Question
Find three consecutive odd numbers whose sum is 147.
✓
Answer
Let the first odd number be x.
Let the second odd number be $(\text{x}+2).$
Let the third odd number be $(\text{x}+4).$
$\therefore\text{x}+(\text{x}+2)+(\text{x}+4)=147$
$\Rightarrow\text{x}+\text{x}+2+\text{x}+4=147$
$\Rightarrow3\text{x}+6=147$
$\Rightarrow3\text{x}=147-6$
$\Rightarrow3\text{x}=141$
$\Rightarrow\text{x}=\frac{141}{3}=47$
Therefore, the first odd number is 47.
Second odd number $=(\text{x}+2)=(47+2)=49$
Third odd number $=(\text{x}+4)=(47+4)$
Let the second odd number be $(\text{x}+2).$
Let the third odd number be $(\text{x}+4).$
$\therefore\text{x}+(\text{x}+2)+(\text{x}+4)=147$
$\Rightarrow\text{x}+\text{x}+2+\text{x}+4=147$
$\Rightarrow3\text{x}+6=147$
$\Rightarrow3\text{x}=147-6$
$\Rightarrow3\text{x}=141$
$\Rightarrow\text{x}=\frac{141}{3}=47$
Therefore, the first odd number is 47.
Second odd number $=(\text{x}+2)=(47+2)=49$
Third odd number $=(\text{x}+4)=(47+4)$
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