## "2014" arithmetic puzzle

For fans of all other kinds of logic puzzles
Serg wrote:

2014 = ((3+3)/3)^(33/3) - 33 - 3/3; (10 digits)

3 used 7 times
Code: Select all
`( 333+3 ) x 3!- 3! / 3` Pat

Posts: 3879
Joined: 18 July 2005

### Re:

Hi, Pat!
Pat wrote:
Serg wrote:

2014 = 2^(22/2) - (2 + 2 + 2)^2 + 2; (9 digits)

2 used 8 times

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`sqrt ( 2^22 )- ( 2+2+2 ) ^2+ 2`

Nice optimization of my example!
Pat wrote:
Serg wrote:
Code: Select all
`[ 2^22 - ( 2+2 )! ] / 2- 22`

Maybe I missed something ... 2^22 is too huge (4194304) Serg
Serg
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Location: Russia

sorry
missed SQRT, and got myself confused
Last edited by Pat on Thu Jan 23, 2014 12:42 pm, edited 1 time in total. Pat

Posts: 3879
Joined: 18 July 2005

### Re:

Hi, Pat!
Pat wrote:
Serg wrote:

2014 = ((3+3)/3)^(33/3) - 33 - 3/3; (10 digits)

3 used 7 times
Code: Select all
`( 333+3 ) x 3!- 3! / 3`

Excellent solution!

Serg
Serg
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Posts: 690
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Location: Russia

2 used 8 times { corrected }
Code: Select all
`sqrt ( 2^22 )- ( 2+2 )! / 2- 22` Pat

Posts: 3879
Joined: 18 July 2005

### Re: "2014" arithmetic puzzle

Hi!
This is summary of all best at the moment results.
Code: Select all
`2014 = (0!+0!) x (((0!+0!+0!)!+0!)!/((0!+0!+0!)!-0!) - 0!);              [999_Springs; 11 digits]2014 = (1+1)^11 - (1+1+1) x 11 - 1;                                      [Serg;        10 digits]2014 = (2x2x2)!/(22-2) - 2;                                              [999_Springs;  7 digits]2014 = 3! x (3!^3) + (3!)! - 3!/3;                                       [999_Springs;  6 digits]2014 = (4+4)!/(4!-4) - sqrt(4);                                          [999_Springs;  5 digits]2014 = 5^5 - 5555/5;                                                     [999_Springs;  7 digits]2014 = 6 x sqrt(6^6) + 6! - (6+6)/6;                                     [999_Springs;  7 digits]2014 = ((7!/7-7x7) x (7+7+7) + 7)/7;                                     [Pat;          9 digits]2014 = (7!+7!)/7 + 7 x (77+7) - 7 - 7;                                   [Serg;         9 digits]2014 = (7!+7!)/7 + 7 x sqrt(7!+7/7) + 77;                                [Serg;         9 digits]2014 = 8!/(8+8+sqrt(8+8)) - sqrt(sqrt(8+8));                             [Pat;          7 digits]2014 = sqrt(9)! x (sqrt(9)!^sqrt(9)) + (sqrt(9)!)! - sqrt(9)!/sqrt(9);   [999_Springs;  6 digits]`

Serg

[Edited. I updated the list by digit "6" variant.]
[Edited. I updated the list by "7", "8", "9" variants, added new Pat's and 999_Springs's results.]
[Edited. I published formula having 12 zeroes.]
[Edited. I published last 999_Springs's results for digits "3", "6" and "9".]
[Edited 1.02.2014. I updated 999_Springs's optimization for "0" formula and added my 2 9-digits formulas for "7".]
Last edited by Serg on Sat Feb 01, 2014 7:05 am, edited 7 times in total.
Serg
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Location: Russia

### Re: "2014" arithmetic puzzle

Hi!
I've found presentation containing 10 digits "6":
2014 = (6x6+6+6) x (6x6+6) -(6+6)/6

Serg
Serg
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Location: Russia

### 6

6 used 9 times
Code: Select all
`6! + 6! + 6!- 6! / 6- 6! / ( 6x6 )- 6` Pat

Posts: 3879
Joined: 18 July 2005

### 7

7 used 9 times

started as
671 x 3 + 1
but that needed 10
so, postponed /7 to the very end
hence the monster
[ 671 x 21 + 7 ] /7
Code: Select all
`[ ( 7! / 7 - 7x7 ) x ( 7+7+7 ) + 7]/ 7` Pat

Posts: 3879
Joined: 18 July 2005

### 8

8 used 7 times
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`8! / [ 8 + 8 + sqrt (8+8) ]- sqrt (sqrt (8+8) )` Pat

Posts: 3879
Joined: 18 July 2005

### 9

9 used 8 times
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`9! / [ 9 x ( 9+9 ) + 9 + 9 ]- ( sqrt ( 9 ) ) !  / sqrt ( 9 )`

Code: Select all
`( 999 + 9 - 9/9 )* ( sqrt ( 9 ) ) !  / sqrt ( 9 )`

3rd version comes from my above 6s,
with 6 rewritten as
( sqrt ( 9 ) ) !
Code: Select all
`  ( sqrt ( 9 ) ) !!   x sqrt ( 9 )- ( sqrt ( 9 ) ) !!   /   ( sqrt ( 9 ) ) !- ( sqrt ( 9 ) ) !!   / ( ( sqrt ( 9 ) ) ! x ( sqrt ( 9 ) ) ! )- ( sqrt ( 9 ) ) !` Pat

Posts: 3879
Joined: 18 July 2005

### Re: "2014" arithmetic puzzle

(2*2^2)!/(22-2)-2

(4+4)!/(4!-4)-sqrt4

5^5-5555/5

(9!*sqrt(9)!*sqrt(9)!)/(9*sqrt(9)!!)-sqrt(9)!/sqrt9

first post in here since forever Last edited by 999_Springs on Fri Jan 24, 2014 10:20 am, edited 1 time in total.
Once upon a time I was a teenager who was active on here 2007-2011
ocean and eleven should have paired up to make a sudoku-solving duo called Ocean's Eleven
999_Springs

Posts: 487
Joined: 27 January 2007
Location: In the toilet, flushing down springs, one by one.

999_Springs wrote:
(2*2*2) ! / (22-2)
- 2

(4+4) ! / (4!-4)
- sqrt4

5^5
- 5555/5

( 9! * sqrt(9)! * sqrt(9)! )
/ ( 9 * sqrt(9)!! )
- sqrt(9)! / sqrt(9)

amazing Pat

Posts: 3879
Joined: 18 July 2005

### Re: 6

Hi, Pat!
Pat wrote:

6 used 9 times
Code: Select all
`6! + 6! + 6!- 6! / 6- 6! / ( 6x6 )- 6`

I considered similar variants, but wasn't manage to implement it, because I didn't find expression 6!/(6x6) = 20. Nice work!

Serg
Serg
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Location: Russia

### Re: 7

Hi!
Pat wrote:

7 used 9 times

started as
671 x 3 + 1
but that needed 10
so, postponed /7 to the very end
hence the monster
[ 671 x 21 + 7 ] /7
Code: Select all
`[ ( 7! / 7 - 7x7 ) x ( 7+7+7 ) + 7]/ 7`

New original idea: (671x21+7)/7 = 2014;

Serg
Serg
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Posts: 690
Joined: 01 June 2010
Location: Russia

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