## "2014" arithmetic puzzle

For fans of all other kinds of logic puzzles

### "2014" arithmetic puzzle

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.]
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

### Re: "2014" arithmetic puzzle

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
Smythe Dakota

Posts: 563
Joined: 11 February 2006

### Re: "2014" arithmetic puzzle

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
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

### Re: "2014" arithmetic puzzle

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
Smythe Dakota

Posts: 563
Joined: 11 February 2006

### Re: "2014" arithmetic puzzle

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

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

Serg
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

### Re: "2014" arithmetic puzzle

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

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

Serg
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

### Re: "2014" arithmetic puzzle

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

Bill Smythe
Smythe Dakota

Posts: 563
Joined: 11 February 2006

### Re: "2014" arithmetic puzzle

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
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

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

Posts: 3879
Joined: 18 July 2005

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)`

Pat

Posts: 3879
Joined: 18 July 2005

### Re:

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
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

### Re:

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
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

### Re: "2014" arithmetic puzzle

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

Bill Smythe
Smythe Dakota

Posts: 563
Joined: 11 February 2006

### Re: "2014" arithmetic puzzle

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.
Serg
2018 Supporter

Posts: 690
Joined: 01 June 2010
Location: Russia

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`

Pat

Posts: 3879
Joined: 18 July 2005

Next