A person starts multiplying consecutive positive integers from 120. How many numbers should he multiply before he will have a result of 4 zeroes?*1 point610115

The number of zeroes at the end of a product is determined by the number of times 10 is a factor in that product. Since 10 is the product of 2 and 5, we need to count the number of times 2 and 5 appear as factors.

In the sequence of numbers, 2 appears more frequently as a factor than 5. Therefore, the number of zeroes at the end of the product is determined by the number of times 5 appears as a factor.

Starting from 120, the numbers that have 5 as a factor are 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200.

Among these, 125, 150, 175, and 200 have an extra 5 as a factor because they are multiples of 25 (5^2).

So, the total number of times 5 appears as a factor is 17 (from the numbers that are multiples of 5) + 4 (from the numbers that are multiples of 25) = 21.

Therefore, the person needs to multiply 21 numbers starting from 120 to get a product with 4 zeroes at the end.