Puzzle
8 minute read

Ponder This Challenge - January 2010 - Make a 5 from 2 and 2

Present a computation whose result is 5, being a composition of commonly used mathematical functions and field operators (anything from simple addition to hyperbolic arc-tangent functions will do), but using only two constants, both of them 2.

It is too easy to do it using round, floor, or ceiling functions, so we do not allow them.

Update 1/11: You can only use 2 instances of the constant 2, so solutions like (2+2+2/2) is illegal. You can not use variables so (x+x+x+x+x)/x is not allowed. You can use square root and power, but squaring will cost you one "2" and other constant powers are not allowed.

Solution

  • We received dozens of nice solutions (many thanks to our innovative solvers).

    Some we did not accept, like using the rounding function: ceil(sqrt((2*2)!)).

    If we allow for the constant e, then we can have ln(ln(e)/ln(sqrt(sqrt(sqrt(sqrt(sqrt(e))))))/ln(2). But allowing for any constant makes the problem too easy, since we can trivially get 5=(i+i+i+i+i)/i. That's why we did not accept the nice solution (2+i)*(2-i). Using esoteric functions we can get solutions like 2 + vercosin(arccos(2)).

    You can get any integer by -log_2(log(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(exp(2))))))))).

    As done sometimes by computer scientists, if we denote lg(x) as log_2(x) then lg(lg(2)/lg(sqrt(sqrt(sqrt(sqrt(sqrt(2))))))) is also a solution.

    And a nice solution for 5 is cos(atan(2))^-2.

Solvers

  • Andrew Marut (01/05/2010 04:01 PM EDT)
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