Daniel is multiplying consecutive counting numbers together, beginning with 1. He finds that 1 x 2 x 3 x 4 x 5 = 120 and that1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 3628800. He continues to multiply consecutive counting numbers together until the resulting product has a string of 20 zeroes at the end of the number, at which point he stops. How many counting numbers does Daniel multiply together to find his final product?

9514 1404 393

  85

Step-by-step explanation:

A number will end in 0 if it has a factor of 10. For a number to end in 20 zeros, it must have 20 factors of 10. That means 2^20 and 5^20 must both be factors of the number.

In any group of sequential numbers, the number of them divisible by 2 will exceed the number divisible by 5, so we only need to have 20 factors of 5 in the group.

Every 5th multiple of 5 will have an additional multiple of 5, so 3 groups of 5 multiples of 5 will have 18 factors of 5. That is, the sequential numbers 1 .. 75 will have a total of 18 factors of 5. For two more factors of 5, we need to have two more multiples of 5: 80 and 85.

The sequential numbers 1 .. 85 will have 20 factors of 5 and at least 20 factors of 2. Hence 85! will end in 20 zeros.

Daniel must multiply together 85 sequential counting numbers to have a product ending in 20 zeros. (It will have a total of 129 digits.)