If the decimal 0.d25d25d25ldots is expressibl | vedclass

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(D) Let $x = 0.d25d25d25\ldots$ $(i)$Multiplying by $1000$,we get $1000x = d25.d25d25\ldots$ (ii)Subtracting $(i)$ from (ii),we get $999x = d25$,where

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$34$

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(D) Let $x = 0.d25d25d25\ldots$ $(i)$Multiplying by $1000$,we get $1000x = d25.d25d25\ldots$ (ii)Subtracting $(i)$ from (ii),we get $999x = d25$,where $d25$ represents the number $100d + 25$.So,$x = \frac{100d + 25}{999}$.Given $x = \frac{n}{27}$,we have $\frac{n}{27} = \frac{100d + 25}{999}$.Multiplying both sides by $27$,we get $n = \frac{100d + 25}{37}$.For $n$ to be an integer,$100d + 25$ must be divisible by $37$.Testing values for $d \in \{1, 2, \ldots, 9\}$:If $d = 9$,$100(9) + 25 = 925$.$925 / 37 = 25$,which is an integer.Thus,$d = 9$ and $n = 25$.Therefore,$d + n = 9 + 25 = 34$.

If the decimal $0.d25d25d25\ldots$ is expressible in the form $n/27$,then $d+n$ must be

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