Question

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?

Answer 1

9514 1404 393

Answer:

  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.)