49. How many divisors do the following integers have: 32 ; $2^{10} ; 2^{m} ; 2^{10} \cdot 3 ; 2^{10} \cdot 3^{2} ; 2^{m} \cdot 3^{n} ; 2^{m} \cdot 3^{n} \cdot 5^{p}$ ?
49. The number $2^{10} \cdot 3^{2}$, for example, has the following divisors:

$1, 2, 2^{2}, 2^{3}, 2^{4}, 2^{3}, 2^{6}, 2^{7}, 2^{8}, 2^{9}, 2^{10}, 3, 2 \cdot 3, 2^{2} \cdot 3, 2^{3} \cdot 3, 2^{4} \cdot 3, 2^{3} \cdot 3, 2^{6} \cdot 3, 2^{7} \cdot 3, 2^{8} \cdot 3, 2^{9} \cdot 3, 2^{10} \cdot 3, 3^{2}, 2 \cdot 3^{2}, 2^{2} \cdot 3^{2}, 2^{3} \cdot 3^{2}, 2^{4} \cdot 3^{2}, 2^{3} \cdot 3^{2}, 2^{6} \cdot 3^{2}, 2^{7} \cdot 3^{2}, 2^{8} \cdot 3^{2}, 2^{9} \cdot 3^{2}, 2^{10} \cdot 3^{2}$.

The number of divisors is $(10+1) \cdot (2+1)=33$, i.e., the product of the exponents of the original divisors, each increased by 1.

The number $2^{m} \cdot 3^{n} \cdot 5^{p}$ has as many divisors as there are numbers of the form $2^{i} \cdot 3^{j} \cdot 5^{k}$, where the exponent $i$ takes on consecutive integer values from 0 to $m$ inclusive, the exponent $j$ from 0 to $n$, and the exponent $k$ from 0 to $p$. Numbers of the form $2^{i} \cdot 3^{j} \cdot 5^{k}$ are formed by combining each of the $m+1$ powers of 2 with each of the $n+1$ powers of 3 and with each of the $p+1$ powers of 5. Therefore, the total number of divisors is $(\boldsymbol{m}+1)(n+1)(p+1)$.