160. Find a number of the form $2^{l} 3^{m}$, if the sum of all its divisors is 403.
160. $\left(2^{l+1}-1\right)\left(3^{m+1}-1\right)=2 \cdot 13 \cdot 31$. This is only possible when $\left\{\begin{array}{l}2^{l+1}-1=31 \\ 3^{m+1}-1=26\end{array}\right.$

Answer. 144.