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# divide $184$ into two parts such that one-third of one part may exceed one-seventh of the other part by $8$.

**Given: **A number $184$.

**
**

**To do: **To divide $184$ into two parts such that one-third of one part may exceed one-seventh of the other part by $8$.

**Solution:**

**
**

Let the two parts be $x$ and $184-x$.

$\therefore \frac{1}{3} x-\frac{1}{7}(184-x)=8$

$\Rightarrow \frac{7x-552+3x}{21}=8$

$\Rightarrow 10x-552=168$

**
**

$\Rightarrow 10x=552+168$

$\Rightarrow 10x=720$

$\Rightarrow x=\frac{720}{10}$

$\Rightarrow x=72$

Thus, the numbers are: $72$ and $184-72=112$.

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