We need to find $k_t$ such that $v_3((2k_tm)!)\ge (k_t-1)m$ suppose $k_t=\frac{3^t+r_t}{m}$ , $r_t {0,1,2,...,m-1}$ $3^t\equiv -r_t \pmod m$ $v_3((2k_tm)!)=\sum_i{[\frac{2km}{3^i}]}\ge \sum_{i=1}^{t}{2*3^{t-i}}=3^t-1 \ge 3^t-(m-r_t)=(k_t-1)m$ suppose $k_t'=k_t-1$ , $2(k_t-1)m=2(3^t+r_t)-2m=2*3^t+2(r_t-m) \le 2*3^t-2$ $v_3((2(k_t-1)m)!)\le (k-2)m$ for the same reason. So we are done.