H5. A two-digit number is divided by the sum of its digits. The result is a number between 2.6 and 2.7 .
Find all of the possible values of the original two-digit number.
Solution
Let the number be ' $a b$ ' (where $a \neq 0$ ), which can be written as $10 a+b$.
The value we are concerned with is $\frac{10 a+b}{a+b}$.
We are given that $2.6\frac{73}{17} a=4 \frac{5}{17} a .
$$

From (2), we can deduce that $b>4 a$, so (since $b$ is a single digit) $a$ can be at most 2 . If $a=1$, combining (1) and (2) gives $4 \frac{5}{17}<b<4 \frac{5}{8}$, which is impossible (since $b$ is a digit and hence an integer).
If $a=2$, combining (1) and (2) gives $8 \frac{10}{17}<b<9 \frac{1}{4}$, which gives $b=9$.
These values of $a$ and $b$ give the number 29 .
We must check that 29 does have the property we are looking for, which we can do by dividing directly:
$$
\frac{29}{11}=2 \frac{7}{11}=2.6363 \ldots,
$$
which is indeed between 2.6 and 2.7 , so the only possible value of the two-digit number is 29 .