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**Soroban** Hello, FlacidCelery!

We have: .$\displaystyle \begin{array}{ccccc}A &=&30 &=& 2\cdot3\cdot5 \\ B &=& 72 &=& 2^3\cdot3^2 \end{array}$

$\displaystyle A$ times $\displaystyle N$ is divisible by $\displaystyle B$: .$\displaystyle \frac{2\cdot3\cdot5\cdot N}{2^3\cdot3^2}$ is an integer.

This reduces to: .$\displaystyle \frac{5N}{12}$ . . . Hence. $\displaystyle N$ is a multiple of 12.

$\displaystyle B$ times $\displaystyle N$ is divisible by $\displaystyle A$: .$\displaystyle \frac{2^3\cdot3^2\cdot N}{2\cdot3\cdot5}$ is an integer.

This reduces to: .$\displaystyle \frac{2^2\cdot3\cdot N}{5}$ . . . Hence, $\displaystyle N$ is a multiple of 5.

The least number which is a multiple of 12 *and* a multiple of 5 is: .$\displaystyle N \:=\:60$