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.
Solution link
Anonymous
(44*(44+4))-4444/44
Anonymous
4444/2=2222
2222-(4x44+44)=2002
2002+4+4= 2010
2010+(4/4)=2011
Anonymous
44*44+(4*4*4)+4+4+4-(4/4)
Anonymous
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.
Anonymous
Least 4s required maybe.
(4^4-4)*(4+4)-(4+4/4)
Anonymous
40320/20 - 5
Anonymous
2011=44^SQRT(4)+4^(4-4/4)+44/4
Anonymous
2011=44^SQRT(4)+4*4*4+44/4
Anonymous
((4^(4 + (4/4)))*(4^1/2)) - 4! - (4*(4^1/2) + (4/4)) - 4
Anonymous
4444 - ((4*4*4*4*4) + (4*4*4*4*4)) - (4*4*4*4) - ((4*4*4) + (4*4*4)) -4/4
Anonymous
444*4 + 44*(4 + 4:4) + 4*4 - 4:4 =
1776 + 220 + 16 - 1 = 2011
Anonymous
=(4^4)*(4+4)-44+4+4-4/4
by tanks
tkruke
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!
Anonymous
...+4+4+4 + 44/4
= 2000 + 11
Is it cheating if I have to use an ellipsis or sigma notation?
Marianne
Yes!
Anonymous
((((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
Anonymous
2011 = (4^4 * 4 * sqrt 4) + (44/4) - 44 -4
By Krani Lupus
Anonymous
sqrt 4 * (4^4 * 4 - 4!) + 44/4
By Krani Lupus
Anonymous
44^((4+4)/4) + 4!(4-4/4) + 4 - 4/4
Anonymous
((4(44/4)-(4/4))*((4((4/44)-(4/4)) + ((44/4)-(4/4)))) + (44/4)
Anonymous
4^4*4*(spuare root of 4)-44+4+4-4/4
10 4's
Anonymous
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
Anonymous
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
bharathsellvan
(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
Anonymous
((4^4) * 4 * sqrt(4)) - (4!) - (4*4) + 4 - (4/4)
Anonymous
[500/((4/4)/4)]+(40/4)+(4/4)
Anonymous
i do belive that 5 and 0 are not the number 4
Anonymous
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
Anonymous
Solution:
= 4444/4 + 4444/4 + 44/4 - 444/4 - 4444/4
= 1111 + 1111 +11 - 111 -111
= 2233 - 222
= 2011
Anonymous
An excellent answer
Choi
Anonymous
I feel there should be a limit on number of times 4 is used. Otherwise, the simplest solution would be to add (4/4) 2011 times ( 4/4+4/4+..................=2011). :-)
Anil Sharma
Anonymous
4+4/4(4^4 x4) - 4+4/4(4x4) - 4 - 4/4
Anonymous
2011 = 4 * 502 + 3
502 = 125 * 4 + 2
125 = 31 * 4 + 1
31 = 4 * 4 * 2 - 1
1 = 4/4
2 = sqrt(4)
3 = 4-4/4
--> 2011 = (((4 * 4 * sqrt(4) - 4/4) * 4 + 4/4) * 4 + sqrt(4)) * 4 + 4 - 4/4
Anonymous
2011 = (444+44+4)*4+44-(4/4)
Also:
2012 = (444+44+4)*4+44
2013 = (444+44+4)*4+44+(4/4)
Anonymous
(4444/4)+(4444+4)-(444/4)-(444/4)+(44/4)
Anonymous
(256 x 16)/2 - 37 = 2011
- 256 = 4 to the power of 4
- 16 = 4 x 4
- 2 = sqrt 4
- 37 = 4 x 4 x sqrt4 + 4 + 4/4
Like I said... I love puzzles with infinite right answers :D
Konstantin
8000=SQRT(SQRT(SQRT(4!-4)^4!)))
2011=(8000+44)/4