4+4+4+4=2!
Math brain teasers require computations to solve.
In "4+4+4+4=1!" it was shown that just using addition, subtraction, multiplication, and division, you can form all the numbers from 0 to 9 using exactly four 4's.
For example: 7 = 4 + 4  4/4
Now if you add the operations square root, factorial, and exponent, what is the first number counting up from zero that can't be formed? How can each number be formed?
You may use parentheses, but you may not combine fours to make 44 or 444.
(An exponent is raising a number to the power of another number such as 4^4 = 256; a factorial is the product of all the integers from 1 to the integer such as 4! = 1x2x3x4 = 24.)
HintThe first number that can't be formed is 39.
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Answer
The first number that can't be formed is 39. This isn't proved, so maybe you can figure out how to form 39.
There are several ways to form most of the numbers, here are some for the numbers 0 to 9 using just addition, subtract, multiplication and division:
0 = 4+444 = (44)*4+4
1 = (4/4)*(4/4) = (4*4)/(4*4) = (4+4)/(4+4) = (4/4)/(4/4)
2 = (4/4)+(4/4) = (4*4)/(4+4) = 4/((4+4)/4)
3 = (4+4+4)/4 = (4*44)/4
4 = (44)*4+4
5 = (4*4+4)/4
6 = 4+((4+4)/4)
7 = (4+4)4/4
8 = 4+4+44 = (4/4)*4+4 = ((4+4)/4)*4 = 4*444
9 = 4+4+4/4
Here are some ways to form the numbers from 10 to 38 using the additional operators:
10 = 4+4+4SQRT(4)
11 = 4!/SQRT(4)4/4
12 = 4*(44/4)
13 = 4!/SQRT(4)+4/4
14 = (4+SQRT(4))+SQRT(4)
15 = 4*44/4
16 = 4*44+4 = 4+4+4+4 = 4*4*4/4
17 = 4*4+4/4
18 = 4*4+4/SQRT(4)
19 = 4!44/4
20 = (4+(4/4))*4
21 = 4!4+4/4
22 = 4!(4+4)/4
23 = 4!4^(44)
24 = 4*4+4+4
25 = (4+4/4)^SQRT(4) = 4!+4^(44) = (4!*4!+4!)/4!
26 = 4!+(4+4)/4
27 = 4!+44/4
28 = 4*(4+4)4 = 4!+4+44
29 = 4!+4+4/4
30 = (4+4)*4SQRT(4) = (4+4/4)!/4 = 4!+SQRT(4)+SQRT(4)+SQRT(4)
31 = ((4+SQRT(4))!+4!)/4! = (SQRT(SQRT(SQRT(4)))^4!SQRT(4))/SQRT(4)
32 = 4*4+4*4 = SQRT(4)^(4+4/4) = 4^4/(4/SQRT(4)) = 4*4*4/SQRT(4) = (4+SQRT(4))^SQRT(4)4
33 = (SQRT(SQRT(SQRT(4)))^4!+SQRT(4))/SQRT(4)
34 = 4*(4+4)+SQRT(4) = (4+SQRT(2))!/4!+4 = 4!+4+4+SQRT(4)
35 = 4! + (4!  SQRT(4)) / SQRT(4)
36 = 4! + 4 + 4 + 4
37 = 4! + (4! + SQRT(4)) / SQRT(4)
38 = 4! + (4 * 4)  SQRT(4)
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Comments
fame
Jan 17, 2009
 1st comment! 
Nerine
Jan 18, 2009
 (2nd comment)
I'm not good at any sort of Maths, it's very confusing, but cool all the same. 
Zag24
Jan 19, 2009
 I can make 39. (But only if you allow infinite progressions.)
4! + 4*4  sqrt(sqrt(sqrt(sqrt(... sqrt(4) ...))))
After taking the square root an infinite number of times, you are left with the number 1.
24 + 16  1 = 39 
Elly_Bellsy
Jan 23, 2009
 thats really confusing!! 
javaguru
Jan 23, 2009
 You can also make 39 if the multifactorial operators are allowed:
4!! * 4!! + (4!! / SQRT(4)) = 39
4!! is the double factorial of 4, which is 4 x 2 = 6. So
6 * 6 + (6 / 2) = 39

javaguru
Jan 23, 2009
 Wow, really lame comment for my last one since 4 x 2 = 8.
You can still make 39 (and probably most if not all of the rest of the numbers up to 100 or so) with multifactorials included.
(4!)!!!!!!!!!!! / (4!  4  4) = 39
24!!!!!!!!!!! / (24  4  4) = 39
624 / 16 = 39
A multifactorial is the product of the number and every nth number counting down from the number where n is the order of the multifactorial. The order is determined by the number of exclamation marks, so 24!!!!!!!!!!! (11 !s) is 24 x (2411) x (242x11) which is 24 x 13 x 2 = 624.
You can see how this allows you to create a wide variety of numbers with a single 4, and an even wider variety with four 4s. 
stil
Jan 26, 2009
 Multifactorials are as different from factorials as horses are from donkeys; they represent a side excursion not invited by the question. 
stil
Jan 26, 2009
 Which is to say if you start using/creating special "cases" of the allowed operations you could have an "infinite" answer. 
javaguru
Jan 26, 2009
 Um, yeah, multifactorials ARE different, which is why I didn't allow them in the problem. I was pointing out that with them included the solution space expands to the point of trivializing the problem.
My comment was not intended to suggest that there is a solution for 39. 
bogdano88
Jan 09, 2013
 You must correct the formulation of the problem, so you can write numbers using 4/4 only once. 4/4 is one unit, and you can add it with oneself to obtain any number. 
Jimbo
Dec 27, 2013
 Well if you add exponent then you should be able to add the inverse operation to exponentiation and that is taking the logarithm. If that is allowed, there is a general solution to the four 4s problem. 
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