## Puzzle: Express 2008 with a minimal number of just one digit.

June 17th, 2008 | Categories: games, general math | Tags:

At the beginning of the year I wondered what interesting facts I could discover about the number 2008 and, with the help of readers of this blog, I came up with a lot more than I expected to. Now that we are over half way through the year I thought it was time to take another look at the integer 2008 with the following puzzle.

Choose any one of the digits from 0 to 9 and attempt to express the number 2008 using only that digit. You can repeat your digit as often as you like and use any of functions that are built into something like Mathematica, MATLAB or SAGE but, as you may expect, kudos will be awarded for using only simple functions and small numbers of repeats.

I spent a few minutes thinking about this problem and so far have only come up with

2008= (2*2^2)*(2^2^2)^2 – 2*2^(2*2) – 2*(2 + 2)

2008 = Ceiling[6*6*6*6 + 6! – Sqrt]

2008 = Ceiling[Gamma[7.7] – (777) + 7 + 7 + 7/7]

but I am sure you can do better – those ceiling functions are pretty ugly for a start. Feel free to post your solutions in the comments section – I look forward to seeing them.

1. 2008 = ( ( (0!+0!+0!)! + 0! )! – (0!+0!)^( 0! + (0!+0!+0!)^(0!+0!) ) / (0!+0!)

(that is, 2008 = (7! – 2^10 ) / 2 )

2. Nice one TwoPi :)

3. For the 2s, you can do 2222 – 222 + 2*2*2 for only ten 2s with simple operations, or if you don’t like concatenation, then

2008 = (2^(2*2*2+2)-2*(2*2*2+2))*2 with eleven 2s.

Or maybe
2008 = (2^(2/.2) – 2*2/.2)*2 with only seven 2s.

The (7! – 2^10)/2 is brilliant, though. I don’t think I would have noticed that cute little relation!

There’s a theorem that if you have one digit greater than 2, and you have !, sqrt, and floor, you can make any positive integer (I think Conway proved it) so in some sense all these games are spoiled if you have floor.

OK, on to 3s:
3^(3*3)/(3*3) – (3+3)*3^3 – 3*3 – 3*3 + 3/3
but that seems like way too many 3s.

OK, two fewer in the same pattern:
3^(3+3)*3 – (3+3)*3^3 – (3+3)*3 + 3/3
and keep factoring,
3^(3+3)*3 – (3+3)*(3^3+3) + 3/3
not too awful, anyway.

Then of course
3^(3!) * 3 – (3!)*(3^3+3) + 3/3
OK, good enough for now.

With 4s, my seven 2s solution again:

2008 = (sqrt(4)^(4/.4) – 4*4/.4)*sqrt(4)

but maybe prettier without the ., and with one fewer 4,
4*((4^4-4)*sqrt(4)-sqrt(4))
rearrange to cut down by 1 sqrt,
sqrt(4) * ((4^4 – 4)*4 – 4)

An ugly first solution:
5^5-(5+5)*(5!-5)+5!/5+5+5-5/5
another perhaps even uglier:
5!*(5+5+5+(5+5)/5)-5*5-5-(5+5)/5

6!+6*6*6*6-6-(6+6)/6

But 7, argh.
[7!/7+7*7*7-7*7-7-(7+7+7)/7]*(7+7)/7
7*7*7*7-7*7*7-7*7-7/7
Now that’s kinda pretty.
Or,
7*7*(7*7-7-7/7)-7/7
Only 9 of them, and only +-*/.

((8*8*8-8)*8)*(8/(8+8))-8

9*9*9*sqrt(9)-9*9-9*9-9-9+9/9, yuck.
9^sqrt(9)*sqrt(9)-(9*9+9)*(9+9)/9+9/9, still yuck.

Better: (9^sqrt(9)-9)*9 – (9!/9 + 9)/9 + 9

I think I did quite well on the “only simple functions” but not so hot on the number of repeats. I’d love to see some better answers!

4. Hi Joshua

Wow, very impressive! Thanks for your contribution – I’ll look into the Conway theorem you mentioned as it sounds very interesting.

Mike