Solution: There are just four zeros at the end of the product of six consecutive natural numbers, so there must be four factors 5 and four factors 2 and 5 in these six numbers, and there are two combinations of 3+ 1 and 4+0;

(1)3+ 1, there must be a multiple of 125.

120~ 125,

125~ 130

245~250,250~255

370~375,375~380

495~500,500~505

745~750,750~755

870~875,875~880

(2) When 4+0, there must be 625.

62 1 ~ 626, 622 ~ 627, 623 ~ 628, 624 ~ 629, 625 ~ 630 (note that this answer contains five 5s, but only four 2s).

To sum up, * * has 12+5 = 17 options.

So the answer is: 17.