To find the number of zeros at the end of $10!$ (10 factorial) when it is written in base 3, we need to consider how many times 3 is a factor in $10!$.

Steps to Solve:

1. Factorize 10!: $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

2. Count the number of factors of 3 in $10!$: We need to determine how many times 3 is a factor in each of the numbers from 1 to 10.

   • $3$ is a factor in: $3$ (once), $6$ (once), and $9$ (twice, since $9 = 3^2$).

   • So, the total number of times 3 appears as a factor in $10!$ is: $1 + 1 + 2 = 4$

3. Number of zeros at the end of $10!$ in base 3: The number of zeros at the end of a number in base 3 is determined by the highest power of 3 that divides $10!$. Since we have four factors of 3 in $10!$, there are 4 zeros at the end of $10!$ when written in base 3.

So, the number of zeros at the end of $10!$ in base 3 is 4.

Would you like more details on this process or have any questions?

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Here are 8 related questions:

1. How do you find the number of trailing zeros in $n!$ for different bases?
2. What is the significance of trailing zeros in different bases?
3. How do you factorize a factorial number to find specific prime factors?
4. How does changing the base affect the number of trailing zeros in $n!$?
5. Can you find the number of trailing zeros in $15!$ when written in base 5?
6. What is the highest power of 2 that divides $20!$?
7. How do trailing zeros relate to the factorization of numbers?
8. What are the general steps for converting a factorial to any base?

Tip: To find the number of trailing zeros in a number, factorize the number and focus on the factors that contribute to the base you're considering.