The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 5 and 7 such that each digit is used at least once, is :

We need to form seven-digit numbers using the digits 1, 2, 3, 5, and 7, with each digit used at least once. Since we have 5 distinct digits and 7 positions, exactly 2 extra positions need to be filled. This leads to two cases.

Case 1: One digit appears 3 times and the remaining four digits appear once each.

We choose which digit repeats 3 times: $${}^{5}C_{1} = 5$$ ways. The number of arrangements of 7 digits where one appears 3 times is $$\frac{7!}{3!} = 840$$.

Total for Case 1: $$5 \times 840 = 4200$$.

Case 2: Two digits each appear 2 times and the remaining three digits appear once each.

We choose which 2 digits repeat: $${}^{5}C_{2} = 10$$ ways. The number of arrangements is $$\frac{7!}{2! \times 2!} = 1260$$.

Total for Case 2: $$10 \times 1260 = 12600$$.

The total number of seven-digit numbers is $$4200 + 12600 = 16800$$.

Hence, the correct answer is Option 3.