<Question>
How many zeroes are at the end of $42!$ (42 factorial)?  (Reminder: The number $n!$ is the product of the integers from 1 to $n$.  For example, $5!=5\cdot 4\cdot3\cdot2\cdot 1= 120$.)
</Question>

<think>
<Plan_1>
Okay, so I need to figure out how many zeros are at the end of 42 factorial, which is written as 42! . I remember that the number of trailing zeros in a factorial is determined by the number of times the number 10 is a factor in that factorial. Since 10 is 2 multiplied by 5, the number of trailing zeros depends on how many pairs of 2s and 5s are in the prime factorization of 42! . But I also recall that in any factorial, there are usually more 2s than 5s, so the number of 5s will be the limiting factor here. So, essentially, I need to count how many times 5 is a factor in the numbers from 1 to 42.

Let me start by recalling the method for counting the number of times a prime number divides a factorial. I think it's called Legendre's formula. The formula is that for any prime p and integer n, the exponent of p in the prime factorization of n! is given by the sum of floor(n/p^k) for k from 1 to infinity, until p^k exceeds n. So, in this case, p is 5, and n is 42. Therefore, I need to compute floor(42/5) + floor(42/25) + floor(42/125) + ... and so on.

Let me compute each term step by step.
First term: floor(42/5). 42 divided by 5 is 8.4, and taking the floor of that gives 8. So that's 8.

Second term: floor(42/25). 25 is 5 squared. 42 divided by 25 is 1.68, so the floor is 1. That's 1.

Third term: floor(42/125).
125 is 5 cubed. But 125 is 125, which is way larger than 42, so 42 divided by 125 is 0.336, and the floor of that is 0. Since this term is 0, and all subsequent terms will also be 0 because higher powers of 5 will only make the denominator larger.
-----

Wait, but let me double-check this to make sure I haven't missed anything. Let's take a smaller example.

<prompt_1.1> Double-check the method using smaller examples (4! and 10! ) to verify the accuracy of the approach. </prompt_1.1>
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But just to be thorough, let me actually factorize 42! and count the number of 2s and 5s. 

Starting with the number of 5s:

<prompt_1.2> Perform a detailed count of factors of 5 and 2 in 42! and verify the number of trailing zeros. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's say 5! is 120, which has one trailing zero. Using the same method: floor(5/5) = 1, floor(5/25) = 0, so total is 1. That's correct. Another example: 10! . The number of trailing zeros should be 2. Let's compute: floor(10/5) = 2, floor(10/25) = 0, so total 2. Correct. And 25! would have floor(25/5) = 5, floor(25/25) = 1, floor(25/125) = 0, so total 6. Let me check 25! actually has 6 trailing zeros. Yes, I think that's right.
</execution_1.1>

<execution_1.2>
Multiples of 5 between 1 and 42: 5,10,15,20,25,30,35,40. That's 8 numbers. Each contributes at least one 5. Then, among these, 25 is 5 squared, so it contributes an extra 5. So that's 8 +1 =9 fives.

For the 2s:

Multiples of 2 between 1 and 42: 2,4,6,....,42. Each contributes at least one 2. Then multiples of 4 contribute an extra 2, multiples of 8 contribute another, and so on.

But since we know that the number of 2s will be more than the number of 5s, so even if we calculate, it's definitely more than 9. So the limiting factor is the number of 5s, which is 9.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me check 2! which should have 0 trailing zeros.
floor(2/5) = 0, so that's correct. 10! is 3628800, which has 2 trailing zeros.

Wait, another check: 15! . The number of trailing zeros should be floor(15/5) + floor(15/25) = 3 + 0 = 3. But 15!
is 1307674368000, which actually has 3 trailing zeros.
-----

Based on execution_1.2:
Therefore, the number of trailing zeros in 42! is 9.
Therefore, the number of trailing zeros in 42! is 9.

I think that's solid. I can't see any mistakes here. Let me just recap once more to make sure.

<prompt_2.1> Recap and confirm the solution for the number of trailing zeros in 42! . </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
The formula is sum of floor(n/5) + floor(n/25) + floor(n/125) + ... For n=42:

42/5 = 8.4 → 8

42/25 = 1.68 → 1

42/125 = 0.336 → 0

So total 8 +1 =9. Correct.

Yes, that seems right. I don't think I missed any steps here. The key is recognizing that each multiple of 5 contributes at least one 5, and multiples of higher powers of 5 contribute additional 5s. Since 25 is 5 squared, it adds an extra one. All other numbers up to 42 won't have higher powers of 5 (like 125, 625, etc. ), so they don't contribute more than that. Therefore, 8 +1 =9 is the total number of 5s in the prime factors of 42! , and hence 9 trailing zeros.
</execution_2.1>
</Execution_2>

<Final_answer>
Therefore, there are $\boxed{9}$ zeros at the end of $42!$.
</Final_answer>
</think>