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Question : 67 of 100
Marks:
+1,
-0
Solution:
(2021)2023
=(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
=2016λ+(5)2023
=7×288λ+(5)2023
=7K+(5)2023......(1)
Now, (5)2023
=(5)2022â‹…5
=(53)674â‹…5
=(125)674â‹…5
=(126−1)674⋅5
=(126−1)674⋅5
=5[‌674C0(126)674+..........−‌674C673(126)+‌674C674]
=5×126[‌674C0(126)673+.......−‌674C673]+5
=5.7.18[‌674C0(126)673+.......−‌674C673]+5
=7λ+5
Replacing (5)2023 in equation (1) with 7λ+5, we get,
(2021)2023=7K+7λ+5
=7(K+λ)+5
∴ Remainer =5
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