Question
Find two consecutive multiples of $3$ whose product is $648$.
✓
Answer
Let the required consecutive multiples of $3$ be $3x$ and $3(x + 1)$.
Then, we have
$3x \times 3(x + 1) = 648$
$\Rightarrow 9x^2 + 9x - 648 = 0$
$\Rightarrow x^2 + x - 72 = 0$
$\Rightarrow x^2 + 9x - 8x - 72 = 0$
$\Rightarrow x(x + 9) - 8(x + 9) = 0$
$\Rightarrow (x + 9)(x - 8) = 0$
$\Rightarrow x + 9 = 0$ or $x - 8 = 0$
$\Rightarrow x = 9$ or $x = 8$
Since x is a positive integer, $x ≠ -9$
$\Rightarrow x = 8$
$\Rightarrow 3x = 3 \times 8 = 24$ and $3(x + 1) = 3(9) = 27$
Hence, the required consecutive multiples of $3$ are $24$ and $27$.
Then, we have
$3x \times 3(x + 1) = 648$
$\Rightarrow 9x^2 + 9x - 648 = 0$
$\Rightarrow x^2 + x - 72 = 0$
$\Rightarrow x^2 + 9x - 8x - 72 = 0$
$\Rightarrow x(x + 9) - 8(x + 9) = 0$
$\Rightarrow (x + 9)(x - 8) = 0$
$\Rightarrow x + 9 = 0$ or $x - 8 = 0$
$\Rightarrow x = 9$ or $x = 8$
Since x is a positive integer, $x ≠ -9$
$\Rightarrow x = 8$
$\Rightarrow 3x = 3 \times 8 = 24$ and $3(x + 1) = 3(9) = 27$
Hence, the required consecutive multiples of $3$ are $24$ and $27$.
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