Question

The remainder, when $19^{200}+23^{200}$ is divided by $49$ , is_______

Answer 1

Correct Answer: 29

The correct answer is 29.
$(21+2{)}^{200}+(21-2{)}^{200}$
$\Rightarrow 2[{}^{100}{C}_{0}21^{200}+200C_{2}21^{198}\cdot 2^2+\dots ..+{}^{200}{C}_{198}21^2{2}^{198}+2^{200}]$
$\Rightarrow 2\left[49I_1+2^{200}\right]=49I_1+2^{201}$
$\text{Now},2^{201}=(8{)}^{67}=(1+7{)}^{67}=49I_2+{}^{67}{C}_0{}^{67}{C}_1\cdot 7=$
$49I_2+470=49I_2+49\times 9+29$
$\therefore \text{Remainder is}29$

Answer 2

Using modular arithmetic, \[ (21 + 2)^{200} + (21 - 2)^{200} \] Applying binomial expansion, \[ 2 \sum_{k=0}^{100} \binom{200}{2k} 21^{198-2k} 2^{2k} \] \[ \Rightarrow 2[49I_1 + 2^{200}] \] Since \( 2^{200} = 49L + 470 \), \[ \text{Remainder} = 49L + 470 \mod 49 = 29 \]