Okay, so I need to figure out how many zeros are at the end of 42 factorial, which is 42!. Hmm, I remember that the number of trailing zeros in a factorial is related to the number of times the number 10 is a factor in that factorial. Since 10 is 2 multiplied by 5, each trailing zero comes from a pair of 2 and 5 in the prime factors of the factorial. But wait, I think the number of 2s is usually more than the number of 5s in a factorial, right? So maybe the number of trailing zeros is determined by the number of 5s in the prime factorization of 42!.

Let me confirm that. If there are more 2s than 5s, then the number of 10s (which contribute to trailing zeros) is limited by the number of 5s. That makes sense because every even number contributes a 2, and there are more even numbers than multiples of 5. So the key is to count how many times 5 is a factor in the numbers from 1 to 42.

How do I count the number of times 5 is a factor? Well, I think you divide 42 by 5 and take the floor of that, which gives the number of multiples of 5. Then, you divide 42 by 25 (since 25 is 5 squared) and take the floor again, because each multiple of 25 contributes an extra 5. Then you divide by 125, but since 125 is greater than 42, that would be zero. So adding those up gives the total number of 5s in the prime factors of 42!.

Let me write that down step by step.

First, divide 42 by 5. 42 ÷ 5 is 8.4. Taking the floor of that gives 8. That means there are 8 multiples of 5 between 1 and 42.

Next, divide 42 by 25. 25 is 5 squared. 42 ÷ 25 is 1.68. Taking the floor gives 1. So there's 1 multiple of 25 in this range.

Then, divide 42 by 125. 125 is 5 cubed, which is 125. 42 ÷ 125 is 0.336. Taking the floor gives 0. Since 125 is larger than 42, there are no multiples of 125 in this range.

Adding those up: 8 + 1 + 0 = 9. So there are 9 factors of 5 in 42!.

Since there are more than 9 factors of 2, the number of trailing zeros should be 9. Let me verify this with an example. For example, 10! is 3628800, which has 2 trailing zeros. Let's check using the same method. 10 ÷ 5 = 2, 10 ÷ 25 = 0.4, so total 2. So 2 trailing zeros, which is correct. Similarly, 15! should have 3 trailing zeros. 15 ÷ 5 = 3, 15 ÷ 25 = 0.6, so total 3. Correct again.

Therefore, applying this method to 42!, the number of trailing zeros should be 8 (from multiples of 5) + 1 (from multiples of 25) = 9.

Wait, but let me check if there are any higher powers of 5 that I missed. 5^3 is 125, which is larger than 42, so no. 5^4 is 625, which is way larger. So no need to consider higher powers.

Therefore, the answer should be 9 trailing zeros in 42!.

But just to be thorough, let me list out the multiples of 5 up to 42:

5, 10, 15, 20, 25, 30, 35, 40. That's 8 numbers. Then, among these, the multiples of 25 contribute an extra 5. The multiples of 25 up to 42 are only 25. So that's 1 extra. So 8 + 1 = 9. Yep, that's correct.

So I think that's the answer. 9 trailing zeros.
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