MCQ
The remainder when $19^{200}+23^{200}$ is divided by $49$ , is  $.........$.
  • A
    $28$
  • B
    $27$
  • $29$
  • D
    $26$

Answer

Correct option: C.
$29$
c
$(21+2)^{200}+(21-2)^{200}$

$\Rightarrow 2\left[{ }^{100} C _0 21^{200}+200 C _2 21^{198} \cdot 2^2+\ldots . .+{ }^{200} C _{198} 21^2\right.$

$\left.2^{198}+2^{200}\right]$

$\Rightarrow 2\left[49 I _1+2^{200}\right]=49 I _1+2^{201}$

Now, $2^{201}=(8)^{67}=(1+7)^{67}=49 I _2+{ }^{67} C _0{ }^{67} C _1 \cdot 7=$ $49 I _2+470=49 I _2+49 \times 9+29$

$\therefore$ Remainder is $29$

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($1$) The tangent to a suitable conic (Column $1$) at $\left(\sqrt{3}, \frac{1}{2}\right)$ is found to be $\sqrt{3} x+2 y=4$, then which of the following options is the only CORRECT combination?

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