Four Fours

Puzzle

Using exactly $4$ instances of the number $4$, it is possible to write an expression for all positive integers from $0$ to $100$, using whatever arithmetical operations are necessary.

Further Results

It is also possible to populate the table for numbers from $101$ to $200$, although for $197$ we appear to need to use the floor function.

Solution

Four Fours: $8$

$8 = \paren {4 \times 4} - \paren {4 + 4}$

Four Fours: $11$

$11 = \dfrac {4!} {\sqrt 4} - \dfrac 4 4$

Four Fours: $13$

$13 = \dfrac {4!} {\sqrt 4} + \dfrac 4 4$

Four Fours: $19$

$19 = 4 \times 4 + \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $23$

$23 = 4! - 4 + \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $30$

$30 = \dfrac 4 {.4} \times \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $31$

$31 = 4! + 4 + \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $32$

$32 = \dfrac {4 \times 4 \times 4} {\sqrt 4}$

Four Fours: $33$

$33 = 4! + \paren {\sqrt {\dfrac 4 {. \dot 4} } }^{\sqrt 4}$

Four Fours: $36$

$36 = \paren {\paren {\sqrt {\dfrac 4 {. \dot 4} } }^{\sqrt 4} } \times 4$

Four Fours: $37$

$37 = 4! + \dfrac {4! + \sqrt 4} {\sqrt 4}$

Four Fours: $40$

$40 = \paren {4 + 4 + \sqrt 4} \times 4$

Four Fours: $45$

$45 = 4! + 4! - \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $51$

$51 = 4! + 4! + \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $53$

$53 = \dfrac {4!} {. \dot 4} - \dfrac 4 4$

Four Fours: $55$

$55 = \dfrac {4!} {. \dot 4} + \dfrac 4 4$

Four Fours: $60$

$60 = 4 \uparrow \sqrt {\dfrac 4 {. \dot 4} } - 4$

Four Fours: $61$

$61 = \dfrac {4!} {.4} + \dfrac {.4} {.4}$

Four Fours: $62$

$62 = \dfrac {4!} {.4} + \dfrac 4 {\sqrt 4}$

Four Fours: $63$

$63 = \dfrac {4!} {.4} + \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $64$

$64 = \paren {4 + 4} \times \paren {4 + 4}$

Four Fours: $65$

$65 = \dfrac {4!} {.4} + \dfrac {\sqrt 4} {.4}$

Four Fours: $67$

$67 = \dfrac {4! + \sqrt 4} {.4} + \sqrt 4$

Four Fours: $68$

$68 = 4 \uparrow \sqrt {\dfrac 4 {. \dot 4} } + 4$

Four Fours: $72$

$72 = 4! \times \paren {4 - \dfrac 4 4}$

Four Fours: $73$

$73 = \dfrac {4! + 4! + \sqrt {. \dot 4} } {\sqrt {. \dot 4} }$

Four Fours: $76$

$76 = \dfrac 4 {.4} \uparrow \sqrt 4 - 4!$

Four Fours: $77$

$77 = \paren {\dfrac 4 {. \dot 4} }^{\sqrt 4} - 4$

Four Fours: $78$

$78 = \dfrac {4! + 4! + 4} {\sqrt {. \dot 4} }$

Four Fours: $79$

$79 = \paren {\dfrac 4 {. \dot 4} }^{\sqrt 4} - \sqrt 4$

Four Fours: $80$

$80 = \dfrac 4 {.4} \times \paren {4 + 4}$

Four Fours: $81$

$81 = \dfrac 4 {. \dot 4} \times \dfrac 4 {. \dot 4}$

Four Fours: $83$

$83 = \paren {\dfrac 4 {. \dot 4} }^{\sqrt 4} + \sqrt 4$

Four Fours: $84$

$84 = \paren {4! + 4} \times \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $85$

$85 = \paren {\dfrac 4 {. \dot 4} }^{\sqrt 4} + 4$

Four Fours: $87$

$87 = 4! \times 4 - \dfrac 4 {. \dot 4}$

Four Fours: $90$

$90 = \dfrac 4 {.4} \times \dfrac 4 {. \dot 4}$

Four Fours: $91$

$91 = 4! \times 4 - \dfrac {\sqrt 4} {.4}$

Four Fours: $93$

$93 = 4! \times 4 - \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $99$

$99 = 4! \times 4 + \sqrt {\dfrac 4 {. \dot 4} }$

Four Fours: $100$

$100 = \dfrac 4 {.4} \times \dfrac 4 {.4}$

$\blacksquare$

This article is complete as far as it goes, but it could do with expansion.
In particular: It is tempting to document Professor Stewart's definitive though whimsical analysis from his Casebook.
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Lemmata

It is a useful exercise to determine which numbers can be made from $1$, $2$ and $3$ instances of $4$, and how this can be done.

$\Box$

Glossary

Symbols used in the Four Fours are defined as follows:

$\ds . \dot 4$

$:=$

$\ds 0.44444 \ldots$

$.4$ recurring, equal to $\dfrac 4 9$

$\ds \sqrt 4$

$:=$

$\ds 2$

square root of $4$

$\ds 4!$

$:=$

$\ds 1 \times 2 \times 3 \times 4$

$4$ factorial

$\ds \map \Gamma 4$

$:=$

$\ds 1 \times 2 \times 3$

gamma function of $4$

$\ds a \uparrow b$

$:=$

$\ds a^b$

Knuth uparrow notation

$\ds \floor x$

$:=$

$\ds \text {largest integer not greater than $x$}$

floor function of $x$

$\ds \map \pi x$

$:=$

$\ds \text {number of primes less than $x$}$

prime-counting function of $x$

Historical Note

As Henry Ernest Dudeney put it in his $58$. - The Two Fours in his $1926$ Modern Puzzles:

Dudeney reports:

I am perpetually receiving inquiries about the old "Four Fours" puzzle.

I published it in $1899$, but have since found that it first appeared in the first volume of Knowledge ($1881$).

It has since been dealt with at some length by various writers.

Martin Gardner locates that original article in Knowledge as being the December $30$th issue.

He then goes on to cite a number of more recent discussions on the subject, including his exposition in his own column in Scientific American for January $1964$.

He finishes with a reference to an article by Donald Ervin Knuth in which it is proved that all positive integers up to $208$ can be expressed with nothing but one $4$, instances of the square root sign, the factorial sign, and the floor function.

Because it is possible to express $4$ using four $4$s, it is hence possible to represent $113$ using four $4$s, although this representation may be somewhat complicated.

Ian Stewart's admittedly whimsical Professor Stewart's Casebook of Mathematical Mysteries from $2014$ delivers a deep analysis of the problem, delivering the final solution as the punchline to a particularly pointless shaggy-dog story.

Sources

1980: Angela Dunn: Mathematical Bafflers (revised ed.) ... (previous) ... (next): $1$. Say it with Letters: Algebraic Amusements: Four Fours

Four Fours

Puzzle

Solution

Sources

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