sinclairdragon428

Each digit must be a factor of 10000, so the only possible digits are 1, 2, 4, 5, and 8. Furthermore, since the prime factorization of 10000 is there must be exactly four digits of 5. Among the remaining three digits, we also need four factors of 2. We have the following cases.

Case 1: The remaining three digits are 1, 2, and 8.

We want to arrange the digits 5, 5, 5, 5, 1, 2, and 8. There are $\frac{7!}{4!} = 210$ ways to arrange these digits.

Case 2: The remaining three digits are 1, 4, and 4.

We want to arrange the digits 5, 5, 5, 5, 1, 4, and 4. There are $\frac{7!}{4! 2!} = 105$ ways to arrange these digits.

Case 3: The remaining three digits are 2, 2, and 4.

We want to arrange the digits 5, 5, 5, 5, 2, 2, and 4. There are $\frac{7!}{4! 2!} = 105$ ways to arrange these digits.

Thus, there are a total of $210 + 105 + 105 = \boxed{420}$ 7-digit numbers.

thank you for all the help!

hectictar had the correct answer :)

C(12, 3) = (12*11*10)/3! = 220

9 * 3/4 = 27/4=6 and 3/4 of candy bar

i checked and it was correct! thanks for your help melody

thanks melody!

thank you for the help, but i got 60 for the answer.

working out if anyone needs it

for a row, there would be 6! arrangements.

however, we divide this by 6 because some arrangements can be rotataed to form each other.

we also divide this by 2 because of reflectional symmetry.

6!/(6*2)=720/12=60