The remainder when 19^{200} + 23^{200} is div | vedclass

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(C) We need to find the remainder of $(19^{200} + 23^{200}) \pmod{49}$.Note that $19 = 21 - 2$ and $23 = 21 + 2$.So,$(21 - 2)^{200} + (21 + 2)^{200} =

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$29$

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(C) We need to find the remainder of $(19^{200} + 23^{200}) \pmod{49}$.Note that $19 = 21 - 2$ and $23 = 21 + 2$.So,$(21 - 2)^{200} + (21 + 2)^{200} = \sum_{k=0}^{200} \binom{200}{k} 21^k (-2)^{200-k} + \sum_{k=0}^{200} \binom{200}{k} 21^k (2)^{200-k}$.Since $21^2 = 441$,which is a multiple of $49$ $(49 	imes 9 = 441)$,all terms with $21^k$ where $k \ge 2$ are divisible by $49$.Thus,the expression is congruent to $\binom{200}{0} (-2)^{200} + \binom{200}{1} 21 (-2)^{199} + \binom{200}{0} (2)^{200} + \binom{200}{1} 21 (2)^{199} \pmod{49}$.$= 2^{200} - 200 	imes 21 	imes 2^{199} + 2^{200} + 200 	imes 21 	imes 2^{199} \pmod{49}$.$= 2 	imes 2^{200} = 2^{201} \pmod{49}$.Now,$2^6 = 64 \equiv 15 \pmod{49}$,$2^7 = 128 \equiv 30 \pmod{49}$,$2^{14} = (128)^2 \equiv 30^2 = 900 = 49 	imes 18 + 18 \equiv 18 \pmod{49}$.Alternatively,$2^{201} = 2^3 	imes (2^6)^{33} = 8 	imes (64)^{33} \equiv 8 	imes (15)^{33} \pmod{49}$.Using $2^{201} = 2 	imes (2^{10})^{20} = 2 	imes (1024)^{20}$. Since $1024 = 49 	imes 20 + 44$,$1024 \equiv 44 \equiv -5 \pmod{49}$.$2^{201} \equiv 2 	imes (-5)^{20} = 2 	imes (25)^{10} = 2 	imes (625)^5$.Since $625 = 49 	imes 12 + 37$,$625 \equiv 37 \equiv -12 \pmod{49}$.$2^{201} \equiv 2 	imes (-12)^5 = 2 	imes (-248832) \equiv 2 	imes 29 = 58 \equiv 9 \pmod{49}$.Wait,re-evaluating: $2^{201} = 8 	imes (2^6)^{33} = 8 	imes (64)^{33} \equiv 8 	imes (15)^{33} \pmod{49}$.$15^2 = 225 = 49 	imes 4 + 29 \equiv 29 \equiv -20 \pmod{49}$.$15^3 = 15 	imes (-20) = -300 = 49 	imes (-7) + 43 \equiv -6 \pmod{49}$.$15^{33} = (15^3)^{11} \equiv (-6)^{11} = -6 	imes (36)^5 = -6 	imes (-13)^5 = -6 	imes (-371293) \equiv 29 \pmod{49}$.$8 	imes 29 = 232 = 49 	imes 4 + 36 \equiv 36 \pmod{49}$.Recalculating: $2^{201} = 2 	imes (2^{10})^{20} = 2 	imes (1024)^{20} \equiv 2 	imes (-5)^{20} = 2 	imes (25)^{10} = 2 	imes (625)^5 \equiv 2 	imes (37)^5 \equiv 2 	imes (-12)^5 = -2 	imes 248832 = -497664 \equiv 29 \pmod{49}$.

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

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