Four fours

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Four fours is a mathematical game. It is often used with older children to explore numbers and mathematical expressions, but many adults have also found it enjoyable.

The goal of four fours is to find the simplest mathematical expression for every whole number from 0 to some maximum, using only common mathematical symbols and the digit four (no other digit is allowed). Most versions of four fours require that each expression have exactly four fours, but some variations require that each expression have the minimum number of fours.

There are many variations of four fours; their primary difference is which mathematical symbols are allowed. Essentially all variations at least allow addition ("+"), subtraction ("−"), multiplication ("*" in ASCII), division ("/"), and parentheses, as well as concatenation (e.g., "44" is allowed). Most also allow the square root operation, factorial ("!"), exponentiation ("^" in ASCII), and the decimal digit ("."). Other operations allowed by some variations include subfactorial, ("!" before the number: !4 equals 9), overline (an infinitely repeated digit), an arbitrary root power, the gamma function (Γ(), where Γ(x) = (x − 1)!), and percent ("%"). Thus 4/4% = 100 and Γ(4)=6. A common use of the overline in this problem is for this value:

.\overline{4} = .4444... = 4/9

Typically the "log" operators are not allowed, since there's a way to trivially create any number using them. Paul Bourke credits Ben Rudiak-Gould with this description of how natural logarithms (ln()) can be used to represent any positive integer n as: n = − ln[ln(sqrt(sqrt(...(sqrt(4))...))) / ln(4)] / ln(4) where the number of nested sqrt() functions is twice n.

Additional variants (usually no longer called "four fours") replace the set of digits ("4, 4, 4, 4") with some other set of digits, say of the birthyear of someone. For example, a variant using "1975" would require each expression to use one 1, one 9, one 7, and one 5.

Here is a set of four fours solutions for the numbers 0 through 20, using typical rules. In a few cases some alternatives are shown, just to illustrate that there are alternatives, but many more alternatives are possible and not shown here:

  • 0 = 44 − 44 = 4 − 4 + 4 − 4
  • 1 = 44/44
  • 2 = 4/4 + 4/4
  • 3 = (4 + 4 + 4)/4
  • 4 = 4×(4 − 4) + 4
  • 5 = (4×4 + 4)/4
  • 6 = 4×.4 + 4.4 = 4 + (4+4)/4
  • 7 = 44/4 − 4 = 4 + 4 − (4/4)
  • 8 = 4 + 4.4 − .4 = 4 + 4 + 4 - 4
  • 9 = 4 + 4 + 4/4
  • 10 = 44/4.4 = 4 + sqrt(4) + sqrt(4) + sqrt(4)
  • 11 = 4/.4 + 4/4
  • 12 = (44 + 4)/4
  • 13 = 4! − 44/4
  • 14 = 4×(4 − .4) − .4
  • 15 = 44/4 + 4
  • 16 = .4×(44 − 4) = 4×4×4 / 4
  • 17 = 4×4 + 4/4
  • 18 = 44×.4 + .4 = 4×4 + 4 / sqrt(4)
  • 19 = 4! − 4 − 4/4
  • 20 = 4×(4/4 + 4)

Note that numbers with values less than one are not usually written with a leading zero. For example, "0.4" is usually written as ".4". This is because "0" is a digit, and in this puzzle only the digit "4" can be used.

A given number will generally have many possible solutions; any solution that meets the rules is acceptable. Some variations prefer the "fewest" number of operations, or prefer some operations to others. Others simply prefer "interesting" solutions, i.e., a surprising way to reach the goal.

Certain numbers, such as 113 and 123, are particularly difficult to solve under typical rules. For 113, Wheeler suggests Γ(Γ(4)) −(4! + 4)/4. For 123, Wheeler suggests the expression:

\sqrt{\sqrt{\sqrt{{\left(\sqrt{4}/.4\right)}^{4!}}}} - \sqrt{4}.

The first printed occurrence of this activity is in "Mathematical Recreations and Essays" by W. W. Rouse Ball published in 1892. In this book it is described as a "traditional recreation".

[edit] Excerpt from the solution to the five fives problem

In the table below, the notation .5... represents the value 5/9 (recurring decimal 5).

139 = ((((5+(5/5)))!/5)-5)
140 = (.5*(5+(5*55)))
141 = ((5)!+((5+(5+.5))/.5))
142 = ((5)!+((55/.5)/5))
143 = ((((5+(5/5)))!-5)/5)
144 = ((((55/5)-5))!/5)
145 = ((5*(5+(5*5)))-5)
146 = ((5)!+((5/5)+(5*5)))
147 = ((5)!+((.5*55)-.5))
148 = ((5)!+(.5+(.5*55)))
149 = (5+(((5+(5/5)))!/5))
150 = (5*(55-(5*5)))
151 = ((5)!+(55-((5)!/5)))
152 = ((((5)!-5)/.5...)-55)
153 = ((5)!+(55/rq((5*.5...))))
154 = (55+(55/.5...))
155 = (5*(55-((5)!/5)))
156 = ((((5)!/.5...)-55)-5)
157 = ((5/(.5...*(.5...-.5)))-5)
158 = ((((5)!/5)+55)/.5)
159 = (((5)!*.5)+(55/.5...))
160 = (((5*5)+55)/.5)

[edit] Excerpt from the solution to the six sixes problem

241 = ((.6+((6+6)*(6+6)))/.6)
242 = ((6*(6+(6*6)))-(6/.6))
243 = (6+((6*(.6*66))-.6))
244 = (.6...*(6+(6*(66-6))))
245 = ((((6)!+((6)!+66))/6)-6)
246 = (66+(6*((6*6)-6)))
247 = (66+((6+((6)!/.6...))/6))
248 = (6*(6+(6*(6-(.6.../6)))))
249 = (.6+(6*(6+((6*6)-.6))))
250 = (((6*(6*6))-66)/.6)
251 = ((6*(6+(6*6)))-(6/6))
252 = (66+(66+((6)!/6)))
253 = ((6/6)+(6*(6+(6*6))))
254 = ((.6...*((6*66)-6))-6)
255 = ((((6*6)+66)/.6)/.6...)
256 = (6*(6*(6-(6/(.6-6)))))
257 = (6+(((6)!+((6)!+66))/6))
258 = ((6)!-(66+(6*66)))
259 = ((((6*6)+((6)!/6))-.6)/.6)
260 = ((66+(((6)!/.6)/6))-6)

[edit] External links

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