Cancelling, and a Cube Trick

1. A careless student reduces $\frac{16}{64}$ to $\frac14$ by “cancelling the sixes”, and is somehow right. He does the same to $\frac{26}{65}$ and $\frac{19}{95}$, again correctly. Are there other two-digit fractions, not merely multiples of these, where striking out a shared digit gives the true value?

2. The same student takes cube roots by adding digits: the cube root of $512$ is $5 + 1 + 2 = 8$, and of $4913$ is $4 + 9 + 1 + 3 = 17$, both correct. Find every number, beyond the trivial $1$, that equals the cube of its own digit sum.

Solution

Part 1. Apart from trivialities like $\frac{11}{11}$, there are exactly four such two-digit fractions: $$\frac{16}{64} = \frac14, \qquad \frac{19}{95} = \frac15, \qquad \frac{26}{65} = \frac25, \qquad \frac{49}{98} = \frac12,$$ where striking out the repeated digit (the $6$, $9$, $6$, $9$ in turn) leaves the correct reduced fraction.

To see there are no others, write such a fraction as $\frac{10a + b}{10b + c}$, where $b$ is the shared digit (the units of the top, the tens of the bottom). Wanting the careless cancellation $\frac{10a+b}{10b+c} = \frac{a}{c}$ to be true, cross-multiply: $$c(10a + b) = a(10b + c), \qquad \text{that is} \qquad 9ac = b(10a - c).$$ Here $a, b, c$ are digits from $1$ to $9$. Running through the few possibilities, this equation holds in exactly four cases, the $(a, b, c)$ being $(1,6,4)$, $(1,9,5)$, $(2,6,5)$ and $(4,9,8)$, which are precisely the four fractions above. So the schoolboy’s luck runs out after these.

Part 2. We want numbers equal to the cube of their digit sum. Running through the cubes and checking, the complete list, beyond $1^3 = 1$, is $$\begin{gathered} 8^3 = 512, \quad 17^3 = 4913, \quad 18^3 = 5832,\\ 26^3 = 17576, \quad 27^3 = 19683, \end{gathered}$$ whose digit sums are $8, 17, 18, 26, 27$ respectively. There are no others: a number with $d$ digits has digit sum at most $9d$, which grows far more slowly than its own cube root, so beyond a point a number always overtakes the cube of its digit sum.