The problem asks us to find the Highest Common Factor (HCF) of two numbers expressed in the form \(2^n - 1\). Specifically, we need the HCF of \(2^{36}-1\) and \(2^{45}-1\). Notice that both numbers are generated using the same base, 2, but with different exponents (36 and 45).
There's a useful mathematical property for finding the HCF of numbers of the form \(a^n - 1\) and \(a^m - 1\). The property states:
HCF(\(a^n - 1\), \(a^m - 1\)) = \(a^{\text{HCF}(n, m)} - 1\)
This means we can find the HCF of the original numbers by first finding the HCF of their exponents and then plugging that result back into the same form (\(a^{\text{HCF}}-1\)).
In our problem, the base is \(a=2\). The exponents are \(n=36\) and \(m=45\).
First, let's find the HCF of the exponents, 36 and 45.
We can list the factors of each number:
The common factors are 1, 3, and 9. The highest common factor is 9.
So, HCF(36, 45) = 9.
Now we use the property HCF(\(a^n - 1\), \(a^m - 1\)) = \(a^{\text{HCF}(n, m)} - 1\).
Substitute \(a=2\) and HCF(36, 45) = 9:
HCF(\(2^{36}-1\), \(2^{45}-1\)) = \(2^{\text{HCF}(36, 45)} - 1\)
= \(2^9 - 1\)
Finally, calculate the value of \(2^9 - 1\).
We know that \(2^9 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 512\).
Therefore, \(2^9 - 1 = 512 - 1 = 511\).
The HCF of \(2^{36}-1\) and \(2^{45}-1\) is 511.
The HCF of \(x\) and \(y\) is \(H\). Consider the following statements in respect of the HCF of \(p=\frac{x^3 + y^3}{x^2-xy+y^2}\) and \(q=\frac{x^3-y^3}{x^2+xy+y^2}\) :
I. The HCF of \(p\) and \(q\) can be \(H\).
II. The HCF of \(p\) and \(q\) can be \(2H\).
Which of the statements given above is/are correct?
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