Question

The sum of three consecutive multiples of 3 is 108. Find the numbers .

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Sum of three consecutive multiples of 3 is 108}$

$\underline{\textbf{To find:}}$

$\textsf{The numbers}$

$\underline{\textbf{Solution:}}$

$\textsf{Let the three consecutive multiples of 3 be}$

$\textsf{3x, 3(x+1) and 3(x+2)}$

$\textsf{As per given data, their sum is 108}$

$\implies\mathsf{3x+3(x+1)+3(x+2)=108}$

$\implies\mathsf{3x+3x+3+3x+6=108}$

$\implies\mathsf{9x+9=108}$

$\implies\mathsf{9x=99}$

$\implies\mathsf{x=\dfrac{99}{9}}$

$\implies\mathsf{x=11}$

$\mathsf{3x=3(11)=33}$

$\mathsf{3(x+1)=3(12)=36}$

$\mathsf{3(x+2)=3(13)=39}$

$\therefore\textbf{The three consecutive multiples of 3 are 33, 36 and 39}$

Answer 2

The number are $33, 36$ and $39.$

Given - sum of three consecutive multiples of $3$ is $108.$

Find the numbers.

Solution-Let the first number be $x$

If $x$ is a multiple of $3$, then the next two multiples are $x+3$ and $x+6$

So, three consecutive multiples of $3$ are $x, x+3, x+6.$

$x+(x+3)+(x+6)=108\\3x+9=108\\3x=99\\x=33$

So,

$x+3=33+x=36\\x+6=33+6=39$

Hence, the three numbers are $33, 36, 39$

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