If n is a natural and $1\le \mathrm{n}\le 100$ , then the number of solutions of $\left[\frac{\mathrm{n}}{2}\right]+\left[\frac{\mathrm{n}}{3}\right]+\left[\frac{\mathrm{n}}{5}\right]=\frac{\mathrm{n}}{2}+\frac{\mathrm{n}}{3}+\frac{\mathrm{n}}{5}$ (where [.] denotes the greatest integer function)

From the given equation, we have $$\begin{gathered} \left(\frac{\mathrm{n}}{2}-\left[\frac{\mathrm{n}}{2}\right]\right)+\left(\frac{\mathrm{n}}{3}-\left[\frac{\mathrm{n}}{3}\right]\right)+\left(\frac{\mathrm{n}}{5}-\left[\frac{\mathrm{n}}{5}\right]\right)=0 \\ \Rightarrow \left\{\frac{\mathrm{n}}{2}\right\}+\left\{\frac{\mathrm{n}}{3}\right\}+\left\{\frac{\mathrm{n}}{5}\right\}=0 \end{gathered}$$

where {.} is a F.P.F. But each of the fraction part functions is positive and their sum is zero. Hence each of the fraction part function is zero. Consequently, each of $\frac{\mathrm{n}}{2},\frac{\mathrm{n}}{3},\frac{\mathrm{n}}{5}$ is an integer. The l.c.m. of 2,3,5 is 30. Therefore we can take n = 30k where k is an integer. Hence the number of solutions such that $1\le \mathrm{n}\le 100$ is = 3 (viz $\mathrm{n}=30,60\mathrm{and} 90)$