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STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. binomial theoram4 Marks
MCQ
Let $N_1=2^{55}+1$ and $N_2=165$.Then
  • A
    $N_1$ and $N_2$ are coprime
  • B
    the $HCF$ (Highest Common Factor) of $N_1$ and $N_2$ is $55$
  • C
    the $HCF$ of $N_1$ and $N_2$ is $11$
  • ✓
    the $HCF$ of $N_1$ and $N_2$ is $33$
✓

Answer

Correct option: D.
the $HCF$ of $N_1$ and $N_2$ is $33$
d
(d)

It is given that, $N_2=165$ $=3 \times 5 \times 11$ and $N_1=2^{55}+1$

As we know that, if $n$ is odd integer then $x^n+y^n$ is divisible by $x+y$.

So, $N_1=2^{55}+1^{55}$ is divisible by $2+1=3$ and $N_1=2^{55}+1^{55}$

$=\left(2^5\right)^{11}+\left(1^5\right)^{11}=(32)^{11}+(1)^{11}$

is divisible by $32+1=33$

$\therefore$ the HCF of $N_1$ and $N_2$ is $33$

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