4. How many zeros does the product of numbers from 1 to 100 inclusive end with?
4. First method of solution. To find out how many zeros the product of numbers from 1 to 100 ends with, we need to determine how many times the factor 5 will appear. Among the factors of the number \(1 \times 2 \times 3 \times \ldots \times 99 \times 100\), 20 numbers are divisible by 5 \((100: 5=20)\).

Of these, 4 numbers are divisible by 25 \((20: 5=4)\). Therefore, all fives appear as a factor 24 times. Among the factors of the number

\(1 \times 2 \times 3 \times \ldots \times 99 \times 100\) there are 50 even numbers, so the factor 2 will appear 24 times for sure. Hence, the product ends with 24 zeros.

Second method of solution (Anna Sonkina, 3a grade student of School No. 24).

Let's factorize all composite numbers into prime factors. We need to find out how many fives are in the factorization. They are in the numbers 5, \(10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100\). There are 10 multiples of 10; two numbers give two fives when factorized. These are 50 and 100. This results in 12 fives. There are also 10 numbers ending in 5, but two numbers give two fives when factorized. These are 25 and 75. This results in 12 fives. In total, there are 24 fives. There are more than 24 twos in the factorization. Therefore, the product ends with 24 zeros. The other numbers in the product do not contribute any zeros.

Answer: 24 zeros.