QJEEM22Ma3fdf: JEE Main 2022 Mathematics - Class 11 JEEM22Ma3fdf Solution: The remainder when (2021) 2023 is divided by 7 is :

= (2016 + 5)²⁰²³ [here 2016 is divisible by 7]

= ²⁰²³C₀ (2016)²⁰²³ + .......... + ²⁰²³C₂₀₂₂ (2016) (5)²⁰²² + ²⁰²³C₂₀₂₃ (5)²⁰²³

= 2016 [²⁰²³C₀ . (2016)²⁰²² + ....... + ²⁰²³C₂₀₂₂ . (5)²⁰²²] + (5)²⁰²³

= 2016 $\lambda$ + (5)²⁰²³

= 7 $\times$ 288 $\lambda$ + (5)²⁰²³

= 7K + (5)²⁰²³ ...... (1)

Now, (5)²⁰²³

= (5)²⁰²² . 5

= (5³)⁶⁷⁴ . 5

= (125)⁶⁷⁴ . 5

= (126 $-$ 1)⁶⁷⁴ . 5

= 5[⁶⁷⁴C₀ (126)⁶⁷⁴ + ......... $-$ ⁶⁷⁴C₆₇₃ (126) + ⁶⁷⁴C₆₇₄]

= 5 $\times$ 126 [⁶⁷⁴C₀(126)⁶⁷³ + ....... $-$ ⁶⁷⁴C₆₇₃] + 5

= 5 . 7 . 18 [⁶⁷⁴C₀(126)⁶⁷³ + ....... $-$ ⁶⁷⁴C₆₇₃] + 5

= 7 $\lambda$ + 5

Replacing (5)²⁰²³ in equation (1) with 7 $\lambda$ + 5, we get,

(2021)²⁰²³ = 7K + 7 $\lambda$ + 5

= 7(K + $\lambda$ ) + 5

$\therefore$ Remainer = 5