2011 in fours
Write the number 2011 using only the digit 4 and any of the operations of addition, subtraction, multiplication, division, exponentiation, taking a square root and factorial. You can use any number composed of the digit 4, even if it's decimal, so 44 and 44.44 are both allowed. You're also allowed to use brackets.
Have fun!
This puzzle was contributed by Paulo Ferro, a maths teacher in Oporto, Portugal. For more of Paulo's puzzles, visit his website in English or Portuguese. If you have a puzzle you think might interest Plus readers, please email us!
Solution link:
2011 in fours - solution 
Comments
An amazing answer
How to get to 2011 using all 4's.
4444 - (444*4) = 2668
2668 - 444 = 2224
2224 - (44*4) = 2048
2048 - 44 = 2004
2004 + 4 + 4 = 2012
2012 - 4/4 = 2011
[500/((4/4)/4)]+(40/4)+(4/4)
[500/((4/4)/4)]+(40/4)+(4/4)
On all fours
((4^4) * 4 * sqrt(4)) - (4!) - (4*4) + 4 - (4/4)
2011
(4^4)*4+(444+444)+(44+44)+(4+4+4)-(4/4)
(256)*(4)+(888)+(88)+(12)-1
1024+888+88+11
2000+11
2011
4*4*4*4*4+4*4*4*4*4-44+4+4-(4\4)
4*4*4*4*4+4*4*4*4*4-44+4+4-(4\4)=
1024+1024-44+4+4-1=
2048-44+4+4-1=2011
2011 in fours (eight fours!)
factorial(4)=24
4*24*24=2304
4*(factorial(4))to_the_power(squareroot(4))=2304
factorial(4)/squareroot(4)=288
(4+4/4)=5
2304-288-5=2011
Solution in eight fours: 4*(factorial(4))to_the_power(squareroot(4))-factorial(4)/squareroot(4)-(4+4/4)
Barry Daniels
Ref: plus.maths card in Xmas 2011 New Scientist
4^4*4*(spuare root of
4^4*4*(spuare root of 4)-44+4+4-4/4
10 4's
2011 in fours
((4(44/4)-(4/4))*((4((4/44)-(4/4)) + ((44/4)-(4/4)))) + (44/4)
twelve fours
44^((4+4)/4) + 4!(4-4/4) + 4 - 4/4
8 fours
sqrt 4 * (4^4 * 4 - 4!) + 44/4
By Krani Lupus
Other solution (10 fours)
2011 = (4^4 * 4 * sqrt 4) + (44/4) - 44 -4
By Krani Lupus
why are people afraid of the decimals?
((((4+4) /.4) * (4/.4)) * (4/.4)) + (44/4) = 2011
broken down:
8 / .4 = 20
4 / .4 = 10
so (20 * 10) * 10 = 2000
44/4 = 11
4+4+4+4+4+4...
...+4+4+4 + 44/4
= 2000 + 11
Is it cheating if I have to use an ellipsis or sigma notation?
Yes!
Yes!
Donald Knuth variant
This puzzle reminds me of a conjecture made by Donald Knuth. I learned about it from the book "Artificial Intelligence" by Russel & Nordvig, and I quote it from there:
Knuth conjectured that, starting with the number 4, a sequence of factorial, square root, and floor operations will reach any desired positive integer. For example, we can reach 5 from 4 as follows:
Floor(Sqr(Sqr(Sqr(Sqr(Sqr((4!)!)))))) = 5
It would be nice to see if anyone could write 2011 using only one 4 and the mentioned functions!
=(4^4)*(4+4)-44+4+4-4/4 by
=(4^4)*(4+4)-44+4+4-4/4
by tanks
Possible solution
444*4 + 44*(4 + 4:4) + 4*4 - 4:4 =
1776 + 220 + 16 - 1 = 2011
4444 - ((4*4*4*4*4) +
4444 - ((4*4*4*4*4) + (4*4*4*4*4)) - (4*4*4*4) - ((4*4*4) + (4*4*4)) -4/4
Another solution
((4^(4 + (4/4)))*(4^1/2)) - 4! - (4*(4^1/2) + (4/4)) - 4
From Bassel
2011=44^SQRT(4)+4^(4-4/4)+44/4
A refinement for less 4s
2011=44^SQRT(4)+4*4*4+44/4
(4+4)!/(4!-4)-(4+4/4)
40320/20 - 5
Searched using Haskell
Least 4s required maybe.
(4^4-4)*(4+4)-(4+4/4)
one possible answer
4*4*4*4*4*((4/4)+(4/4))-4*4*((4/4)+(4/4))-4-(4/4)=2011.
4*4*4*4*4*2-4*4*2-4-1=2011.
2048-32-5=2011.
44*44+(4*4*4)+4+4+4-(4/4)
44*44+(4*4*4)+4+4+4-(4/4)
(44*(44+4))-4444/44
(44*(44+4))-4444/44
Solution
4444/2=2222
2222-(4x44+44)=2002
2002+4+4= 2010
2010+(4/4)=2011