For how many positive integers n does n! end in exactly 5 0s?
To determine the number of positive integers n for which n! ends in exactly 5 zeros, we need to analyze the numbers that contribute to these zeros.
To have a trailing zero at the end of n!, we need to have a factor of 10. As 10 equals 2 × 5, we need to count the number of 2's and 5's present in n!.
Since even numbers contribute factors of 2, we focus on counting the number of factors of 5. This is because the number of factors of 5 will be lower than the number of factors of 2 in the factorization of n!.
Now, let's see how many multiples of 5 are present in the numbers from 1 to n. There will be n/5 multiples of 5, n/5² multiples of 5², n/5³ multiples of 5³, and so on, until n/5^k becomes less than 1. The sum of these multiples will give us the total number of factors of 5.
Therefore, the number of positive integers n for which n! ends in exactly 5 zeros can be determined by finding the largest power of 5 that divides n!.
Let's calculate it step by step:
1. Find the largest power of 5 that divides n!
To find the largest power of 5, we divide n by increasing powers of 5 until the power becomes larger than n.
For example, for n = 100, we have:
- Largest power of 5 ≤ 100: 5^1 = 5 (5 divides 100)
- Largest power of 5 ≤ 100: 5^2 = 25 (25 divides 100)
- Largest power of 5 ≤ 100: 5^3 = 125 (125 is larger than 100, so we stop here)
2. Count the number of multiples of 5 in the factorization of n!
The number of multiples of 5 in the factorization of n! is given by:
- n/5 + n/5² + n/5³ + ... until the term becomes less than 1
Finally, the number of positive integers n for which n! ends in exactly 5 zeros will be the count obtained from step 2.
Computing this count will give you the answer to the original question.