# 24 with One Digit

This page summarizes a tweeter.com exchange started by Jim Wilder with the question:

Using the same digit three times, can you make \(24\)?

This problem happened to have four solutions: \(22 + 2\), \(3^{3} - 3\), \(4! + 4-4\), and \(8 + 8 + 8\).

The problem can be expanded in several ways. For example, Ben Vitale has observed that in base 5, \((2\times 22)_{5} = (44)_{5} = (24)_{10}\).

Returning to base \(10\), there are nine nonzero digits each of which, in principle, could alone provide a representation of \(24\). What is the least number of digits required to obtain all nine representation? Seven is definitely more than enough, because a single formula works for each of the nine digits:

For each \(n = 1, 2, ..., 9\),

\(\displaystyle\frac{nn + nn + n + n}{n} = 24,\)

where \(nn = 10n + n\).

For example, \(\displaystyle\frac{77 + 77 + 7 + 7}{7} = 24\).

Mr Drake - a math teacher from Australia - reduced the number of allowed digits to five by coming up with different formulas for different digits:

\( \begin{align} 24 &= (11+1)\cdot (1+1) \\ &= 22+2^{2/2} \\ &= 3\cdot (3\cdot 3-3/3) \\ &= 4\cdot (4+4)-(4+4) \\ &= 5\cdot 5-5^{5-5} \\ &= (6+6)\cdot (6+6)/6 \\ &= ((7+7+7+7)/7)! \\ &= (8+8+8)\cdot 8/8 \\ &= ((9\cdot 9-9)\cdot \sqrt{9})/9 \end{align} \)

What can be done with four digits? Here are some - probably partial - results:

\( \begin{align} 24 &= 3\cdot 3\cdot 3 - 3 \\ &= 4\cdot 4+4+4 \\ &= 5\cdot 5 - 5/5 \\ &= 6+6+6+6 \\ &= 7+7+7/.7 \\ &= 9+9 +\sqrt{9} +\sqrt{9} \\ \end{align} \)

If the *floor function* - \(\lfloor x\rfloor\) - is allowed then the list can be easily expanded. For example,

\( \begin{align} 24 &= 22 + 2\cdot\lfloor\sqrt{2}\rfloor \\ &= 8+8+8\cdot \lfloor\sqrt{\sqrt{8}}\rfloor \\ \end{align} \)

Mr Drake has completed the list. His expression with four \(1\)s also includes the floor function:

\( \begin{align} 24 &= (\lfloor\sqrt{11}\rfloor + 1\cdot 1)! \end{align} \)

In his turn, Ben Vitale found a simpler expression and also incorporated an example with the digit \(0\):

\( \begin{align} 24 &= (1 + 1 + 1 + 1)! \\ &= (0! + 0! + 0! + 0!)! \end{align} \)

Next, Mr Drake noticed that with the ceiling function - \(\lceil x\rceil\) - there is an universal formula that works for any three digits:

\( \begin{align} 24 &= \Bigg\lceil\sqrt{\frac{nn}{n}}\Bigg\rceil! \end{align} \)

A combination of both functions let Mr. Drake to a solution of the problem with just two digits:

\( \begin{align} 24 &= \Big\lceil\sqrt{11}\Big\rceil ! \\ &= (2+2)! \\ &= \lceil 3.3\rceil ! \\ &= (\sqrt{4}\cdot\sqrt{4})! \\ &= 5!/5 \\ &= (\big\lfloor\sqrt{6}\big\rfloor\cdot\big\lfloor\sqrt{6}\big\rfloor)! \\ &= (\big\lfloor\sqrt{7}\big\rfloor\cdot\big\lfloor\sqrt{7}\big\rfloor)! \\ &= (\big\lfloor\sqrt{8}\big\rfloor\cdot\big\lfloor\sqrt{8}\big\rfloor)! \\ &= \Big\lceil\sqrt{9.9}\Big\rceil! \\ \end{align} \)

As an extreme example of making do with a single digit, Mr. Drake found that

\( \begin{align} 24 &=\Bigg\lceil\sqrt{\sqrt{\Big\lfloor\sqrt{\sqrt{(3!)!}}\Big\rfloor !}}\Bigg\rceil ! \end{align} \)

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