r's complement practise

Question

10's complement of $5690$

10's complement of $(5690)_8$

8's complement of $(6250)_8$

8's complement of $(6250)_{16}$

Comments

1. 9999-5690=4309+1=4310

2. $(5690)_{8}$ is illegal as 9 cannot be present in base 8.

3. 7777-6250 = 1527 + 1 = $(1530)_{8}$

4. $(6250)_{16} = (061120)_{8}$

777777 - 061120 = 716657 + 1 = $(716660)_{8}$

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Thanks a lot for quick reply :)

Small doubts 

1 is same 

What if 2nd is $(5660)_8$

convert it into base 10 then do same as 1 ??

3 is same 

To find any r's complement then base should be same as r ryt ??

That's why you converted 4th ??

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8's complement of $(5660)_{8}$ = $(2120)_{8}$

Here 7+1 = 0 not 8 as it is in base 8. No need to convert to base 10. Just subtract the highest digit possible in that base from each digit and then add 1 to get r's complement.

r's complement of a number = $r^{n}-y$

Reference on complementary number system:

https://en.wikipedia.org/wiki/Method_of_complements#Numeric_complements

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7777-6250 = 1527 + 1 = (1530)8 is it correct ??

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@air1ankit

Yes the addition is correct as it is in base 8, not base 10