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
$x \times(x+2)=483$
$\Rightarrow x^2+2 x-483=0$
$\Rightarrow x^2+23 x-21 x-483=0$
$\Rightarrow x(x+23)-21(x+23)=0$
$\Rightarrow(x+23)(x-21)=0$
$\Rightarrow x+23=0 \text { or } x-21=0$
$\Rightarrow x=-23 \text { or } x=21$
$\text { Since } x \text { is a positive integer, } x \neq-23$
$\Rightarrow x=21$
$\Rightarrow x+2=21+2=23$
Hence, the required consecutive positive odd integers are $21$ and $23.$
Then, we have
$x \times(x+2)=483$
$\Rightarrow x^2+2 x-483=0$
$\Rightarrow x^2+23 x-21 x-483=0$
$\Rightarrow x(x+23)-21(x+23)=0$
$\Rightarrow(x+23)(x-21)=0$
$\Rightarrow x+23=0 \text { or } x-21=0$
$\Rightarrow x=-23 \text { or } x=21$
$\text { Since } x \text { is a positive integer, } x \neq-23$
$\Rightarrow x=21$
$\Rightarrow x+2=21+2=23$
Hence, the required consecutive positive odd integers are $21$ and $23.$
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