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From a natural number $$3$$ is subtracted, then the result is divided by $$4$$ and the outcome is increased by $$4$$. The whole result is then divided by $$5$$, and the final resulting number is $$2$$. Then the natural number taken in the beginning is
Let the number be $$n$$. The operations give $$\frac{\frac{n-3}{4}+4}{5}=2$$, so $$\frac{n-3}{4}=6$$ and $$n=27$$. Since $$27=3^3$$, the number is a perfect cube.
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