If 10 ndivides 6 23 × 75 9× 105 2, then what is the largest value of n?
20
Understanding Divisibility by \(10^n\)
To find the largest value of \(n\) such that \(10^n\) divides a given number, we need to analyze the prime factorization of the number. The number 10 can be factored into its prime components as \(10 = 2 \times 5\). Therefore, \(10^n = (2 \times 5)^n = 2^n \times 5^n\).
For \(10^n\) to divide a number, the prime factorization of that number must contain at least \(n\) factors of 2 and at least \(n\) factors of 5. The largest possible value of \(n\) is limited by the minimum of the highest power of 2 and the highest power of 5 present in the prime factorization of the given number.
Analyzing the Given Expression
The given expression is \(6^{23} \times 75^9 \times 105^2\). To find the highest power of 10 that divides this expression, we first need to find the prime factorization of each base number: 6, 75, and 105.
- Prime factorization of 6: \(6 = 2 \times 3\)
- Prime factorization of 75: \(75 = 3 \times 25 = 3 \times 5^2\)
- Prime factorization of 105: \(105 = 5 \times 21 = 5 \times 3 \times 7\)
Substituting and Combining Prime Factors
Now, substitute these prime factorizations back into the original expression:
\(\left(2 \times 3\right)^{23} \times \left(3 \times 5^2\right)^9 \times \left(3 \times 5 \times 7\right)^2\)
Using the exponent rule \((ab)^m = a^m b^m\) and \((a^p)^q = a^{pq}\), we expand the terms:
\(2^{23} \times 3^{23} \times 3^9 \times (5^2)^9 \times 3^2 \times 5^2 \times 7^2\)
\(2^{23} \times 3^{23} \times 3^9 \times 5^{18} \times 3^2 \times 5^2 \times 7^2\)
Next, combine the terms with the same base by adding their exponents:
- Base 2: The only term with base 2 is \(2^{23}\). So, the power of 2 is 23.
- Base 3: The terms with base 3 are \(3^{23}\), \(3^9\), and \(3^2\). Their exponents add up: \(23 + 9 + 2 = 34\). So, the power of 3 is 34 (\(3^{34}\)).
- Base 5: The terms with base 5 are \(5^{18}\) and \(5^2\). Their exponents add up: \(18 + 2 = 20\). So, the power of 5 is 20 (\(5^{20}\)).
- Base 7: The only term with base 7 is \(7^2\). So, the power of 7 is 2.
The prime factorization of the entire expression is \(2^{23} \times 3^{34} \times 5^{20} \times 7^2\).
Finding the Highest Power of 10
For \(10^n = 2^n \times 5^n\) to divide \(2^{23} \times 3^{34} \times 5^{20} \times 7^2\), the power of 2 in \(10^n\) must be less than or equal to the power of 2 in the expression, and the power of 5 in \(10^n\) must be less than or equal to the power of 5 in the expression.
- Power of 2 in the expression is 23. So, \(n \le 23\).
- Power of 5 in the expression is 20. So, \(n \le 20\).
For both conditions to be true, \(n\) must be less than or equal to the minimum of 23 and 20.
Largest value of \(n = \min(\text{power of 2, power of 5}) = \min(23, 20) = 20\).
Thus, the largest value of \(n\) for which \(10^n\) divides the given expression is 20.
Step-by-Step Calculation Summary
| Step | Description | Calculation/Result |
|---|---|---|
| 1 | Prime factorize bases | \(6=2\times3\), \(75=3\times5^2\), \(105=3\times5\times7\) |
| 2 | Substitute into expression | \((2\times3)^{23} \times (3\times5^2)^9 \times (3\times5\times7)^2\) |
| 3 | Expand exponents | \(2^{23}\times3^{23}\times3^9\times5^{18}\times3^2\times5^2\times7^2\) |
| 4 | Combine like bases | \(2^{23}\times3^{23+9+2}\times5^{18+2}\times7^2 = 2^{23}\times3^{34}\times5^{20}\times7^2\) |
| 5 | Identify powers of 2 and 5 | Power of 2 is 23, Power of 5 is 20 |
| 6 | Find largest \(n\) | \(n = \min(23, 20) = 20\) |
Revision Table: Prime Factorization and Exponents
| Concept | Explanation | Example |
|---|---|---|
| Prime Factorization | Breaking down a composite number into its prime number factors. | \(12 = 2 \times 2 \times 3 = 2^2 \times 3\) |
| Exponents | Indicates how many times a base number is multiplied by itself. | \(5^3 = 5 \times 5 \times 5 = 125\) |
| Exponent Rule: \((a \times b)^m\) | \((a \times b)^m = a^m \times b^m\) | \((2 \times 3)^4 = 2^4 \times 3^4\) |
| Exponent Rule: \((a^p)^q\) | \((a^p)^q = a^{p \times q}\) | \((5^2)^3 = 5^{2 \times 3} = 5^6\) |
| Exponent Rule: \(a^p \times a^q\) | \(a^p \times a^q = a^{p+q}\) | \(7^2 \times 7^5 = 7^{2+5} = 7^7\) |
| Divisibility by \(10^n\) | A number is divisible by \(10^n\) if its prime factorization contains at least \(n\) factors of 2 and at least \(n\) factors of 5. The highest such \(n\) is \(\min(\text{power of 2, power of 5})\). | \(1200 = 12 \times 100 = (2^2 \times 3) \times (10^2) = 2^2 \times 3 \times (2 \times 5)^2 = 2^2 \times 3 \times 2^2 \times 5^2 = 2^4 \times 3^1 \times 5^2\). Power of 2 is 4, Power of 5 is 2. Max power of 10 is \(\min(4, 2) = 2\). So, \(1200\) is divisible by \(10^2=100\). |
Additional Information: Finding Powers of Primes
The process used here is a fundamental concept in number theory, specifically when dealing with powers of primes in the prime factorization of large numbers or expressions. To find the exponent of a prime factor \(p\) in the prime factorization of an integer \(N\), you typically perform the prime factorization of \(N\). If \(N\) is given in factored form like \(a^x \times b^y \times c^z\), and \(a, b, c\) are composite, you first factorize \(a, b, c\) into their primes and then combine the exponents for each unique prime factor.
For example, if you needed to find the power of 2 in \(18^5\):
\(18 = 2 \times 3^2\)
\(18^5 = (2 \times 3^2)^5 = 2^5 \times (3^2)^5 = 2^5 \times 3^{10}\).
The power of 2 is 5.
This technique is crucial for problems involving divisibility by powers of 10, or finding the number of trailing zeros in a factorial (which is determined by the power of 5). The key is always to break down the numbers into their basic prime building blocks.
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select the correct answer using the code given below:
