The New Sudoku Players' Forum

[Pat]

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

3 used 7 times

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( 333+3 ) x 3!

- 3! / 3



[Serg]

Hi, Pat!

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!

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[ 2^22 - ( 2+2 )! ] / 2

- 22



Maybe I missed something ... 2^22 is too huge (4194304)

Serg

[Pat]

sorry
missed SQRT, and got myself confused

[Serg]

Hi, Pat!

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

[Pat]

2 used 8 times { corrected }

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sqrt ( 2^22 )

- ( 2+2 )! / 2

- 22



[Serg]

Hi!
This is summary of all best at the moment results.

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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".]

[Serg]

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

Serg

[Pat]

6 used 9 times

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6! + 6! + 6!

- 6! / 6
- 6! / ( 6x6 )
- 6



[Pat]

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

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[
 ( 7! / 7 - 7x7 ) x ( 7+7+7 )
 + 7
]

/ 7



[Pat]

8 used 7 times

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8! / [ 8 + 8 + sqrt (8+8) ]

- sqrt (sqrt (8+8) )



[Pat]

9 used 8 times

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9! / [ 9 x ( 9+9 ) + 9 + 9 ]

- ( sqrt ( 9 ) ) !
  / sqrt ( 9 )



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( 999 + 9 - 9/9 )

* ( sqrt ( 9 ) ) !
  / sqrt ( 9 )



3rd version comes from my above 6s,
with 6 rewritten as
( sqrt ( 9 ) ) !

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  ( sqrt ( 9 ) ) !!   x sqrt ( 9 )

- ( sqrt ( 9 ) ) !!   /   ( sqrt ( 9 ) ) !
- ( sqrt ( 9 ) ) !!   / ( ( sqrt ( 9 ) ) ! x ( sqrt ( 9 ) ) ! )

- ( sqrt ( 9 ) ) !


[999_Springs]

(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

[Pat]

(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

[Serg]

Hi, Pat!

6 used 9 times

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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]

Hi!

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