A friend presented an easy puzzle to me:
"Given the digits 2 3 4 5 6 , insert 4 mathematical operators between them, do not change the order and get 10 as a result."
My solution was (2^3+4)*5/6 to which my friend reacted by:
"Exponentiation is not permitted, use just the 4 arithmetical operations (or less) and not more than one pair of brackets."

I've found a solution and dare you to find it or maybe find other solutions as well.

Exponentiation - superscripting to the right - is noted as not being permitted, but nothing was said about subscripting to represent different bases:

2 + 3₄ + 5₆ = 10   (note: 3₄ = 3 and  5₆ = 5)
23₄ + 5 - 6 = 10   (note: 23₄ = 2*4¹ + 3*4⁰ = 11)

As exponentiation is not permitted, I have assumed 'iterated exponetiation', i.e., tetration, is also excluded, else the following could be a solution:
(2^^3 + 4) / 5 + 6  = 10 (note: 2^^3 = 2²²= 16,
^^ represents a single operator for tetration in ASCII notation)

Posted by Dej Mar on 2013-01-06 07:36:10