# Fours challenge

I came across this great mathematical challenge in early 2015. **Writing equations using only four 4s and mathematical symbols, what number can you get up to?** For example *8 = 4 + 4 + 4 - 4*. Can you reach 10 or 20, or 100?

I played around with this a lot for several weeks, with alternating phases of intense concentration, frustration and delight. I eventually got stuck on 113. Please give it a try and only reveal the answers after you've thought (and scribbled) long and hard about it. The answers are grouped together according to the new techniques you'll need to discover.

### Show 0 to 2

You can get up to 2 using just + - and division or fractions. 3 needs an extra symbol.

\(0 = 4 + 4 - 4 - 4\)

\(1 = 4 - 4 + \frac44\)

\(2 = \frac44 + \frac44\)

### Show 3 to 10

3 is easy enough to get to by doing \(3 = 4 - \frac44\), but that only uses three 4s and you have to use all four for this challenge. So you need a way of making one 4 into two 4s. You can use the square root of 4 for that, \(\sqrt4 = 2\). This will be used all over the place in the numbers to come.

\(3 = 2 + 2 - 1 = \sqrt4 + \sqrt4 - \frac44\)

\(4 = 4 - 4 + \sqrt4 + \sqrt4\)

\(5 = \sqrt4 + \sqrt4 + \frac44\)

\(6 = 4 + \sqrt4 + \sqrt4 - \sqrt4\)

\(7 = 4 + 4 - \frac44\)

\(8 = 4 + 4 + 4 - 4\)

\(9 = 4 + 4 + \frac44\)

\(10 = 4 + 4 + 4 - \sqrt4\)

### Show 11 to 18

11 is tricky. You can get to 12 with three 4s, but it takes two more 4s to subtract 1. And you can get to 8 in two 4s but adding 3 takes three more 4s. So you need a new symbol. A decimal point can get you to 10 with just two 4s, because dividing 4 by 0.4 equals 10. And from there you can also get to \(5 = \frac{\sqrt4}{.4}\) which will help you with 13.

\(11 = 10 + 1 = \frac4{.4} + \frac44\)

\(12 = 4 + 4 + \sqrt4 + \sqrt4\)

\(13 = 8 + 5 = 4 + 4 + \frac{\sqrt4}{.4}\)

\(14 = 4 + 4 + 4 + \sqrt4\)

\(15 = 4 * 4 - \frac44\)

\(16 = 4 * 4 + 4 - 4\)

\(17 = 4 * 4 + \frac44\)

\(18 = 4 * 4 + 4 - \sqrt4\)

### Show 19 to 30

19 is not too bad, but requires another symbol. You have to introduce factorials. The factorial of 4 is 4 * 3 * 2 * 1 = 24. And it is written with an exclamation point. So \(4! = 24\). This is very useful and can quickly get you up to much bigger numbers. Note that for 23, you have to find a convenient place to squeeze in an extra 4. For 31 you'll need to make a creative leap.

\(19 = 24 - 4 - 1 = 4! - 4 - \frac44 \)

\(20 = 4 * 4 + \sqrt4 + \sqrt4\)

\(21 = 4 * 4 + \frac{\sqrt4}{.4}\)

\(22 = 4 * 4 + 4 + \sqrt4\)

\(23 = 4! - \frac{\sqrt4 + \sqrt4}4\)

\(24 = 4! + 4 - \sqrt4 - \sqrt4\)

\(25 = 4! + \frac{\sqrt4 + \sqrt4}4\)

\(26 = 4! + 4 - 4 + \sqrt4\)

\(27 = 4! + 4 - \frac44\)

\(28 = 4! + 4 + 4 - 4\)

\(29 = 4! + 4 + \frac44\)

\(30 = 4! + 4 + 4 - \sqrt4\)

### Show 31 to 40

I found 31 very difficult, but it can be approached similarly to the shortcut for 10. Mathematicians have a short-hand way of writing infinitely repeating decimals. For example, one third as a decimal is 0.333333333... but can be written as just \(0.\overline3\). Similarly \(\frac49 = 0.\overline4\). And if you divide 4 by that you get \(9 = \frac4{.\overline4}\)! (That exclamation point was a real one, not a factorial.) You can use that 9 to get from 22 up to 31. You can also use it to get from 48 (which is 2 * 4!) down to 39. (Recently I also thought of another way to do 31 as \(\frac{5! + 4}4\).)

\(31 = 24 - 2 + 9 = 4! - \sqrt4 + \frac4{.\overline4}\)

\(32 = 4! + 4 + \sqrt4 + \sqrt4 \)

\(33 = 4! + \frac{\sqrt4 + \sqrt4}{.\overline4}\)

\(34 = 4! + 4 + 4 + \sqrt4 \)

\(35 = 4! + \sqrt4 + \frac4{.\overline4}\)

\(36 = 4! + 4 + 4 + 4 \)

\(37 = 4! + 4 + \frac4{.\overline4}\)

\(38 = 4! + 4 + \frac{\sqrt4}{.4}\)

\(39 = 48 - 9 = 4! * \sqrt4 - \frac4{.\overline4}\)

\(40 = 4! + 4 * \sqrt4 * \sqrt4 \)

### Show 41 to 50

41 is awkward. You can get to 48 with just two 4s, but you can't make up the difference 7 out of the remaining two 4s. But you can get to 36 in just two 4s. Notice that the square root of \(\sqrt{.\overline4} = \sqrt{\frac49} = \frac23\). This is not so useful in itself, but if you divide something by \(\frac23\), its the same as multiplying by \(\frac32\). And you can multiply it by 4!! (The first exclamation point is a factorial in this case.) So 4! times \(\frac32\) gets you to 36 and adding 5 more arrives at 41. Most of the other numbers around here can be reached straight from 48.

\(41 = 36 + 5 = \frac{4!}{\sqrt{.\overline4}} + \frac{\sqrt4}{.4}\)

\(42 = 4! * \sqrt4 - 4 - \sqrt4\)

\(43 = 4! * \sqrt4 - \frac{\sqrt4}{.4}\)

\(44 = 4! * \sqrt4 - \sqrt4 - \sqrt4\)

\(46 = 4! * \sqrt4 - 4 + \sqrt4\)

\(47 = 4! * \sqrt4 - \frac44\)

\(48 = 4! * \sqrt4 + 4 - 4\)

\(49 = 4! * \sqrt4 + \frac44\)

\(50 = 4! * \sqrt4 + 4 - \sqrt4\)

### Show 51 to 54

Numbers ending in 1 seem to be the trickiest. 51 isn't too difficult though. Not if you use the square root of 9 to get to 3 with just two 4s, \(3 = \sqrt{\frac4{.\overline4}}\) and then add that to 48.

\(51 = 48 + 3 = 4! * \sqrt4 + \sqrt {\frac4{.\overline4}}\)

\(52 = 4! * \sqrt4 + \sqrt4 + \sqrt4 \)

\(53 = 4! * \sqrt4 + \frac{\sqrt4}{.4}\)

\(54 = 4! * \sqrt4 + 4 + \sqrt4\)

### Show 55 to 60

55 is unexpectedly difficult. Even though it isn't prime, it is 7 away from 48 and so is hard to get to. However, you can use a similar technique as for 36 and divide 24 by \(0.\overline4 = \frac49\), which is the same as multiplying it by \(\frac94\), and thereby get to its neighbour 54. 59 looks particularly impressive but is really just 54 + 5.

\(55 = 54 + 1 = \frac{4!}{.\overline4} + \frac44\)

\(56 = 4! * \sqrt4 + 4 + 4\)

\(57 = 4! * \sqrt4 + \frac4{.\overline4}\)

\(58 = 4! * \sqrt4 + \frac4{0.4}\)

\(59 = 54 + 5 = \frac{4!}{.\overline4} + \frac{\sqrt4}{.4}\)

\(60 = 4 * 4 * 4 - 4\)

### Show 61 to 66

To get to 61, you have to get to 60. Fortunately you can do that from 24 fairly easily using almost the same technique as above. Notice that \(0.4 = \frac25\). If you divide 24 by this, it's the same as multiplying by \(\frac52\) and you get 60. You can get to lots of other numbers around here from either 60 or \(64 = 4 * 4 * 4\).

\(61 = 60 + 1 = \frac{4!}{.4} + \frac44\)

\(62 = \frac{4!}{.4} + 4 - \sqrt4\)

\(63 = \frac{4!}{.4} + \sqrt{\frac4{.\overline4}}\)

\(64 = 4 * 4 * \sqrt4 * \sqrt4\)

\(65 = \frac{4!}{.4} + \frac{\sqrt4}{.4}\)

\(66 = 4 * 4 * 4 + \sqrt4\)

### Show 67 to 70

67 needs another new technique. You got to 60 by multiplying 24 * \(\frac52\). You can get to 65 by multiplying 26 * \(\frac52\), and 26 is just 24 + 2. Similarly 70 is 28 * \(\frac52\). I found 71 to be the hardest number of all, so good luck!

\(67 = 65 + 2 = \frac{4! + \sqrt4}{.4} + \sqrt4 \)

\(68 = 4 * 4 * 4 + 4\)

\(69 = \frac{4! + \sqrt2}{.4} + 4 \)

\(70 = \frac{4! + \sqrt2 + \sqrt2}{.4} \)

### Show 71 to 72

71 took me ages, about ten times longer than any other number, and I very nearly gave up. Then I had an inspiration. I realised that *96 - 25 = 71*. (That wasn't the inspriation.) You can get to 96 in two with *96 = 4 * 4!* but it takes at least three 4s to reach 25. But what if they can be combined somehow and save a 4? You can do it using negative powers to square and invert a fraction. For example \(2^{-2} = \frac14\) and \(.4^{-2} = \frac25^{-2} = \frac52^2 = \frac{25}4 = 6.25\). And from there \(\frac{96}4 - \frac{25}4 = \frac{71}4\) and then you just need to multiply the whole thing by 4. Or in other words \(4 * (24 - 6.25) = 71\). I am so pleased with that.

\(71 = 4 * (24 - 6.25) = 4 * (4! - .4^{-\sqrt4}) \)

\(72 = 36 * 2 = \frac{4!}{.\overline4} * \frac4{\sqrt4} \)

### Show 73 to 76

It's amazing. I spent weeks on 71 and in the space of less than a day I found TWO ways to get there. And the second method, discovered shortly after the one above, also works for 73. It involves fractions again. You can make \(\frac23\) as the square root of \(\sqrt{.\overline4}\). That's an old trick, but now you can add that to 24 + 24 to get \(48 \frac23\) which is the same as \(\frac{72}3 + \frac{72}3 + \frac23 = \frac{146}3\). 146 thirds is not a number that comes up very often in day to day life. But if you divide that by \(\frac23\) the thirds go away and you are left with 146 over 2 which equals 73. And can get to 71 by doing \(\frac{72}3 + \frac{72}3 - \frac23 = \frac{142}3\). I'm really pleased with that too, though it came a lot quicker than the first 71 method above.

\(73 = \frac {24 + 24 + \frac23} {\frac23} = \frac {4! + 4! + \sqrt{.\overline4}} {\sqrt{.\overline4}} \)

\(74 = \frac{4! + 4}{.4} + 4 \)

\(75 = \frac{4! + 4 + \sqrt4}{.4} \)

\(76 = \frac{4!}{.\overline4} * \sqrt4 + 4 \)

### Show 77 to 102

Prime numbers are the hardest as you can't multiply straight to them. But odds like 77 can be a bit difficult too. For 77 I used a method for getting to 81 with just three 4s by making \(9^2\). That covers some of the late 70s and early 80s. Many of the other numbers in the 80s and 90s use \(4! * 4 = 96\) as their basis, as shown below.

\(77 = 81 - 4 = \frac4{.\overline4}^{\sqrt4} - 4\)

\(78 = 54 + 24 = \frac{4!}{.\overline4} + (\sqrt4 + \sqrt4)!\)

\(79 = \frac4{.\overline4}^{\sqrt4} - \sqrt4\)

\(80 = 96 - 16 = 4! * 4 - 4 * 4\)

\(81 = \frac4{.\overline4} * \frac4{.\overline4}\)

\(82 = 60 + 24 - 2 = \frac{4!}{.4} + 4! + \sqrt4 \)

\(83 = \frac4{.\overline4}^{\sqrt4} + \sqrt4 \)

\(84 = 96 - 12 = 4! * 4 - \frac{4!}{\sqrt4}\)

\(85 = \frac4{.\overline4}^{\sqrt4} + 4\)

\(86 = 96 - 10 = 4! * 4 - \frac4{.4} \)

\(87 = 96 - 9 = 4! * 4 - \frac4{.\overline4} \)

\(88 = 96 - 8 = 4! * 4 - 4 - 4 \)

\(89 = 65 + 24 = \frac{4! + \sqrt2}{.4} + 4! \)

\(90 = 96 - 6 = 4! * 4 - 4 - \sqrt4\)

\(91 = 96 - 5 = 4! * 4 - \frac{\sqrt4}{.4}\)

\(92 = 96 - 4 = 4! * 4 - \sqrt4 - \sqrt4 \)

\(93 = 96 - 3 = 4! * 4 - \sqrt {\frac4{.\overline4}}\)

\(94 = 96 - 2 = 4! * 4 - 4 + \sqrt4 \)

\(95 = 96 - 1 = 4! * 4 - \frac44 \)

\(96 = 4! * 4 + 4 - 4 \)

\(97 = 96 + 1 = 4! * 4 + \frac44 \)

\(98 = 96 + 2 = 4! * 4 + 4 - \sqrt4 \)

\(99 = 96 + 3 = 4! * 4 + \sqrt {\frac4{.\overline4}}\)

\(100 = 96 + 4 = 4! * 4 + \sqrt4 + \sqrt4 \)

\(101 = 96 + 5 = 4! * 4 + \frac{\sqrt4}{.4}\)

\(102 = 96 + 6 = 4! * 4 + 4 + \sqrt4\)

### Show 103 to 106

103 is interesting. It needs the number 44, which I've avoided up till now, but I'm quite pleased to be forced to use it. In the very early days, I used 44 to get to 11, but then decided to get there by other means as it seemed a bit like cheating. But it's no more cheating than using \(0.4\) or \(0.\overline4\).

\(103 = 99 + 4 = \frac{44}{.\overline4} + 4\)

\(104 = 96 + 8 = 4! * 4 + 4 + 4 \)

\(105 = 96 + 9 = 4! * 4 + \frac4{.\overline4} \)

\(106 = 96 + 10 = 4! * 4 + \frac4{.\overline4} \)

### Show 107 to 112

107 and 109 can use the same technique as 73 but relying on \(.\overline4 = \frac49\) this time. You can add up \(24 + 24 - \frac49 = \frac{216}9 + \frac{216}9 - \frac49 = \frac{428}9\). Dividing this by \(\frac49\) gets rid of the ninths and leaves you with 107. Adding the \(\frac49\) instead of substracting it arrives at 109. A lot of the other numbers around here can be based on 5 factorial. \(5! = 1 * 2 * 3 * 4 * 5 = 120\), and you can produce a 5 using only two 4s. Which gets you up to 113, another difficult prime, and where I am currently stuck. I might take a break from this for a while (last updated 19 March 2015).

\(107 = \frac {24 + 24 - \frac49} {\frac49} = \frac {4! + 4! - {.\overline4}} {.\overline4} \)

\(108 = 96 + 12 = 4! * 4 + \frac{4!}{\sqrt4}\)

\(109 = \frac {24 + 24 + \frac49} {\frac49} = \frac {4! + 4! + {.\overline4}} {.\overline4} \)

\(110 = 5! - 10 = \frac{\sqrt4}{.4}! - \frac4{.4}\)

\(111 = 5! - 9 = \frac{\sqrt4}{.4}! - \frac4{.\overline4}\)

\(112 = 5! - 8 = \frac{\sqrt4}{.4}! - 4 - 4 \)

I hope you have enjoyed this challenge. And please tell me if you can figure out 113.