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2011 in fours

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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!

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4444/2=2222
2222-(4x44+44)=2002
2002+4+4= 2010
2010+(4/4)=2011

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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.

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2011=44^SQRT(4)+4^(4-4/4)+44/4

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((4^(4 + (4/4)))*(4^1/2)) - 4! - (4*(4^1/2) + (4/4)) - 4

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4444 - ((4*4*4*4*4) + (4*4*4*4*4)) - (4*4*4*4) - ((4*4*4) + (4*4*4)) -4/4

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444*4 + 44*(4 + 4:4) + 4*4 - 4:4 =
1776 + 220 + 16 - 1 = 2011

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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!

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...+4+4+4 + 44/4

= 2000 + 11

Is it cheating if I have to use an ellipsis or sigma notation?

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sqrt 4 * (4^4 * 4 - 4!) + 44/4
By Krani Lupus

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44^((4+4)/4) + 4!(4-4/4) + 4 - 4/4

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((4(44/4)-(4/4))*((4((4/44)-(4/4)) + ((44/4)-(4/4)))) + (44/4)

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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

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(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

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((4^4) * 4 * sqrt(4)) - (4!) - (4*4) + 4 - (4/4)

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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

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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

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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

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

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(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

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8000=SQRT(SQRT(SQRT(4!-4)^4!)))
2011=(8000+44)/4

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