Problem 7.3. In the example with fractions, some two-digit natural numbers have been replaced by the letters $A$ and $B$.

$$
\frac{A-5}{A}+\frac{4}{B}=1
$$

(a) (2 points) What is the smallest value that $A$ can take?

(b) (2 points) What is the largest value that $B$ can take?
# Answer:

(a) (2 points) 15.

(b) (2 points) 76.

Solution. According to the condition,

\[
1=\frac{A-5}{A}+\frac{4}{B}=1-\frac{5}{A}+\frac{4}{B}
\]

from which we obtain \(\frac{A}{5}=\frac{B}{4}\) and \(4A=5B\). It follows that for some integer \(k\), the equalities \(A=5k\) and \(B=4k\) hold.

Since \(B=4k \geqslant 10\), we get \(k \geqslant 3\) and \(A=5k \geqslant 5 \cdot 3=15\). The value \(A=15\) is possible if \(B=12\).

Since \(A=5k \leqslant 99\), we get \(k \leqslant 19\) and \(B=4k \leqslant 4 \cdot 19=76\). The value \(B=76\) is possible if \(A=95\).