smc

Note that $$\begin{array}{rl}18& =2\cdot 3^2,\\ 180& =2^2\cdot 3^2\cdot 5,\\ 45& =3^2\cdot 5\\ 15& =3\cdot 5.\end{array}$$ Let $n=2^a\cdot 3^b\cdot 5^c.$ It follows that: 1. From the least common multiple condition, we have $$\mathrm{lcm}(n,18)=\mathrm{lcm}(2^a\cdot 3^b\cdot 5^c,2\cdot 3^2)=2^{\max (a,1)}\cdot 3^{\max (b,2)}\cdot 5^{\max (c,0)}=2^2\cdot 3^2\cdot 5,$$ from which $a=2,b\in \{0,1,2\},$ and $c=1.$ 2. From the greatest common divisor condition, we have $$\gcd (n,45)=\gcd (2^2\cdot 3^b\cdot 5,3^2\cdot 5)=2^{\min (2,0)}\cdot 3^{\min (b,2)}\cdot 5^{\min (1,1)}=3\cdot 5,$$ from which $b=1.$ Together, we conclude that $n=2^2\cdot 3\cdot 5=60.$ The sum of its digits is $6+0=\boxed{6}.$