Final answer:
To determine the number of trailing zeroes in 1000!, one must calculate the number of times 1000 can be divided by 5 repeatedly until the quotient is less than 5. Adding up these quotients gives the total number of zeroes, which is 249 for 1000!.
Step-by-step explanation:
The number of zeroes at the end of the factorial of a number, like 1000! (1000 factorial), is determined by the number of times the number can be divided by 10, which in turn is determined by the number of times it can be divided by 5 and 2, since 10 = 5 × 2. However, since every second integer is divisible by 2, there are always more factors of 2 than 5, so the number of zeroes is dictated by the number of 5's present in the numbers that make up the factorial.
To find the number of times 1000 can be divided by 5, we divide 1000 by 5 to count the number of factors of 5, then divide the result by 5 to count the number of factors of 5² (since each factor of 5² contributes an additional factor of 5), and so on until the quotient becomes zero. This is represented by:
8 ÷ 5 = 1.6, which we consider as 1 since we are counting whole numbers.
Summing up these results (200 + 40 + 8 + 1), we get 249. Therefore, 1000! has 249 zeroes at the end.