Tricky Factorial

Let $1000! = A*14^k$ , in which A is not divided by 14. Then k will be the number of trailing zeros of 1000! in base 14.

In the expansion of 1000! there are $\lfloor \frac{1000}{7} \rfloor=142$ numbers divide by $7$ , $\lfloor\frac{1000}{49}\rfloor=20$ number divide by $7^2$ and $\lfloor\frac{1000}{343}\rfloor=2$ number divide by $7^3$ , so the factoring of 1000! will have $7^{162}$ . Also we could easily see that $2^{162}|1000!$ , so we have k = $\boxed{162}$