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: $0$
- $0 = \paren {4 + 4} - \paren {4 + 4}$
Four Fours: $1$
- $1 = \dfrac {4 + 4} {4 + 4}$
Four Fours: $2$
- $2 = \dfrac 4 4 + \dfrac 4 4$
Four Fours: $3$
- $3 = \dfrac {4 \times 4 - 4} 4$
Four Fours: $4$
- $4 = 4 - 4 + \sqrt 4 + \sqrt 4$
Four Fours: $5$
- $5 = \dfrac {4 \times 4 + 4} 4$
Four Fours: $6$
- $6 = \dfrac {4 + 4} 4 + 4$
Four Fours: $7$
- $7 = 4 + 4 - \dfrac 4 4$
Four Fours: $8$
- $8 = \paren {4 \times 4} - \paren {4 + 4}$
Four Fours: $9$
- $9 = 4 + 4 + \dfrac 4 4$
Four Fours: $10$
- $10 = 4 + 4 + 4 - \sqrt 4$
Four Fours: $11$
- $11 = \dfrac {4!} {\sqrt 4} - \dfrac 4 4$
Four Fours: $12$
- $12 = \paren {4 - \dfrac 4 4} \times 4$
Four Fours: $13$
- $13 = \dfrac {4!} {\sqrt 4} + \dfrac 4 4$
Four Fours: $14$
- $14 = 4 + 4 + 4 + \sqrt 4$
Four Fours: $15$
- $15 = 4 \times 4 - \dfrac 4 4$
Four Fours: $16$
- $16 = 4 + 4 + 4 + 4$
Four Fours: $17$
- $17 = 4 \times 4 + \dfrac 4 4$
Four Fours: $18$
- $18 = 4 \times 4 + \dfrac 4 {\sqrt 4}$
Four Fours: $19$
- $19 = 4 \times 4 + \sqrt {\dfrac 4 {. \dot 4} }$
Four Fours: $20$
- $20 = \paren {4 + \dfrac 4 4} \times 4$
Four Fours: $21$
- $21 = 4! - 4 + \dfrac 4 4$
Four Fours: $22$
- $22 = 4! - \dfrac {4 + 4} 4$
Four Fours: $23$
- $23 = 4! - 4 + \sqrt {\dfrac 4 {. \dot 4} }$
Four Fours: $24$
- $24 = 4 \times 4 + 4 + 4$
Four Fours: $25$
- $25 = \dfrac {4! \times 4 + 4} 4$
Four Fours: $26$
- $26 = 4! + \dfrac {4 + 4} 4$
Four Fours: $27$
- $27 = 4! + 4 - \dfrac 4 4$
Four Fours: $28$
- $28 = 4! + 4 + 4 - 4$
Four Fours: $29$
- $29 = 4! + 4 + \dfrac 4 4$
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: $34$
- $34 = 4! + 4 + 4 + \sqrt 4$
Four Fours: $35$
- $35 = 4! + \dfrac {4 + .4} {.4}$
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: $38$
- $38 = 4! + 4! - \dfrac 4 {.4}$
Four Fours: $39$
- $39 = 4! + 4! - \dfrac 4 {. \dot 4}$
Four Fours: $40$
- $40 = \paren {4 + 4 + \sqrt 4} \times 4$
Four Fours: $41$
- $41 = \dfrac {4 \times 4 + .4} {.4}$
Four Fours: $42$
- $42 = 4! + 4! - 4 - \sqrt 4$
Four Fours: $43$
- $43 = 4! + 4! - \dfrac {\sqrt 4} {.4}$
Four Fours: $44$
- $44 = \dfrac {4 + .4} {.4} \times 4$
Four Fours: $45$
- $45 = 4! + 4! - \sqrt {\dfrac 4 {. \dot 4} }$
Four Fours: $46$
- $46 = 4! + 4! - \dfrac 4 {\sqrt 4}$
Four Fours: $47$
- $47 = 4! + 4! - \dfrac 4 4$
Four Fours: $48$
- $48 = 4! + 4! + 4! - 4!$
Four Fours: $49$
- $49 = 4! + 4! + \dfrac 4 4$
Four Fours: $50$
- $50 = 4! + 4! + \dfrac 4 {\sqrt 4}$
Four Fours: $51$
- $51 = 4! + 4! + \sqrt {\dfrac 4 {. \dot 4} }$
Four Fours: $52$
- $52 = 4! + 4! + \sqrt 4 + \sqrt 4$
Four Fours: $53$
- $53 = \dfrac {4!} {. \dot 4} - \dfrac 4 4$
Four Fours: $54$
- $54 = 4! + 4! + 4 + \sqrt 4$
Four Fours: $55$
- $55 = \dfrac {4!} {. \dot 4} + \dfrac 4 4$
Four Fours: $56$
- $56 = 4! + 4! + 4 + 4$
Four Fours: $57$
- $57 = 4! + 4! + \dfrac 4 {.\dot 4}$
Four Fours: $58$
- $58 = 4! + 4! + \dfrac 4 {.4}$
Four Fours: $59$
- $59 = \dfrac {4!} {.4} - \dfrac 4 4$
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: $66$
- $66 = \dfrac {4!} {.4} + \dfrac {4!} 4$
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: $69$
- $69 = \dfrac {4! + \sqrt 4} {.4} + 4$
Four Fours: $70$
- $70 = \dfrac {4!} {.4} + \dfrac 4 {.4}$
Four Fours: $71$
- $71 = \dfrac {4! + 4 + .4} {.4}$
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: $74$
- $74 = 4! + 4! + 4! + \sqrt 4$
Four Fours: $75$
- $75 = \dfrac {4! + 4 + \sqrt 4} {.4}$
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: $82$
- $82 = \dfrac {4!} {. \dot 4} + 4! + 4$
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: $86$
- $86 = 4! \times 4 - \dfrac 4 {.4}$
Four Fours: $87$
- $87 = 4! \times 4 - \dfrac 4 {. \dot 4}$
Four Fours: $88$
- $88 = 4! \times 4 - 4 - 4$
Four Fours: $89$
- $89 = \dfrac {4! + \sqrt 4} {.4} + 4!$
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: $92$
- $92 = 4! \times 4 - \sqrt 4 - \sqrt 4$
Four Fours: $93$
- $93 = 4! \times 4 - \sqrt {\dfrac 4 {. \dot 4} }$
Four Fours: $94$
- $94 = 4! \times 4 - \dfrac 4 {\sqrt 4}$
Four Fours: $95$
- $95 = 4! \times 4 - \dfrac 4 4$
Four Fours: $96$
- $96 = 4! + 4! + 4! + 4!$
Four Fours: $97$
- $97 = 4! \times 4 + \dfrac 4 4$
Four Fours: $98$
- $98 = 4! \times 4 + \dfrac 4 {\sqrt 4}$
Four Fours: $99$
- $99 = 4! \times 4 + \sqrt {\dfrac 4 {. \dot 4} }$
Four Fours: $100$
- $100 = \dfrac 4 {.4} \times \dfrac 4 {.4}$
$\blacksquare$
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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