If `0 lt x lt 1000 and [x/2]+[x/3]+[x/5]=31/30x`, (where `[.]` denotes the greatest integer function then number of possible values of x.

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If 0ltxlt1000 and [x/2]+[x/3]+[x/5]=31/30x , where [x] is the greatest interger less than or equal to x, the number of possible values of x is

If [x+[2x]] lt3 , where [.] denotes the greatest integer function, then x is

If [log_2 (x/[[x]))]>=0 . where [.] denotes the greatest integer function, then :

If [log_(2)((x)/([x]))]>=0, where [.] denote the greatest integer function,then

Lt_(x to oo) ([x])/(x) (where[.] denotes greatest integer function )=

If 0 lt x lt 1000 and [(x)/(2)]+[(x)/(3)]+[(x)/(5)]=(31)/(30)x, where [x] is the greatest integer less than or equal to x ,the number of possible values of x is

Lt_(xto2) [x] where [*] denotes the greatest integer function is equal to

If f(x)={|1-4x^2|,0lt=x<1 and [x^2-2x],1lt=x<2 where [.] denotes the greatest integer function, then

If 0 lt x lt 1000 and [(x)/(2)] +[(x)/(3)]+[(x)/(5)] =(31)/(30)x , where [x] is the greatest integer less than or equal to x, the number of possible values of x is