2011 in fours

in
2011

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!

Comments

i love open ended questions like this...

(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

another solution

(4444/4)+(4444+4)-(444/4)-(444/4)+(44/4)

2011 in 4's

2011 = (444+44+4)*4+44-(4/4)
Also:
2012 = (444+44+4)*4+44
2013 = (444+44+4)*4+44+(4/4)

Using only the number 4 (as opposed to digit)

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

4+4/4(4^4 x4) - 4+4/4(4x4) -

4+4/4(4^4 x4) - 4+4/4(4x4) - 4 - 4/4

2011 in 4

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

Solution: = 4444/4 + 4444/4 +

Solution:
= 4444/4 + 4444/4 + 44/4 - 444/4 - 4444/4
= 1111 + 1111 +11 - 111 -111
= 2233 - 222
= 2011

An excellent answer Choi

An excellent answer
Choi

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)

uh oh

i do belive that 5 and 0 are not the number 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