Binomial Theorem Question 21
Question 21 - 01 February - Shift 1
The remainder when $19^{200}+23^{200}$ is divided by 49 , is_______
Show Answer
Answer: (29)
Solution:
Formula: Divisibility concept
$(21+2)^{200}+(21-2)^{200}$
$\Rightarrow 2[{ }^{100} C_0 21^{200}+200 C_2 21{ }^{198} \cdot 2^{2}+\ldots . .+{ }^{200} C _{198} 21^{2}.$
$.2^{198}+2^{200}]$
$\Rightarrow 2[49 I_1+2^{200}]=49 I_1+2^{201}$
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
