Problem 8.6.1. Even natural numbers $a$ and $b$ are such that $\operatorname{GCD}(a, b) + \operatorname{LCM}(a, b) = 2^{23}$. How many different values can $\operatorname{LCM}(a, b)$ take?
Answer: 22.

Solution. Note that LCM $(a, b) \vdots$ GCD $(a, b)$, therefore

$$
2^{23}=\text { GCD }(a, b)+\text { LCM }(a, b) \vdots \text { GCD }(a, b) .
$$

From this, it follows that GCD $(a, b)$ is a natural divisor of the number $2^{23}$.

At the same time, GCD $(a, b) \neq 1$ (since $a$ and $b-$ are even numbers), and GCD $(a, b) \neq 2^{23}$ (since GCD $\left.(a, b)=2^{23}-\operatorname{LCM}(a, b)<2^{23}\right)$. Thus, GCD $(a, b)$ takes one of the values $2,2^{2}, 2^{3}, \ldots, 2^{22}$.

Note that all these values are possible: to make GCD $(a, b)$ equal to $2^{k}$ for $1 \leqslant k \leqslant 22$, it is sufficient to choose $a=2^{k}$ and $b=2^{23}-2^{k}$. In this case, $b \vdots a$, so GCD $(a, b)=a$ and LCM $(a, b)=b$, that is, GCD $(a, b)+$ LCM $(a, b)=a+b=2^{23}$.