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
⇒ x = 8
⇒ 3x = 3 × 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
⇒ x = 8
⇒ 3x = 3 × 8 = 24 and 3(x + 1) = 3(9) = 27
Hence, the required consecutive multiples of 3 are 24 and 27.
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