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• The $(r-1)$’s complement of a number can be found using formula $(r^{n}-r^{-m}-N)$ where $r$ is base of the number $N$ having $n$ digits and $m$ digits in integral an fraction part respectively. We have been given some decimal numbers as shown below:
$(i) 325$                    $(ii) 325.893$
$(iii) –819$                 $(iv) –517.67$
How many $(r – 1)’s$ complement of above decimal numbers can be calculated using mentioned formula?

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i think number is given in decimal so 9's complement will be unique for every number.
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yes, but (i) and (iv) maynot be decimal.
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What is ans. given??

(iii)180  and (iv)482.32 ??

integer part n digits

fraction part m digits

N= positive given number.

$\left ( i \right ) 325$

$10^{3}-10^{0}-325=999-325=674$

$\left ( ii \right )325.893$

$10^{3}-10^{-3}-325.893=999.999-325.893=674.217$

Similar for $(iii)$ and $(iv)$ too.-ve willnot have any extra effect on numbers

Ref: Page 27 here

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