Euler is haunting (2)

Since we want $5\phi(n) = 2n$ , we must have $n$ divisible by $5$ . Thus $n = 5^bm$ where $b,m \ge 1$ and $5,m$ are coprime, and hence $2 \times 5^bm \; = \; 2n \; = \; 5\phi(n) \; = \; 5\times 4\times 5^{b-1} \times \phi(m)$ and so $2\phi(m) = m$ . This means that $m$ is even, and so $m = 2^ak$ where $a,k \ge 1$ and $k,10$ are coprime, and hence $2^ak \; = \; m \; = \; 2\phi(m) \; = \; 2\times 2^{a-1} \times \phi(k)$ so that $\phi(k) = k$ . Thus we obtain $k=1$ , and so $n = 2^a5^b$ for $a,b \ge 1$ . There are therefore $\boxed{19}$ such values of $n$ , namely $10,20,40,50,80,100,160,200,250,320,400,500,640,800,1000,1250,1280,1600,2000$

HINT : Try and look at what could be the prime factors of the number.!