Question
Find two consecutive positive odd integers whose product is 483.
✓
Answer
Let the required consecutive positive odd integers be x and (x + 2).
Then, we have
$x \times (x + 2) = 483$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x + 23)(x - 21) = 0$
$\Rightarrow x + 23 = 0 or x - 21 = 0$
$\Rightarrow x = -23 or x = 21$
Since x is a positive integer, $x \neq -23$
$\Rightarrow x = 21$
$\Rightarrow x + 2 = 21 + 2 = 23$
Hence, the required consecutive positive odd integers are 21 and 23.
Then, we have
$x \times (x + 2) = 483$
$\Rightarrow x^2 + 2x - 483 = 0$
$\Rightarrow x^2 + 23x - 21x - 483 = 0$
$\Rightarrow x(x + 23) - 21(x + 23) = 0$
$\Rightarrow (x + 23)(x - 21) = 0$
$\Rightarrow x + 23 = 0 or x - 21 = 0$
$\Rightarrow x = -23 or x = 21$
Since x is a positive integer, $x \neq -23$
$\Rightarrow x = 21$
$\Rightarrow x + 2 = 21 + 2 = 23$
Hence, the required consecutive positive odd integers are 21 and 23.
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