Answered

If $64 n=2^7 \cdot 3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2$, explain why $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ is a factor of $n$.



Choose the correct explanation below.

ARewrite $64 n =(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11) \cdot\left(3^3 \cdot 5^4 \cdot 7^5 \cdot 11\right)$. Thus $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ is a factor of $n$.
BRewriting 64 as $2^6$ and dividing leaves $n=2 \cdot 3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2$. This can be rewritten as $n =(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11) \cdot\left(3^3 \cdot 5^4 \cdot 7^5 \cdot 11\right)$. Thus $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ is a factor of $n$.
CRewriting 64 as $2^6$ and dividing leaves $n =3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2$. This can be rewritten as $n =(3 \cdot 5 \cdot 7 \cdot 11) \cdot\left(3^3 \cdot 5^4 \cdot 7^5 \cdot 11\right)$. Thus $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ is a factor of $n$.
D$2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ is not a factor of $n$.

Verified Answer :

  • Rewrite \(64\) as \(2^6\) and solve for \(n\) by dividing the original equation by \(2^6\), yielding \(n = 2 \cdot 3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2\).
  • Factor out \(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11\) from the expression for \(n\), which results in \(n = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11) \cdot (3^3 \cdot 5^4 \cdot 7^5 \cdot 11)\).
  • Confirm that this demonstrates \(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11\) is a factor of \(n\).
  • The correct choice is \(\boxed{B}\).

Explanation

  1. Express 64 as a Power of 2

The given equation is \(64 n = 2^7 \cdot 3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2\). We need to determine if \(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11\) is a factor of \(n\). First, note that \(64\) can be expressed as a power of 2, specifically \(64 = 2^6\). Substituting this into the original equation, we get:
\[2^6 \cdot n = 2^7 \cdot 3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2\]

  1. Solve for \(n\)

To isolate \(n\), we divide both sides of the equation by \(2^6\):
\[n = \frac{2^7 \cdot 3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2}{2^6}\]
Using the laws of exponents, \(\frac{2^7}{2^6} = 2^{7-6} = 2^1 = 2\). Thus, the expression for \(n\) becomes:
\[n = 2 \cdot 3^4 \cdot 5^5 \cdot 7^6 \cdot 11^2\]

  1. Verify the Factor and Select the Correct Option

To show that \(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11\) is a factor of \(n\), we try to write \(n\) as a product involving this expression. We can rewrite the prime powers in \(n\) to isolate one of each:
\[n = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11) \cdot (3^{4-1} \cdot 5^{5-1} \cdot 7^{6-1} \cdot 11^{2-1})\]
Simplifying the exponents:
\[n = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11) \cdot (3^3 \cdot 5^4 \cdot 7^5 \cdot 11)\]
Since \(n\) can be written as the product of \((2 \cdot 3 \cdot 5 \cdot 7 \cdot 11)\) and another integer, it is clear that \((2 \cdot 3 \cdot 5 \cdot 7 \cdot 11)\) is indeed a factor of \(n\). This corresponds to option B.

Examples

In cryptography, understanding the factors of a large number is essential for creating secure encryption keys. Just as we factored out \(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11\) from the expression for \(n\), mathematicians and computer scientists look for prime factors of very large composite numbers to determine the strength of security systems. This process of identifying factors is the foundation of RSA encryption, which keeps your online transactions and communications safe!

Related Questions