How many zeros are there in the product 1 50 × 2 49 × 3 48 × .... × 50 1 ?
262
Finding the Number of Trailing Zeros in the Product
To determine the number of trailing zeros in any integer or a product of integers, we need to find how many times 10 is a factor in its prime factorization. Since $\(10 = 2 \times 5\)$, the number of factors of 10 is limited by the number of pairs of prime factors 2 and 5 available. In most cases, the total count of the prime factor 5 is less than the total count of the prime factor 2. Therefore, the number of trailing zeros is equal to the total exponent of 5 in the prime factorization of the number or product.
The given product is $\(P = 1^{50} \times 2^{49} \times 3^{48} \times \dots \times 50^1\)$. This product can be written in a more compact form using product notation as $\(P = \prod_{k=1}^{50} k^{51-k}\)$.
Calculating the Total Exponent of Prime Factor 5
The total exponent of a prime \(p\) in a product of the form $\(\prod_{k=1}^{n} k^{a_k}\)$ is calculated by summing the contributions from each term. The contribution of a term $\(k^{a_k}\)$ to the total power of prime \(p\) is \(v_p(k) \times a_k\), where \(v_p(k)\) represents the exponent of the prime \(p\) in the prime factorization of the base \(k\). Thus, the total exponent of \(p\) in the product is $\(\sum_{k=1}^{n} v_p(k) a_k\)$.
In this problem, we are interested in the prime \(p=5\), the upper limit of the product is \(n=50\), and the exponent for each base \(k\) is \(a_k = 51-k\).
We need to calculate the sum $\(\sum_{k=1}^{50} v_5(k) (51-k)\)$. The term \(v_5(k)\) is non-zero only for values of \(k\) that are multiples of 5. These multiples of 5 within the range 1 to 50 are 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50.
Let's list these bases and calculate the exponent of 5 contributed by each term:
| Base \(k\) | Prime Factorization of \(k\) | \(v_5(k)\) (Exponent of 5 in \(k\)) | Exponent of term (\(a_k = 51-k\)) | Contribution to Total Power of 5 (\(v_5(k) \times a_k\)) |
|---|---|---|---|---|
| 5 | \(5^1\) | 1 | \(51-5 = 46\) | \(1 \times 46 = 46\) |
| 10 | \(2 \times 5^1\) | 1 | \(51-10 = 41\) | \(1 \times 41 = 41\) |
| 15 | \(3 \times 5^1\) | 1 | \(51-15 = 36\) | \(1 \times 36 = 36\) |
| 20 | \(2^2 \times 5^1\) | 1 | \(51-20 = 31\) | \(1 \times 31 = 31\) |
| 25 | \(5^2\) | 2 | \(51-25 = 26\) | \(2 \times 26 = 52\) |
| 30 | \(2 \times 3 \times 5^1\) | 1 | \(51-30 = 21\) | \(1 \times 21 = 21\) |
| 35 | \(7 \times 5^1\) | 1 | \(51-35 = 16\) | \(1 \times 16 = 16\) |
| 40 | \(2^3 \times 5^1\) | 1 | \(51-40 = 11\) | \(1 \times 11 = 11\) |
| 45 | \(3^2 \times 5^1\) | 1 | \(51-45 = 6\) | \(1 \times 6 = 6\) |
| 50 | \(2 \times 5^2\) | 2 | \(51-50 = 1\) | \(2 \times 1 = 2\) |
The total power of 5 in the product is the sum of these contributions:
\(\text{Total power of 5} = 46 + 41 + 36 + 31 + 52 + 21 + 16 + 11 + 6 + 2\)
\(\text{Total power of 5} = 87 + 36 + 31 + 52 + 21 + 16 + 11 + 6 + 2\)
\(\text{Total power of 5} = 123 + 31 + 52 + 21 + 16 + 11 + 6 + 2\)
\(\text{Total power of 5} = 154 + 52 + 21 + 16 + 11 + 6 + 2\)
\(\text{Total power of 5} = 206 + 21 + 16 + 11 + 6 + 2\)
\(\text{Total power of 5} = 227 + 16 + 11 + 6 + 2\)
\(\text{Total power of 5} = 243 + 11 + 6 + 2\)
\(\text{Total power of 5} = 254 + 6 + 2\)
\(\text{Total power of 5} = 260 + 2\)
\(\text{Total power of 5} = 262\)
So, the total exponent of the prime factor 5 in the product is 262.
Considering the Total Exponent of Prime Factor 2
To find the exact number of zeros, we also need the total exponent of the prime factor 2 in the product, which is $\(\sum_{k=1}^{50} v_2(k) (51-k)\)$.
The term \(v_2(k)\) is non-zero for all even values of \(k\). There are 25 even numbers between 1 and 50. Moreover, many of these even numbers are multiples of 4, 8, 16, 32, contributing multiple factors of 2 (\(v_2(k) \ge 2\)).
Since the exponents \(51-k\) are positive for all \(k\) from 1 to 50, every factor of 2 in the bases \(k\) contributes to the total power of 2. Because there are many more multiples of 2 than multiples of 5, and higher powers of 2 appear more frequently than higher powers of 5, the total exponent of 2 in the product will be significantly larger than the total exponent of 5.
Calculating the exact total power of 2 is more involved, but it's clear it will be greater than 262.
Final Number of Zeros
The number of trailing zeros in the product is the minimum of the total exponent of the prime factor 2 and the total exponent of the prime factor 5.
Number of zeros = \(\min(\text{Total power of 2, Total power of 5})\)
Number of zeros = \(\min(\text{Total power of 2, 262})\)
Since we've established that the Total power of 2 is greater than 262, the minimum value is 262.
Thus, there are 262 trailing zeros in the given product $\(1^{50} \times 2^{49} \times 3^{48} \times \dots \times 50^1\)$.
Revision Table: Calculating Trailing Zeros
| Step | Description |
|---|---|
| 1 | Understand that trailing zeros come from factors of 10, which are pairs of 2 and 5. |
| 2 | Identify the need to find the total exponent of prime factors 2 and 5 in the product. |
| 3 | Use the formula $\(\sum v_p(k) a_k\)$ to calculate the total exponent of a prime \(p\), where \(v_p(k)\) is the exponent of \(p\) in \(k\) and \(a_k\) is the exponent of the term \(k\). |
| 4 | Calculate the total exponent of 5 by summing \(v_5(k) \times (51-k)\) for all \(k\) from 1 to 50 (only multiples of 5 contribute). |
| 5 | Recognize that the total exponent of 2 will be greater than that of 5. |
| 6 | The number of zeros is the minimum of the total exponents of 2 and 5. |
Additional Information: Understanding \(v_p(k)\)
The function \(v_p(k)\), also known as the p-adic valuation of \(k\), gives the exponent of the highest power of the prime \(p\) that divides \(k\). For example:
- \(v_5(10)\): \(10 = 2 \times 5^1\), so \(v_5(10) = 1\).
- \(v_5(25)\): \(25 = 5^2\), so \(v_5(25) = 2\).
- \(v_5(50)\): \(50 = 2 \times 25 = 2 \times 5^2\), so \(v_5(50) = 2\).
- \(v_5(12)\): \(12 = 2^2 \times 3\), 5 is not a factor, so \(v_5(12) = 0\).
When calculating the total power of a prime in a product like this, we sum the \(v_p(k)\) value for each base \(k\) multiplied by the exponent of that base $\((51-k)\)$. This method correctly accounts for all prime factors within each base \(k\) and their overall impact on the final product's prime factorization.
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