31x}{30} \\ 150x = 186x \\ = 186 - 150\\ = 36$

hope it helps