Question

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Answer 1

We know successive multiples of 9 differ by 9, i.e. they have a common difference of 9.

Let the three consecutive multiples of 9 be $\displaystyle\sf{x-9,\ x}$ and $\displaystyle\sf {x+9.}$

Sum of these three multiples is 999.

$\displaystyle\longrightarrow\sf{(x-9)+x+(x+9)=999}$

$\displaystyle\longrightarrow\sf{x-9+x+x+9=999}$

$\displaystyle\longrightarrow\sf{3x=999}$

$\displaystyle\longrightarrow\sf{x=\dfrac {999}{3}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {x=333}}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {x-9=324}}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {x+9=342}}}$

Hence the three multiples are 324, 333 and 342.

• 324 = 9 × 36

• 333 = 9 × 37

• 342 = 9 × 38

• 324 + 333 + 342 = 999

Answer 2

$\rm Let \ us \ take \ three \ consecutive \ multiples \ of \ nine \ as \ \bf{x} \ , \ \bf{(x+9)} \ \rm and \ \bf{(x+18)}\ .\\\\\rm \longrightarrow x+(x+9)+(x+18)=999\\\\\rm\longrightarrow 3x+27=999\\\\\rm\longrightarrow 3x=972\\\\\rm\longrightarrow\underline{\bf x=324}$

$\rm x+9=324+9=333\\\\\rm x+18 = 324+18=342$

The consecutive multiples of 9 are 324 , 333 and 342 .