Problem 3. The sum of the natural numbers $a$ and $b$ is equal to 10. What is the greatest, and what is the smallest value of $26 \cdot a + 27 \cdot b ?$
Solution. First method. From $a+b=10$ we obtain the following cases: $\quad a=9, b=1 ; \quad a=8, b=2 ; \quad a=7, b=3 ; \quad a=6, b=4 ; \quad a=5, b=5$; $a=4, b=6 ; a=3, b=7 ; a=2, b=8$ and $a=1, b=9$. In this case, we have

$$
\begin{aligned}
& 26 \cdot 9+27 \cdot 1=261,26 \cdot 8+27 \cdot 2=262,26 \cdot 7+27 \cdot 3=263 \\
& 26 \cdot 6+27 \cdot 4=264,26 \cdot 5+27 \cdot 5=265,26 \cdot 4+27 \cdot 6=266 \\
& 26 \cdot 3+27 \cdot 7=267,26 \cdot 2+27 \cdot 8=268 \text { and } 26 \cdot 1+27 \cdot 9=269
\end{aligned}
$$

Thus, the smallest value of $26 \cdot a+27 \cdot b$ is 261, and the largest value is 269.

Second method. We have

$$
\begin{aligned}
26 \cdot a+27 \cdot b & =26 \cdot a+(26+1) \cdot b \\
& =26 \cdot a+26 \cdot b+b \\
& =26(a+b) \\
& =26 \cdot 10+b=260+b
\end{aligned}
$$

The smallest value is obtained when $b$ is the smallest, i.e., when $b=1$, and this is the number 261. The largest value is obtained when $b$ is the largest, i.e., when $b=9$, and this is the number 269.