The New Sudoku Players' Forum

[Serg]

Hi, people!
Years ago, soviet scientific popular magazine "Science and Life" ("Nauka i zhizn" in russian) every year declared competition. You should find next year representation by some arithmetic formula containing one decimal digit only (unlimited number of times) and arithmetic operations signs ("+", "-", "*" or "x", "/", "^" (power sign), "!", root sign, etc.). For example, 2014 = 2222 - 222 + 2^(2^2) - 2. So, you should compose 9 formulas (the first must contain "1" digits only, the second must contain "2" digits only, up to formula containing "9" digits only) to enter competition. In the fall of the year competitor was determined for each digit category (9 independent competitions). Less digits in the formula, the better result.

Any takers for "2014" number composing?

Serg

[Edited. I found an error in my example of "2014" representation by "2" digits. Fixed.]

[Smythe Dakota]

Why not zero also? You have exclam "!" (factorial) in your list of acceptable operations, so 0 is always possible.

So, construct not nine, but ten, solutions, one for each digit 0 through 9.

Bill Smythe

[Serg]

Hi, Bill!

Why not zero also? You have exclam "!" (factorial) in your list of acceptable operations, so 0 is always possible.

So, construct not nine, but ten, solutions, one for each digit 0 through 9.

Bill Smythe

You are right from the formal point of view. 0! = 1, so some variants are possible. For example: 2014 = 0! + 0! + ... + 0! (2014 repetitions). But there are too few variants to construct such formula. So, I think "0" should not be used.

These are my results for digits "1" and "2":

2014 = (1 + 1)^11 - (1 + 1 +1)x11 - 1; (10 digits)
2014 = 2^(22/2) - (2 + 2 + 2)^2 + 2; (9 digits)

Can you find formulas having less digits?

Serg

[Smythe Dakota]

I still think zero can be exciting. In your example for 1's, you have eight 1's (each of which could be written 0!) and two 11's. My entry for expressing 11 with 0's is:

11 = (0!+0!+0!+0!)! / (0!+0!) - 0!

So your 2014 with ten 1's becomes 2014 with 22 zeroes.

So let's start with a simpler contest: What is the shortest way to represent 11 with zeroes? Can anybody beat my seven?

Bill Smythe

[Serg]

Hi, Bill!
Here is another way of presenting "11" by 7 digits "0":

11 = (0!+0!+0!)^(0!+0!) + 0! + 0!

Serg

[Serg]

Hi!
Here is the way of presenting "11" by 6 digits "0":

11 = sqrt((0!+0!+0!+0!+0!)! + 0!)

Serg

[Smythe Dakota]

Well, I guess you did list "root sign" as one of the acceptable symbols, so you escape being razzed on this one.

Bill Smythe

[Serg]

Hi!
Summary presentation for "0" digit:
2014 = (0!+0!)^sqrt((0!+0!+0!+0!+0!)!+0!) - (0!+0!+0!)*sqrt((0!+0!+0!+0!+0!)!+0!) - 0!; (18 digits)

Some variants for digits "3" and "4":

2014 = ((3+3)/3)^(33/3) - 33 - 3/3; (10 digits)
2014 = 44*(44+sqrt(4)) - 4 - 4 - sqrt(4); (8 digits)

Serg

[Pat]

2014
0 used 15 times

Code: Select all


(0!+0!) ^ [ (0!+0!+0!)! x (0!+0!) - 0! ]

+ 0!
+ 0!

- (0!+0!+0!)! ^ (0!+0!)



[Pat]

2014 =
44 x (44 + sqrt(4))
- 4
- 4
- sqrt(4)

2014
4 used 6 times

Code: Select all


4! x sqrt(4) x [ 44 - sqrt(4) ]

- sqrt(4)



[Serg]

Hi, Pat!

2014
0 used 15 times

Code: Select all


(0!+0!) ^ [ (0!+0!+0!)! x (0!+0!) - 0! ]

+ 0!
+ 0!

- (0!+0!+0!)! ^ (0!+0!)



Congratulations! Your formula with zeroes is the best at the moment.

Serg

[Serg]

Hi, Pat!

2014 =
44 x (44 + sqrt(4))
- 4
- 4
- sqrt(4)

2014
4 used 6 times

Code: Select all


4! x sqrt(4) x [ 44 - sqrt(4) ]

- sqrt(4)



Your formula for "4" digit is very impressive! I hardly beleived that my result with 8 digits can be beaten.
I see you are really human (despite your post months ago) . I think this puzzle is very difficult to solve by computer assistance.

Serg

[Smythe Dakota]

I'm glad everybody is having such fun with zeroes. I told you so!

Bill Smythe

[Serg]

Hi!
Here are 2 10-digits variants of "5"-digits presentations of 2014:
2014 = ((5 + 5)/5)^(55/5) - 5!/5 - 5 - 5; (10 digits)
2014 = 5^5 - (5! + 5)x(5 + 5) + 5! - 5!/5 - 5; (10 digits)

It is 9-digits variant:
2014 = 5x(5! + 5! + 5! - 5) + 5! + 5! - 5/5; (9 digits)

Serg

[Edited. I noticed the way to remove one "5" from the 3-rd example.]

[Pat]

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

2 used 8 times

Code: Select all


sqrt ( 2^22 )

- ( 2+2+2 ) ^2

+ 2



Code: Select all


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

- 22