- Break down each expression into its fundamental prime components.
- For \(12!\), aggregate the powers of each prime factor to reach the result \(2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11\), which corresponds to option D.
- Simplify other expressions by rewriting bases as prime powers: \(6^5 \cdot 22 \cdot 169^{15} = 2^6 \cdot 3^5 \cdot 11 \cdot 13^{30}\), \(113\) is prime, and \(1000^8 = 2^{24} \cdot 5^{24}\).
- The correct choice for the prime factorization of \(12!\) is \(\boxed{D}\).
Explanation
- General Approach to Prime Factorization
To find the prime factorization of a number, we express it as a product of powers of prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. We will address each part of the problem step by step.
- Factorizing 12! (Part a)
For the product \(P = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \cdot 11 \cdot 12\), which is 12 factorial (\(12!\)), we decompose each factor into prime components:
- \(1 = 1\)
- \(2 = 2\)
- \(3 = 3\)
- \(4 = 2^2\)
- \(5 = 5\)
- \(6 = 2 \cdot 3\)
- \(7 = 7\)
- \(8 = 2^3\)
- \(9 = 3^2\)
- \(10 = 2 \cdot 5\)
- \(11 = 11\)
- \(12 = 2^2 \cdot 3\)
Now, we collect the exponents for each prime base (2, 3, 5, 7, 11):
- Prime base 2: \(1+0+0+2+0+1+0+3+0+1+0+2 = 10 \implies 2^{10}\)
- Prime base 3: \(0+0+1+0+0+1+0+0+2+0+0+1 = 5 \implies 3^5\)
- Prime base 5: \(0+0+0+0+1+0+0+0+0+1+0+0 = 2 \implies 5^2\)
- Prime base 7: \(0+0+0+0+0+0+1+0+0+0+0+0 = 1 \implies 7^1\)
- Prime base 11: \(0+0+0+0+0+0+0+0+0+0+1+0 = 1 \implies 11^1\)
Combining these, we get \(2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11\). This corresponds to Option D.
- Factorizing 6^5 22 169^15 (Part b)
To factorize \(6^5 \cdot 22 \cdot 169^{15}\), we break down the bases into their prime factors:
- \(6 = 2 \cdot 3\), so \(6^5 = (2 \cdot 3)^5 = 2^5 \cdot 3^5\)
- \(22 = 2 \cdot 11\)
- \(169 = 13^2\), so \(169^{15} = (13^2)^{15} = 13^{30}\)
Putting them together:
\((2^5 \cdot 3^5) \cdot (2 \cdot 11) \cdot 13^{30} = 2^{5+1} \cdot 3^5 \cdot 11 \cdot 13^{30} = 2^6 \cdot 3^5 \cdot 11 \cdot 13^{30}\)
- Factorizing 113 (Part c)
The number 113 is a prime number. We verify this by checking if it is divisible by any prime number up to its square root (\(\sqrt{113} \approx 10.63\)). The primes to check are 2, 3, 5, and 7:
- \(113 / 2\) is not an integer.
- \(113 / 3 = 37.66...\)
- \(113 / 5 = 22.6...\)
- \(113 / 7 = 16.14...\)
Since 113 is not divisible by any of these, it is a prime number. Its prime factorization is simply 113.
- Factorizing 1000^8 (Part d)
To factorize \(1000^8\), we first factor the base 1000:
- \(1000 = 10^3 = (2 \cdot 5)^3 = 2^3 \cdot 5^3\)
Now we raise this to the power of 8:
- \((2^3 \cdot 5^3)^8 = 2^{3 \cdot 8} \cdot 5^{3 \cdot 8} = 2^{24} \cdot 5^{24}\)
- Final Results
The correct prime factorization for \(12!\) is \(2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11\), which matches Option D. The prime factorizations are:
A\(\boxed{2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11}\)
B\(\boxed{2^6 \cdot 3^5 \cdot 11 \cdot 13^{30}}\)
C\(\boxed{113}\)
D\(\boxed{2^{24} \cdot 5^{24}}\)
And the correct choice is \(\boxed{D}\).
Examples
Prime factorization is essential in cryptography, where large numbers are used to secure data. By representing complex numbers as products of primes, computers can efficiently perform modular arithmetic, which is the foundation of RSA encryption. Understanding these structures allows engineers to build secure systems for everything from online banking to private messaging, ensuring that only the intended recipients can read protected information.