For each digit from 1 to 9, use that digit three times, with arithmetic operations, to make 3.

arithmetic/digits/equations/make.three New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

For each digit from 1 to 9, use that digit three times, with arithmetic
operations, to make 3.

For example:

1+1+1
2/2+2
3-3+3
4-4/4

That's all I can do, anyone else know more?

-- Wolf303

[ Editor: problem statement reconstructed from the replies; the
original question is not in the collected material. ]

Show solution

Chang W. Chen wrote:

> 1^0 + 1^0 + 1^0 = 3
> 2^0 + 2^0 + 2^0 = 3
> 3^0 + 3^0 + 3^0 = 3
> 4^0 + 4^0 + 4^0 = 3
> ............... = 3
> ............... = 3
> ............... = 3
> ............... = 3
> 9^0 + 9^0 + 9^0 = 3
>
> would that work?

I don't think so since you are using zeroes and they are not in the
list.  How about --

sqrt ((1-.1)/.1)
sqrt ((2-.2)/.2)
sqrt ((3-.3)/.3)
sqrt ((4-.4)/.4)
sqrt ((5-.5)/.5)
sqrt ((6-.6)/.6)
sqrt ((7-.7)/.7)
sqrt ((8-.8)/.8)
sqrt ((9-.9)/.9)

-- Mike Conner ("RS")

This is pretty darn good.

7 minus .7 is 6.3
6.3 divided by 7/10 is the same as 6.3 x 10 / 7 or 63/7 or 9
and the square root of 9 is sure 3!  Very good!

-- Dave Gates <davegates@davegates.com>, January 1998