Determine two consecutive multiples of 3 whose product is 270.

Given:

The product of two consecutive multiples of 3 is 270.

 To do:

We have to find the numbers.

Solution:

Let the numbers be $3x$ and $3x+3$.

According to the question,

$3x(3x+3)=270$

$9x^2+9x=270$

$9(x^2+x)=9\times30$

$x^2+x=30$

$x^2+x-30=0$

Solving for $x$ by factorization method, we get,

$x^2+6x-5x-30=0$

$x(x+6)-5(x+6)=0$

$(x+6)(x-5)=0$

$x+6=0$ or $x-5=0$

$x=-6$ or $x=5$

Considering positive values of $x$, we get,

$3x=3(5)=15$ and $3x+3=3(5)+3=15+3=18$

The required numbers are $15$ and $18$. 

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Updated on: 10-Oct-2022

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