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Answer» x=[x]+{x} [x]=x-{x} `[x/2]+[x/3]+[x/5]=(31x)/30` `x/2-{x/2}+x/3-{x/3}+x/5-{x/5}=(31x)/30` `x/2+x/3+x/5-[(x/2)+(x/3)+(x/5)]=(31x)/30` `(3x)/30=t` `t=0` x is multiple of 2,3,5 LCM of(2,3,5)=30 x=30m `0<30m<1000` n=1to33 33 possibles values of x. |
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