Let the unknown number be represented by the variable $x$. The problem states that the sum of its half ($\frac{x}{2}$), one-third ($\frac{x}{3}$), and one-fifth ($\frac{x}{5}$) exceeds the number itself by $12$. This relationship can be written as the following algebraic equation:
$ \left( \frac{x}{2} + \frac{x}{3} + \frac{x}{5} \right) - x = 12 $
To find the value of $x$, we first need to combine the fractional terms. The least common multiple (LCM) of the denominators $2$, $3$, and $5$ is $30$. We rewrite each fraction using this common denominator:
Substitute these equivalent fractions back into the equation:
$ \left( \frac{15x}{30} + \frac{10x}{30} + \frac{6x}{30} \right) - x = 12 $
Combine the numerators of the fractions:
$ \frac{31x}{30} - x = 12 $
To subtract $x$, express it with the same denominator ($x = \frac{30x}{30}$):
$ \frac{31x}{30} - \frac{30x}{30} = 12 $
Simplify the expression on the left side:
$ \frac{x}{30} = 12 $
Isolate $x$ by multiplying both sides of the equation by $30$:
$ x = 12 \times 30 $
$ x = 360 $
The number is $360$.
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