Let N 1-255-1 and N 2-165- ThenA- N 1 and N 2 are coprimeB- the HCF Highest Common Factor of N 1 and N 2 is 55C- the HCF of N 1 and N 2 is 11D- the HCF of N 1 and N 2 is 33

The correct option is D the HCF of N_1 and N_2 is 33Given that N_1=2^55+1 and N_2=165=3×5×11 2^2n+1+1 is divisible by 3. So, 2^55 is divisible by 3. nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Binary Operation

Let N 1=255+1...

Question

Let $N_{1} = 2^{55} + 1$ and $N_{2} = 165.$ Then

N_{1}

and

N_{2}

are coprime

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the HCF (Highest Common Factor) of

N_{1}

and

N_{2}

55

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the HCF of

N_{1}

and

N_{2}

11

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the HCF of

N_{1}

and

N_{2}

33

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Solution

The correct option is D the HCF of $N_{1}$ and $N_{2}$ is $33$
Given that $N_{1} = 2^{55} + 1$ and $N_{2} = 165 = 3 \times 5 \times 11$

$2^{2 n + 1} + 1 is divisible by 3$ . So, $2^{55}$ is divisible by $3.$
$2^{55}$ ends with $8$ . So, $2^{55} + 1$ ends with $9$ and thus not divisible by $5$ .

$2^{55} mod (11) = (32)^{11} mod 11 = 10$
$\Rightarrow 2^{55} = 11 k + 10 \Rightarrow N_{1} = 11 k + 11$ ,
so, $N_{1}$ is divisible by $11$

So, the HCF of $N_{1}$ and $N_{2}$ is $33$ .

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Let N1=255+1 and N2=165. Then

Let $N_{1} = 2^{55} + 1$ and $N_{2} = 165.$ Then