100! in base 10 when written in base 14 will end with 16 zeroes. (200)! in base 10 when written in base 34 will end with how many
zeroes?

To solve how many zeros \((200)! \) in base 34 ends with, we need to determine how many factors of 34 are present in the prime factorization of \( (200)! \). 

### Step 1: Factorize 34
The base 34 can be factored into prime factors as:
\[
34 = 2 \times 17
\]
So, we need to determine how many factors of 2 and 17 are present in the prime factorization of \( (200)! \).

### Step 2: Count the factors of 2 in \( (200)! \)
To count the number of factors of 2 in \( (200)! \), we use Legendre's formula, which counts the total number of a prime factor \( p \) in \( n! \):
\[
\text{Number of factors of } p = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots
\]
For \( p = 2 \) and \( n = 200 \):
\[
\left\lfloor \frac{200}{2} \right\rfloor = 100, \quad \left\lfloor \frac{200}{4} \right\rfloor = 50, \quad \left\lfloor \frac{200}{8} \right\rfloor = 25, \quad \left\lfloor \frac{200}{16} \right\rfloor = 12
\]
\[
\left\lfloor \frac{200}{32} \right\rfloor = 6, \quad \left\lfloor \frac{200}{64} \right\rfloor = 3, \quad \left\lfloor \frac{200}{128} \right\rfloor = 1
\]
Summing these, the total number of factors of 2 in \( (200)! \) is:
\[
100 + 50 + 25 + 12 + 6 + 3 + 1 = 197
\]

### Step 3: Count the factors of 17 in \( (200)! \)
Now, we count the number of factors of 17 in \( (200)! \). Again, using Legendre's formula:
\[
\left\lfloor \frac{200}{17} \right\rfloor = 11, \quad \left\lfloor \frac{200}{17^2} \right\rfloor = 0
\]
So, the total number of factors of 17 in \( (200)! \) is:
\[
11
\]

### Step 4: Determine the limiting factor
To determine the number of trailing zeros, we find how many complete factors of \( 34 = 2 \times 17 \) are present. The limiting factor is the smaller of the two counts from Step 2 and Step 3. Therefore, the number of trailing zeros in \( (200)! \) when written in base 34 is:
\[
\min(197, 11) = 11
\]

### Final Answer:
The number of trailing zeros in \( (200)! \) when written in base 34 is \( \boxed{11} \).

Would you like a detailed explanation on any part of this solution or more insights?

Here are some follow-up questions for further exploration:
1. How would the number of zeros change if the factorial was calculated for a different number?
2. What is the general process for finding trailing zeros of any factorial in an arbitrary base?
3. How do trailing zeros relate to prime factorization in various bases?
4. How would the problem change if the base was a prime number?
5. Can we find a faster way to compute trailing zeros for large factorials in different bases?

**Tip:** When working with factorials in different bases, always factorize the base into prime factors and then use Legendre's formula to determine the limiting prime factor for trailing zeros.

Learn how to determine the number of trailing zeros in (200)! when written in base 34 using Legendre's formula and prime factorization. Understand the mathematical process involved in converting factorials to different bases. Suitable for advanced high school students and anyone interested in number theory.

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