Question 150 of 347MediumMCQ
The remainder when (2021)2023 is divided by 7 is:
Solutions · 1
Official 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
= 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
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