179. Find $N=p^{3}$, if $\varphi(N)=100$.

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179. From the equation $p^{3}\left(1-\frac{1}{p}\right)=100$ we get $p^{2}(p-1)=$ $=25 \cdot 4$, from which it is clear that $p=5$.

We can also do it this way: $p^{3}-p^{2}-100=0, p^{3}-125-\left(p^{2}-25\right)=$ $=0,(p-5)\left(p^{2}+4 p+20\right)=0$, from which $p=5$.

Answer. 125.