149. Find two numbers if their sum is 60, and the sum of their greatest common divisor and least common multiple is 84.

Chapter IV.

## THE NUMBER OF ALL DIVISORS OF A GIVEN NUMBER AND THEIR SUM

The number of all divisors of a given number $N=P_{1}^{\lambda_{1}} P_{2}^{\lambda_{2}} \ldots \ldots P_{n}^{\lambda}$, including one and the given number, is

$$
\left(\lambda_{1}+1\right)\left(\lambda_{2}+1\right) \ldots\left(\lambda_{n}+1\right)
$$

This can be proven as follows:

Consider the product:

$$
\begin{gathered}
\left(p_{1}^{\lambda_{1}}+p_{1}^{\lambda_{1}-1}+\ldots+1\right)\left(p_{2}^{\lambda_{2}}+\right. \\
\left.+p_{2}^{\lambda_{2}-1}+\ldots+1\right) \ldots\left(p_{n}^{\lambda_{n}}+p_{n}^{\lambda_{n}-1}+\ldots+1\right)
\end{gathered}
$$

If we expand the brackets, we get a polynomial whose terms represent all the divisors of the given number $N$. Clearly, the number of these terms is $\left(\lambda_{1}+1\right)\left(\lambda_{2}+1\right) \ldots\left(\lambda_{n}+1\right)$. The product (1) in expanded form is the sum of all divisors of the number $N$, and this sum is equal to:

$$
\frac{p_{1}^{\lambda_{1}+1}-1}{p_{1}-1} \cdot \frac{p_{2}^{\lambda_{2}+1}-1}{p_{2}-1} \cdot \ldots \cdot \frac{p_{n}^{\lambda_{n} n+1}-1}{p_{n}-1}
$$
149. Answer. 24 and 36.