Denote $n=p_1^{\alpha_1}...p_k^{\alpha_k}.5^m$ where $k>0$ since $d(5^m)=m+1\neq 5^{m-1}$ \[(\alpha_1+1)...(\alpha_k+1)(m+1)=5^{m-1}.p_1^{\alpha_1}...p_k^{\alpha_k}\]If $p_i>2,$ then $\frac{m+1}{5^{m-1}}=\Pi{\frac{p_i^{\alpha_i}}{(\alpha_i+1)}}\geq \Pi{\frac{3^{\alpha_i}}{\alpha_i+1}}>1\implies m=1\implies 2=m+1|n$ which contradicts with $p_i>2$. Hence $p_1=2$. \[(\alpha_1+1)...(\alpha_k+1)(m+1)=5^{m-1}.2^{\alpha_1}...p_k^{\alpha_k}\]$p_i^{\alpha_i}>\alpha_i+1$ thus, $(\alpha_1+1)(m+1)\geq 2^{\alpha_1}.5^{m-1}$ If $\alpha_1\geq 2,$ then $m=1\implies 2(\alpha_1+1)\geq 2^{\alpha_1}\implies 3\geq \alpha_1$. $i)\alpha_1=3,m=1$ Then \[(\alpha_2+1)...(\alpha_k+1)=p_2^{\alpha_2}...p_k^{\alpha_k}\]But $p_i^{\alpha_i}>\alpha_i+1$ hence $k=1\implies \boxed{n=40}$ $ii)\alpha_1=2,m=1$ Then \[3(\alpha_2+1)...(\alpha_k+1)=2p_2^{\alpha_2}...p_k^{\alpha_k}\]\[\Pi{\frac{p_i^{\alpha_i}}{\alpha_i+1}}=\frac{3}{2}\]We have $p_i^{\alpha_i}\geq 3^{\alpha_i}\geq \frac{3}{2}.(\alpha_i+1)$ so $\frac{3}{2}=LHS\geq (\frac{3}{2})^{k-1}\implies k=1,2$. If $k=1,$ then $n=20$ which doesnot hold the condition. So $k=2$. $\frac{p_2^{\alpha_2}}{\alpha_2+1}=\frac{3}{2}\implies p_2=3$ and $\alpha_2=1$ which gives that $\boxed{n=60}$ $iii)\alpha_1=1$ \[(\alpha_2+1)...(\alpha_k+1)(m+1)=5^{m-1}p_2^{\alpha_2}...p_k^{\alpha_k}\]Since $p_i^{\alpha_i}>\alpha_i+1$ we get $m+1>5^{m-1}\implies m=1$. \[2(\alpha_2+1)...(\alpha_k+1)=p_2^{\alpha_2}...p_k^{\alpha_k}\]But $2|p_2^{\alpha_2}...p_k^{\alpha_k}$ is impossible.$\blacksquare$

Sol:- Let $n=(5^k)t$ such that $gcd(5,t)=1$. Then $5d(n)=5(k+1)d(t)=(5^k)t \implies (k+1)d(t)=(5^{k-1})t.$ Since $d(t) \leq t$, $k=1$ is forced.Thus $2d(t)=t.$ If $t$ has an odd prime divisor $p>3$ such that $t=(p^r)v$ and $gcd(p,v)=1$. Then $2(r+1)d(v)=2d(t)=t=(p^r)v$. But $d(v)\leq v$ and $2(r+1)<p^r$ , contradicting the equality. Thus let $t=(2^g)(3^h)$. $2d(t)=t \iff 2(g+1)(h+1)=(2^g)(3^h)$. Note that the only solutions are $(g,h)=(2,1);(3,0)$ since otherwise $h>1$ makes RHS larger. Thus $n=60,40$ are the only solutions.

Easy: $5<p_i$ cant divide $n$ So===>>> $n=2^a3^b5^c$ $(a+1)(b+1)(c+1)=2^a3^b5^c$ There is a bound for $a,b,c<4$ Checking,checking...$n=40,60$...