if $\frac{m}{n},\frac{m+1}{n+2},\frac{m+2}{n+4},\frac{m+3}{n+6}$be positive integers, find the minimum of the m let $m$ be a positive integer,and $n$ be an  positive integers,such $$\dfrac{m}{n},\dfrac{m+1}{n+2},\dfrac{m+2}{n+4},\dfrac{m+3}{n+6}$$be positive integers,find the minimum of the $m$.
It seem use CRT to solve it? But I can't find it so far

If $\frac{m}{n}, \frac{m+1}{n+2}, \frac{m+2}{n+4}, \frac{m+3}{n+6}$ are all integers, so are $\frac{2m}{n}, \frac{2m+2}{n+2}, \frac{2m+4}{n+4}, \frac{2m+6}{n+6}$, and so are $\frac{2m-n}{n}, \frac{2m-n}{n+2}, \frac{2m-n}{n+4}, \frac{2m-n}{n+6}$. Hence $2m-n$ is a multiple of $\mbox{lcm}(n, n+2, n+4, n+6)$.
By Sridhar's comment, $n$ has to be odd, and by looking at the least common multiples of small $n$'s, the minimal l.c.m. is at $n=1$, hence $2m-1 \geq 105$, and $m\geq 53$. From Oscar's post we see that actually $m=53$ is the minimal solution!
