"2014" arithmetic puzzle

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"2014" arithmetic puzzle

Postby Serg » Mon Jan 20, 2014 11:56 am

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.]
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Re: "2014" arithmetic puzzle

Postby Smythe Dakota » Mon Jan 20, 2014 7:16 pm

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.

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Re: "2014" arithmetic puzzle

Postby Serg » Tue Jan 21, 2014 11:38 am

Hi, Bill!
Smythe Dakota wrote: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
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Re: "2014" arithmetic puzzle

Postby Smythe Dakota » Tue Jan 21, 2014 12:26 pm

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?

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Re: "2014" arithmetic puzzle

Postby Serg » Tue Jan 21, 2014 12:40 pm

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

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

Serg
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Re: "2014" arithmetic puzzle

Postby Serg » Tue Jan 21, 2014 12:55 pm

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

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

Serg
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Re: "2014" arithmetic puzzle

Postby Smythe Dakota » Wed Jan 22, 2014 3:33 am

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

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Re: "2014" arithmetic puzzle

Postby Serg » Wed Jan 22, 2014 10:36 am

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
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Postby Pat » Wed Jan 22, 2014 12:23 pm


    2014
    0 used 15 times
    Code: Select all

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

    + 0!
    + 0!

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

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Postby Pat » Wed Jan 22, 2014 12:46 pm

Serg (8 digits) wrote:

    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)

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Re:

Postby Serg » Wed Jan 22, 2014 1:02 pm

Hi, Pat!
Pat wrote:

    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
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Re:

Postby Serg » Wed Jan 22, 2014 1:08 pm

Hi, Pat!
Pat wrote:
Serg (8 digits) wrote:

    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
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Re: "2014" arithmetic puzzle

Postby Smythe Dakota » Thu Jan 23, 2014 12:38 am

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

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Re: "2014" arithmetic puzzle

Postby Serg » Thu Jan 23, 2014 11:12 am

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.]
Last edited by Serg on Thu Jan 23, 2014 12:28 pm, edited 1 time in total.
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Postby Pat » Thu Jan 23, 2014 11:43 am

Serg wrote:

    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

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