What is the largest power of 10 that divides the product 29 × 28 × 27 × ...2 × 1 ?
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Finding the Largest Power of 10 Dividing 29 Factorial
The problem asks for the largest power of 10 that divides the product of integers from 1 to 29, which is denoted as \(29!\). To find the largest power of 10 that divides a number, we need to determine how many times 10 is a factor in its prime factorization.
The number 10 is the product of the prime numbers 2 and 5 (\(10 = 2 \times 5\)). Therefore, the number of times 10 is a factor in \(29!\) is limited by the number of factors of 5 or the number of factors of 2, whichever is smaller. In factorials, the number of factors of 5 is always less than or equal to the number of factors of 2.
Calculating the Exponent of 5 in 29!
To find the number of times a prime \(p\) divides \(n!\), we use Legendre's formula:
\( \text{Exponent of } p \text{ in } n! = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots \)
We need to find the exponent of the prime 5 in \(29!\). Here, \(n=29\) and \(p=5\).
- First term: \( \left\lfloor \frac{29}{5} \right\rfloor = \lfloor 5.8 \rfloor = 5 \) (This counts numbers like 5, 10, 15, 20, 25 which contribute at least one factor of 5)
- Second term: \( \left\lfloor \frac{29}{5^2} \right\rfloor = \left\lfloor \frac{29}{25} \right\rfloor = \lfloor 1.16 \rfloor = 1 \) (This counts numbers like 25 which contribute an additional factor of 5)
- Third term: \( \left\lfloor \frac{29}{5^3} \right\rfloor = \left\lfloor \frac{29}{125} \right\rfloor = \lfloor 0.232 \rfloor = 0 \) (Higher powers of 5 will also result in 0)
The sum of these terms gives the total number of factors of 5 in \(29!\):
\( \text{Exponent of 5 in } 29! = 5 + 1 + 0 = 6 \)
Calculating the Exponent of 2 in 29!
Similarly, we find the exponent of the prime 2 in \(29!\). Here, \(n=29\) and \(p=2\).
- \( \left\lfloor \frac{29}{2} \right\rfloor = \lfloor 14.5 \rfloor = 14 \)
- \( \left\lfloor \frac{29}{2^2} \right\rfloor = \left\lfloor \frac{29}{4} \right\rfloor = \lfloor 7.25 \rfloor = 7 \)
- \( \left\lfloor \frac{29}{2^3} \right\rfloor = \left\lfloor \frac{29}{8} \right\rfloor = \lfloor 3.625 \rfloor = 3 \)
- \( \left\lfloor \frac{29}{2^4} \right\rfloor = \left\lfloor \frac{29}{16} \right\rfloor = \lfloor 1.8125 \rfloor = 1 \)
- \( \left\lfloor \frac{29}{2^5} \right\rfloor = \left\lfloor \frac{29}{32} \right\rfloor = \lfloor 0.90625 \rfloor = 0 \)
The sum of these terms gives the total number of factors of 2 in \(29!\):
\( \text{Exponent of 2 in } 29! = 14 + 7 + 3 + 1 + 0 = 25 \)
Determining the Largest Power of 10
To form a factor of 10, we need one factor of 2 and one factor of 5. Since the number of factors of 5 (which is 6) is less than the number of factors of 2 (which is 25), the number of times we can form the factor 10 is limited by the number of factors of 5.
The largest power of 10 that divides \(29!\) is \(10^{\min(\text{Exponent of 2}, \text{Exponent of 5})}\).
\( \min(25, 6) = 6 \)
So, the largest power of 10 that divides \(29!\) is \(10^6\).
The question asks for the exponent of 10, which is 6.
The final answer is 6.
Revision Table
| Concept | Explanation | Application to 29! |
|---|---|---|
| Largest power of 10 dividing N | Find the number of factors of 10 in the prime factorization of N. \(10 = 2 \times 5\). The number of 10s is the minimum of the exponents of 2 and 5. | Find exponents of 2 and 5 in \(29!\). |
| Legendre's Formula | Calculates the exponent of a prime \(p\) in \(n!\) as \( \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \). | Used to find exponents of 2 and 5 in \(29!\). |
| Exponent of 5 in 29! | Calculated as \( \lfloor\frac{29}{5}\rfloor + \lfloor\frac{29}{25}\rfloor = 5 + 1 = 6 \). | There are 6 factors of 5 in \(29!\). |
| Exponent of 2 in 29! | Calculated as \( \lfloor\frac{29}{2}\rfloor + \lfloor\frac{29}{4}\rfloor + \lfloor\frac{29}{8}\rfloor + \lfloor\frac{29}{16}\rfloor = 14 + 7 + 3 + 1 = 25 \). | There are 25 factors of 2 in \(29!\). |
| Largest power of 10 | Minimum of the exponents of 2 and 5. | \( \min(25, 6) = 6 \). The largest power is \(10^6\). |
Additional Information on Factorials and Divisibility
Understanding the prime factorization of factorials is crucial for solving problems related to divisibility, especially finding the number of trailing zeros.
- Trailing Zeros: The number of trailing zeros in an integer is equal to the largest power of 10 that divides it. This is exactly what we calculated in the problem: the exponent of 10 in the factorial.
- Why 5 is the Limiting Factor: In the product \(n! = 1 \times 2 \times \dots \times n\), factors of 2 appear much more frequently than factors of 5. Every even number contributes a factor of 2, while only multiples of 5 contribute a factor of 5 (and multiples of 25 contribute two factors of 5, etc.). Therefore, when calculating the number of pairs of \(2 \times 5\), the number of 5s will always be the bottleneck.
- Generalizing Legendre's Formula: The formula \( \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \) works for any prime \(p\). We use it for 5 and 2 to find the power of 10, but it could be used for any other prime (e.g., 3, 7, 11) to find its exponent in the factorial. The summation continues only as long as \(p^k \le n\).
This method provides a systematic way to determine the divisibility of large factorial numbers by powers of 10 or any other prime-based number.
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