Hamilton 2018 Q5

Let the number be ‘\(ab\)’ (where \(a \neq 0\)), which can be written as \(10a + b\).

The value we are concerned with is \(\dfrac{10a + b}{a + b}\).

We are given that \(2.6 < \dfrac{10a + b}{a + b} < 2.7\).

Since \(a + b\) must be positive (since both \(a\) and \(b\) are positive), we can multiply through by \((a + b)\) without changing the direction of the inequality signs, to give:

The first of these two inequalities leads to

which can be rearranged to give

The second inequality leads to

which can be rearranged to give

From (2), we can deduce that \(b > 4a\), so (since \(b\) is a single digit) \(a\) can be at most 2.

If \(a = 1\), combining (1) and (2) gives \(4\tfrac{5}{17} < b < 4\tfrac{5}{8}\), which is impossible (since \(b\) is a digit and hence an integer).

If \(a = 2\), combining (1) and (2) gives \(8\tfrac{10}{17} < b < 9\tfrac{1}{4}\), 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:

which is indeed between 2.6 and 2.7, so the only possible value of the two-digit number is 29.