Henry Ernest Dudeney/Puzzles and Curious Problems/103 - Expressing Twenty-Four/Solution

by : $103$

 * Expressing Twenty-Four
 * In a book published in America was the following:
 * "Write $24$ with three equal digits, none of which is $8$.
 * (There are two solutions to this problem.)"


 * Of course, the answers given are $22 + 2 = 24$, and $3 ^3 - 3 = 24$.
 * Readers who are familiar with the old "Four Fours" puzzle, and others of the same class,
 * will ask why there are supposed to be only these solutions.
 * With which of the remaining digits is a solution equally possible?

Solution
For any digit $n$, we have:


 * $\dfrac n {\cdotp \dot n} = \dfrac n {n / 9} = 9$

where the notation $\cdotp \dot n$ is used to denote $\cdotp nnnnnn \ldots$

Hence:


 * $\sqrt {\dfrac n {\cdotp \dot n} } = 3$

This gives us:
 * $\paren {1 + \sqrt {\dfrac 1 {\cdotp \dot 1} } }! = \paren {1 + 3}! = 4! = 24$

where $!$ denotes the factorial sign.

We have an expression for each of $2$ and $3$.

Then:
 * $\paren {4 + 4 - 4}! = 4! = 24$

as one of many such similar and straightforward expressions.

We continue:
 * $\paren {5 - \dfrac 5 5}! = 4! = 24$

To accomplish $6$, we use the fact that or any digit $n$, we have:
 * $\dfrac n {\cdotp n} = \dfrac n {n / 10} = 10$

Hence:
 * $\paren {\dfrac 6 {\cdotp 6} - 6}! = \paren {10 - 6}! = 24$

Using the above technique, we see:


 * $\paren {7 - \sqrt {\dfrac 7 {\cdotp \dot 7} } }! = \paren {7 - 3}! = 4! = 24$

$8$ is simple:
 * $8 + 8 + 8 = 24$

$9$ falls to the factorial technique again:


 * $\paren {\sqrt 9 + \dfrac 9 9}! = \paren {3 + 1}! = 4! = 24$