The lowest number that is a multiple of both 60 and the integer n is 180. Find the smallest possible value of n.

$$180=2^2.3^2.5$$ $$60=2^2.3.5$$ so $n$ must have the same primes as $180$ $$n=2^i3^j5^k$$ where $i,j\leq 2$ and $k\leq 1$

$60$ is not a multiple of $n$, otherwise their $lcm$ would have been $60$. Therefore, we must have $$n\equiv 0 \pmod {3^2}$$ It follows the smallest $n$: $$n=9$$

$$\operatorname{lcm}(60,n) = 180$$

$$\operatorname{lcm}(2^2 \cdot 3^1 \cdot 5^1, 2^{n_2} \cdot 3^{n_3} \cdot 5^{n_5}) = 2^2 \cdot 3^2 \cdot 5^1$$

$$ 2^{\max(2,n_2)} \cdot 3^{\max(1,n_3)} \cdot 5^{\max(1,n_5)} = 2^2 \cdot 3^2 \cdot 5^1$$

\begin{align} \max(2,n_2) &= 2\\ \max(1,n_3) &= 2 \\ \max(1,n_5) &= 1 \end{align}

The smallest possible exponents are thus

\begin{align} n_2 &= 0\\ n_3 &= 2 \\ n_5 &= 0 \end{align}

So $ n = 2^0 \cdot 3^2 \cdot 5^0 = 9$.

$n$ is the smallest number that divides 180 but not 60. (If it divided both, 180 couldn't be the least common multiple because $60 < 180$.)

The primes in $60$ are $(2, 2, 3, 5)$ and the ones in $180$ are $(2, 2, 3, 3, 5)$.

If $n$ divides $180$, its primes are contained in $(2, 2, 3, 3, 5)$.

The smallest list contained in $(2, 2, 3, 3, 5)$ but not in $(2, 2, 3, 5)$ is $(3, 3)$, and $3 \cdot 3 = 9$.