# Find two consecutive multiples of 3 whose product is 648.

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,
$3x×3\left(x+1\right)=648\phantom{\rule{0ex}{0ex}}⇒9\left({x}^{2}+x\right)=648\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x=72\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-72=0$

x = 8           (Neglecting the negative value)

When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27

Hence, the required multiples are 24 and 27.

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