# Four 4s

1. Nov 7, 2005

### Joffe

Have you ever heard of that challenge where you have to make each number from four 4s? This list is the best I have been able to come up with, will you help me fill in the gaps?
Any solution that is simpler than one here can take its place, for example if you come up with a solution to 33 that doesnt use a decimal symbol it is simpler.
EDIT: An underline indicates repetition: (i.e: .4 = .44444 / 4/9).
Code (Text):
1 = 4*4/(4*4)
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
7 = 4+4-4/4
8 = 4+4+4-4
9 = 4/4+4+4
10 = (4*4+4!)/4
11 = (4+4!)/4+4
12 = (4-4/4)*4
13 = (4+4!+4!)/4
14 = 4!/4+4+4
15 = 4*4-4/4
16 = 4*4+4-4
17 = 4*4+4/4
18 = (4*4!-4!)/4
19 = 4!-(4+4/4)
20 = (4/4+4)*4
21 = 4!+4/4-4
22 = 4!-(4+4)/4
23 = 4!-4^(4-4)
24 = 4*4+4+4
25 = 4!+(4/4)^4
26 = 4!+4!/4-4
27 = 4!+4-4/4
28 = (4+4)*4-4
29 = 4/4+4!+4
30 = (4*4!+4!)/4
31 = (4+4!)/4+4!
32 = 4^4/(4+4)
33 = (4-.4)/.4+4!
34 = 4!/4+4+4!
35 = (4.4/.4)+4!
36 = (4+4)*4+4
37 = 4/.[U]4[/U]+4+4!
38 = 44-4!/4
39 = (4*4-.4)/.4
40 = (4^4/4)-4!
41 = (4*4+.4)/.4
42 = 4!+4!-4!/4
43 = 44-4/4
44 = 4*4+4+4!
45 = (4!/4)!/(4*4)
46 = (4!-4)/.4 - 4
47 = 4!+4!-4/4
48 = (4*4-4)*4
49 = 4!+4!+4/4
50 = (4*4+4)/.4
51 = 4!/.4-4/.[U]4[/U]
52 = 44+4+4
53 = 44+4/.[U]4[/U]
54 = (4!/4)^4/4!
55 = (4!-.4)/.4-4
56 = 4!+4!+4+4
57 = 4/.[U]4[/U]+4!+4!
58 = (4^4-4!)/4
59 = 4!/.4-4/4
60 = 4*4*4-4
61 = 4!/.4+4/4
62 = (4!+.4+.4)/.4
63 = (4^4-4)/4
64 = 4^(4-4/4)
65 = 4^4+4/4
66 = (4+4!)/.4-4
67 = (4+4!)/.[U]4[/U]+4
68 = 4*4*4+4
69 = (4+4!-.4)/.4
70 = (4^4+4!)/4
71 = (4!+4.4)/.4
72 = (4-4/4)*4!
73 = ([sup].4[/sup]√4+.[u]4[/u])/.[u]4[/u]
74 = (4+4!)/.4+4
75 = (4!/4+4!)/.4
76 = (4!-4)*4-4
77 = (4!-.[u]4[/u])/.[u]4[/u]+4!
78 = (4!x.[u]4[/u]+4!)/.[u]4[/u]
79 = ([sup].4[/sup]√4-.4)/.4
80 = (4*4+4)*4
81 = (4/4-4)^4
82 = 4!/.[U]4[/U]+4!+4
83 = (4!-.4)/.4+4!
84 = (4!-4)*4+4
85 = (4/.4+4!)/.4
86 = (4-.4)x4!-.4
87 = 4!x4-4/.[U]4[/U]
88 = 4^4/4+4!
89
90 = (4!/4)!/(4+4)
91
92 = (4!-4/4)*4
93
94 = (4+4!)/.4 + 4!
95 = 4!*4-4/4
96 = 4!*4+4-4
97 = 4!*4+4/4
98 = (4!+.4)*4+.4
99 = (4!+4!-4)/.[U]4[/U]
100 = 4*4/(.4*.4)
Thanks:
• ceptimus: 71
• LarrrSDonald: 51,53,82,87,79

Last edited: Nov 7, 2005
2. Nov 7, 2005

### ceptimus

71 = (4! + 4.4) / .4

...and if we can use square roots:

37 = (sqrt(4) + 4!)/sqrt(4) + 4!
51 = (4! - sqrt(4)) / .4 - 4
53 = sqrt(4) / .4 + 4! + 4!
57 = (4! - .4) / .4 - sqrt(4)
67 = (sqrt(4) + 4!) / .4 + sqrt(4)
78 = 4 * (4! - 4) - sqrt(4)
79 = (4! - sqrt(4)) / .4 + 4!
82 = 4 * (4! - 4) + sqrt(4)
89 = (sqrt(4) + 4!) / .4 + 4!
91 = 4 * 4! - sqrt(4) / .4

3. Nov 7, 2005

### Joffe

Thanks for 71 ceptimus!
The others I am not as interested in because (as I probably should have stated) I am not considering square roots because they require a 2, I would however consider 4th roots so long as the 4 is counted. No triggonometric functions either, basically whatever techniques I have used so far.

4. Nov 7, 2005

### ArielGenesis

if we can use zero as if in decimal, why couldn't we use 2 as if in square root as well as trigonometry function and, most probabaly also included, logarithm. If you are about using arithmetich signs only, then decimal should not be used.

5. Nov 7, 2005

### ArielGenesis

and also exclemation mark !, as 4! should include other number such 1, 2 and 3

6. Nov 7, 2005

### LarrrSDonald

If you're ok with $$.\bar{4} = \frac{4}{9}$$ then you have:

$$51 = \frac{4!}{4}-\frac{4}{.\bar{4}}$$
$$53 = 44+\frac{4}{.\bar{4}}$$
$$82 = \frac{4!}{.\bar{4}}+4!+4$$
$$87 = 4!*4-\frac{4}{.\bar{4}}$$

Possibly others, I haven't had time to seach around that much.

7. Nov 7, 2005

### RandallB

Should be:
$$51 = \frac{4!}{.4}-\frac{4}{.\bar{4}}$$

8. Nov 7, 2005

### LarrrSDonald

You're right, lost a decimal point. Sorry 'bout that.
[EDIT] If you're willing to consider 4th root, then perhaps you'd consider $$\sqrt[.4]{4} = 32$$ and thus
$$79 = \frac{(\sqrt[.4]{4} - .4)}{.4}$$

Last edited: Nov 7, 2005
9. Nov 7, 2005

### Joffe

LarrrSDonald: I like your ideas! Your repetition idea got me a whole heap of answers including 37! And this latest one, 79 is great lateral thinking, Keep em coming.

ArielGenesis: If I were to include square root signs then it might as well be called the four 4 or 2s challenge, it would make it far too easy. Using decimals is still bending a little (hence the reason to try to try to eliminate some of the solutions that use them) but since it does not make a whole number it keeps the game challenging. And if I were to include trig I could make combinations like: sin-1 ( cos ( 4 ) ) = 86, which is clearly too convenient.

EDIT: only 3 remain!

Last edited: Nov 7, 2005
10. Nov 7, 2005

### Alkatran

4 != 0/4

I'd go for 4 = 4! - 4*4 - 4

11. Nov 7, 2005

### Joffe

(4-4)/4 + 4
= 0/4 + 4
= 0 + 4
= 4

Perhaps you misinterpreted my grouping symbols.

12. Nov 8, 2005

### RandallB

BUT you didn't say what 37 was! Maken us work for it eh?
Did you use:
$$37 = \frac{{4}*{4}+{.\bar{4}}}{.\bar{4}}$$

13. Nov 8, 2005

### ceptimus

He's got 37 = 4 / .4 + 4 + 4! in the table above

(he used an underline to show recurring)

14. Nov 22, 2005

### Joffe

Indeed, here are two more:

89 = (4-.4-4%)/4%

91 = (4-.4+4%)/4%

93 is all that remains!

15. Nov 22, 2005

### ceptimus

I can only do it with the forbidden square root.

$$93 = 4 \times 4! - \sqrt{\frac{4}{.\bar{4}}}$$

16. Nov 27, 2005

### croxbearer

If logarithm is allowed:

93 = 4 x 4! - log(4/.4%)

17. Nov 28, 2005

### Joffe

There is only one problem with that, there is an implied 10. I would definantly accept log in base 4 and maby the natural logarithm of an expression but that (albeit quite novel) just seems to violate in my opinion.

If I don't ever find a better solution I think I will settle for that though croxbearer.

18. Nov 21, 2008

### Markii

Another way to get 93 using an exponent of 0 is 4! x 4 - 4 - 4°

19. Nov 10, 2010

### Fahim555

Yo, I got 89!

20. Aug 5, 2011

### Fang_Lou

5 = (4x4+4)/4
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