Question

The remainder, when 19²⁰⁰ + 23²⁰⁰ is divided by 49, is___. Please solve it using negative remainder concept if possible.

Answer 1

\(19^{200} + 23^{200} = (21-2)^{200} + (21 + 2)^{200}\)

\(= 2(^{200}C_0 \,21^{200} + \,^{200}C_2\,21^{198}\,2^2 + ....+ \,^{200}C_{198}\,21^2 \,2^{198}+ \,^{200}C_{200}\,2^{200})\)

\(= 2((7\times 3)^{200} + \,^{200}C_2(7\times 3)^{198}\times 4 +....+\,^{200}C_{198}(3\times 7)^2 \,2^{198}+2^{200})\)

\(= 49q + 2^{201}\)    \((\because q = 2(7^{198}.3^{200} + 4^2\,^{200}C_2 \, 7^{196}\,3^{198}+....+ \,^{200}C_{198}.3^2\,2^{198}))\)

\(= 49 q + 8^{67}\)

\(= 49q+ (7 + 1)^{67}\)

\(= 49 q + (^{67}C_0\,7^{67} + \,^{67}C_1\,7^{66}+\,^{67}C_2\,7^{65} + .... + ^{67}C_{65}7^2 + \,^{67}C_{66}\, 7+\,^{67}C_{67})\)

\(= 49q + 49 (^{67}C_0\,7^{65} + \,^{67}C_1\,7^{64} + \,^{67}C_2\,7^{63} + ....+ \,^{67}C_{65}) + 67 \times 7 + 1\)

\(= 49r + 47\)   where \(r = q + \,^{67}C_{0}\,7^{65} + \,^{67}C_1 \, 7^{64} + ....+ ^{67} C_{65}\)

\(= 49r + 441 + 29\)

\(= 49(r + 9) + 29\)

\(= 49k + 29, k = r + 9\)

Hence, when 19²⁰⁰ + 23²⁰⁰ is divided by 49, it leaves remainder 29.