If p is the product of the first n consecutive positive integers, and the last six digits of p are...

1. Translate the problem requirements: We need to find the smallest value of n such that \(\mathrm{n!}\) (the product of first n positive integers) ends with exactly six zeros. This means \(\mathrm{n!}\) must be divisible by \(\mathrm{10^6}\).
2. Identify the source of trailing zeros: Trailing zeros come from factors of 10, and each factor of 10 requires one factor of 2 and one factor of 5. Since there are always more factors of 2 than 5 in factorials, the number of trailing zeros equals the number of times 5 appears as a factor.
3. Count factors of 5 systematically: Use the formula that counts factors of 5 in \(\mathrm{n!}\): the number of multiples of 5, plus multiples of 25, plus multiples of 125, etc., since higher powers of 5 contribute additional factors.
4. Test values around the answer choices: Since we need exactly 6 trailing zeros, calculate the number of factors of 5 for values near the given options until we find the smallest n that gives us at least 6 factors of 5.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking. We have \(\mathrm{p = n!}\), which means p is the product of the first n positive integers (like \(\mathrm{1×2×3×4×...×n}\)). The problem tells us that the "last six digits of p are 0" - this means p ends with exactly six zeros, like ...000000.

For p to end with six zeros, it must be divisible by 1,000,000, which is \(\mathrm{10^6}\). So we need to find the smallest value of n such that \(\mathrm{n!}\) is divisible by \(\mathrm{10^6}\).

Process Skill: TRANSLATE - Converting the phrase "last six digits are 0" into the mathematical requirement that \(\mathrm{n!}\) must be divisible by \(\mathrm{10^6}\)

2. Identify the source of trailing zeros

Where do trailing zeros come from? Let's think about this with simple examples:

• \(\mathrm{10 = 2 × 5}\) gives us one trailing zero
• \(\mathrm{100 = 10 × 10 = (2×5) × (2×5)}\) gives us two trailing zeros
• \(\mathrm{1000 = 10^3}\) gives us three trailing zeros

The pattern is clear: each trailing zero comes from a factor of 10, and each factor of 10 requires exactly one factor of 2 and one factor of 5.

Now here's the key insight: in any factorial like \(\mathrm{n!}\), there are always many more factors of 2 than factors of 5. For example, in \(\mathrm{10! = 1×2×3×4×5×6×7×8×9×10}\), we have factors of 2 from 2, 4, 6, 8, 10 (and 4 and 8 contribute even more), but only factors of 5 from 5 and 10.

Since factors of 2 are abundant and factors of 5 are scarce, the number of trailing zeros equals the number of times 5 appears as a factor in \(\mathrm{n!}\).

Process Skill: INFER - Recognizing that factors of 5 are the limiting factor in creating trailing zeros

3. Count factors of 5 systematically

To count factors of 5 in \(\mathrm{n!}\), we need to be systematic. Some numbers contribute one factor of 5, while others contribute more:

• Numbers like 5, 10, 15, 20 each contribute one factor of 5
• Numbers like 25, 50, 75 each contribute two factors of 5 (since \(\mathrm{25 = 5^2}\))
• Numbers like 125, 250 each contribute three factors of 5 (since \(\mathrm{125 = 5^3}\))

So for any n, the total number of factors of 5 in \(\mathrm{n!}\) is:

• Count of multiples of 5 up to n, PLUS
• Count of multiples of 25 up to n, PLUS
• Count of multiples of 125 up to n, PLUS...

Mathematically, this is: \(\mathrm{\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \lfloor n/625 \rfloor + ...}\)

4. Test values around the answer choices

Since we need exactly 6 factors of 5, let's test the given answer choices systematically:

For \(\mathrm{n = 20}\):
factor count = \(\mathrm{\lfloor 20/5 \rfloor + \lfloor 20/25 \rfloor = 4 + 0 = 4}\) factors of 5
This gives only 4 trailing zeros - not enough.

For \(\mathrm{n = 25}\):
factor count = \(\mathrm{\lfloor 25/5 \rfloor + \lfloor 25/25 \rfloor + \lfloor 25/125 \rfloor = 5 + 1 + 0 = 6}\) factors of 5
This gives exactly 6 trailing zeros - this works!

Let's verify that \(\mathrm{n = 24}\) doesn't work:
For \(\mathrm{n = 24}\):
factor count = \(\mathrm{\lfloor 24/5 \rfloor + \lfloor 24/25 \rfloor = 4 + 0 = 4}\) factors of 5
This gives only 4 trailing zeros - not enough.

So \(\mathrm{n = 25}\) is indeed the smallest value that gives us at least 6 trailing zeros.

Process Skill: APPLY CONSTRAINTS - Testing boundary values to find the exact minimum

4. Final Answer

The least possible value of n such that \(\mathrm{n!}\) ends with six zeros is \(\mathrm{n = 25}\).

This corresponds to answer choice B.

Common Faltering Points

Errors while devising the approach

Students often confuse "last six digits are 0" with "at least six trailing zeros" or "exactly six zeros somewhere in the number." The phrase specifically means the number ends with exactly six consecutive zeros (like ...000000), which translates to the number being divisible by \(\mathrm{10^6}\). Some students might think it means the number has six zeros total or that it could have more than six trailing zeros.

When determining trailing zeros, students sometimes try to count both factors of 2 and factors of 5 separately, not realizing that in any factorial \(\mathrm{n!}\), there are always more factors of 2 than factors of 5. This leads them to unnecessarily complicate their approach by tracking both types of factors instead of focusing solely on counting factors of 5.

Students often try to manually list multiples of 5 (like 5, 10, 15, 20, 25...) without realizing that some numbers contribute multiple factors of 5. They might count 25 as contributing only one factor of 5 instead of two (since \(\mathrm{25 = 5^2}\)), leading to an incomplete counting strategy.

Errors while executing the approach

When applying the formula \(\mathrm{\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + ...}\), students frequently make errors with the floor function or forget to include higher powers of 5. For example, when calculating for \(\mathrm{n=25}\), they might compute \(\mathrm{\lfloor 25/5 \rfloor = 5}\) but forget to add \(\mathrm{\lfloor 25/25 \rfloor = 1}\), getting 5 total factors instead of the correct 6.

Students often make basic arithmetic mistakes when computing \(\mathrm{\lfloor n/5 \rfloor}\), \(\mathrm{\lfloor n/25 \rfloor}\), etc. For instance, they might incorrectly calculate \(\mathrm{\lfloor 25/25 \rfloor}\) as 0 instead of 1, or make division errors that lead to wrong factor counts.

Errors while selecting the answer

After finding that \(\mathrm{n=25}\) gives exactly 6 trailing zeros, students might immediately select this answer without verifying that smaller values (like \(\mathrm{n=24}\)) don't work. The question asks for the "least possible value," so they need to confirm that 25 is indeed the minimum, not just a value that satisfies the condition.