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The remainder left out when 8^(2n)""(62)...

The remainder left out when `8^(2n)""(62)^(2n+1)` is divided by 9 is (1) 0 (2) 2 (3) 7 (4) 8

A

0

B

2

C

7

D

8

Text Solution

Verified by Experts

The correct Answer is:
B
`8^(2n)-(62)^(2n+1)=(1+63)^(n)-(63-1)^(2n+1)`
`= (1+63)^(n) + (1-63)^(2n+1)`
`=[1+.^(n)C_(1)63+.^(n)C_(2)(63)^(2)+"......"+(63)^(n)]`
`+[1-.^((2x+1))C_(2)(63)^(2)+"...."-63^(2n+1)]`
`= 2+63K`
Hence, the remainder is 2.
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