Four Fours problem: Any number with One Four

The four fours puzzle is a recreational puzzle that asks: can you create an expression using four fours and some arithmetic operations that evaluates to some number n? The problem often specifies some operations such as addition, subtraction, multiplication, division, exponentiation, concatenation (e.g. 44), negation, decimal point, repeating decimal, square root, nth root, and factorial.

To get numbers larger than 100 (for example 113) you must use some additional operations. Different people like to introduce operations such as percent (see David Bevan’s site) or the gamma function (see Dave Wheeler’s site). Pete Karsanow’s site has a number of good links to other solutions sites.

Some people have demonstrated that it is possible to get any number with three fours if you introduce the logarithm function (for example see here). (Thus the logarithm function is generally disallowed as an operation in the four fours problem.) Paul Bourke has a contribution by “whetstone” that shows how to get any number with two fours if you use the logarithm function and the percent sign.

This page shows that it is possible to create any number with one four using trigonometric functions. In short:

n = sec(atan(…(sec(atan(4))…)) (n²-4² times), for n>4

n = tan(asec(…(tan(asec(4))…)) (4²-n² times), for n=0, 1, 2, 3

n = 4, for n = 4

Numbers larger than four

Using the trigonometric relation for the tangent and secant functions

tan²(u)+1 = sec²(u)

you can then derive that

sec(atan(x)) = sqrt(x²+1)

(This does not depend on whether you are using degrees or radians!) Therefore

sec(atan(sec(atan(x)))) = sqrt((sqrt(x²+1))²+1) = sqrt(x²+2)

sec(atan(…(sec(atan(x))…)) (k times) = sqrt(x²+k)

Thus repeated application of sec(atan()) will create the next largest square root in a sequence. For example, the table shows a formula that evaluates to 5 using just one 4 and the secant and arctangent functions.

x	sec(atan(x))
4	4.123105626	sec(atan(4))
4.123105626	4.242640687	sec(atan(sec(atan(4))))
4.242640687	4.358898944	sec(atan(sec(atan(sec(atan(4))))))
4.358898944	4.472135955	sec(atan(sec(atan(sec(atan(sec(atan(4))))))))
4.472135955	4.582575695	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4))))))))))
4.582575695	4.69041576	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4))))))))))))
4.69041576	4.795831523	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4))))))))))))))
4.795831523	4.898979486	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4))))))))))))))))
4.898979486	5	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(4))))))))))))))))))

The last line shows the expression for 5 that uses just one four.

For any integer n greater than 4, there is a k that satisfies the equation above (specifically k = n²-4²). This means that the number of times you need to apply sec(atan()) is n²-4² for any n greater than 4. (It takes a lot of operations, but it is possible!)

n = sec(atan(…(sec(atan(4))…)) (n²-4² times), for n>4

Numbers less than 4

To create an expression for the numbers 3, 2, 1, and 0 using one four, the same expression

tan²(u) = sec²(u)-1

yields the formula

tan(asec(x)) = sqrt(x²-1)

So repeated application of tan(asec()) gives the next lowest square root in a sequence.

n = tan(asec(…(tan(asec(4))…)) (4²-n² times), for n=0, 1, 2, 3

The following table shows the expression for 3, 2, 1, and 0.

x	tan(asec(x))
4	3.872983346	tan(asec(4))
3.872983346	3.741657387	tan(asec(tan(asec(4))))
3.741657387	3.605551275	tan(asec(tan(asec(tan(asec(4))))))
3.605551275	3.464101615	tan(asec(tan(asec(tan(asec(tan(asec(4))))))))
3.464101615	3.31662479	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))
3.31662479	3.16227766	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))
3.16227766	3	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))
3	2.828427125	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))
2.828427125	2.645751311	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))
2.645751311	2.449489743	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))))
2.449489743	2.236067977	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))))))
2.236067977	2	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))))))))
2	1.732050808	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))))))))))
1.732050808	1.414213562	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))))))))))))
1.414213562	1	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))))))))))))))
1	0	tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(tan(asec(4))))))))))))))))))))))))))))))))

Any number using one zero

Some people like to do the problem using combinations of different numbers besides four, for example, four zeros. Using the approach above, you can create any number using one of any digit, for example zero. The table below shows the expressions for 0, 1, 2, and 3 using one zero.

x	sec(atan(x))
0	1	sec(atan(0))
1	1.414213562	sec(atan(sec(atan(0))))
1.414213562	1.732050808	sec(atan(sec(atan(sec(atan(0))))))
1.732050808	2	sec(atan(sec(atan(sec(atan(sec(atan(0))))))))
2	2.236067977	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0))))))))))
2.236067977	2.449489743	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0))))))))))))
2.449489743	2.645751311	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0))))))))))))))
2.645751311	2.828427125	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0))))))))))))))))
2.828427125	3	sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(sec(atan(0))))))))))))))))))

The number of operations sec(atan()) you need to create an expression that evaluates to n is n².

n = sec(atan(…(sec(atan(0))…)) (n² times), for n>0

Other trigonometric functions

The same approach is possible using the cosecant and cotangent functions.

cot²(u)+1 = csc²(u)

Therefore

csc(acot(x)) = sqrt(x²+1)

cot(acsc(x)) = sqrt(x²-1)

So repeated applications of csc(acot()) give the next largest square root in a sequence, similar to sec(atan()) above.

Back to fractal designs using pattern blocks.