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.

(In reply to More solutions by Dej Mar)

Whatever you added is interesting and welcome, but way out of
the scope of the  basic  arithmetic operations. explicitly  spelled out in the puzzle.

Whenever any constraint is lifted or any new operator allowed,
new solutions will pop up.

Assuming 2 pairs of brackets are admissible -
then (2-3*4)*(5-6) would qualify.

Assuming 2 pairs of brackets and 5 arith. operators-
then (2*3+4)*(-5+6) fits the requirements.

I remember the FIVE FIVES PUZZLE, where by expanding the list of allowable operations  we've ended up with over 1000   entries.

Still the main point demonstrated by this "easy" problem is that the most obvious and the most elegant straightforward solution is not necessarily the first  one that you find.

 

Edited on January 6, 2013, 10:32 am

Posted by Ady TZIDON on 2013-01-06 10:04:49