If I place 1 coin on the bottom lefthand square of a chess board and then 2 on second ,4 on the third and keep doubling the number of coins until the chess board is covered how many coins will i need ? ( assuming of course it's physically possible to do this!). Give your approximate answer in scientific for form.

Question

If  I place 1 coin on the bottom lefthand square of a chess board and then 2 on second ,4 on the third and keep doubling the number of coins until the chess board is covered how many coins will i need ? ( assuming of course it's physically possible to do this!). Give your approximate answer in scientific for form.

turingtest · Answer

numbering the squares 0 to 63 we need $$\sum_{n=0}^{63}2^n$$whatever that is...

anonymous · Answer

Would the factor be 10^18, 10^19... somewhere there?

anonymous · Answer

yeah  10^19

anonymous · Answer

i wonder if there's enough coins in UK to do this!!?

anonymous · Answer

1.8446744 * 10^19

jamesj · Answer

@TT: That sum is $  2^{64}−1 $.
@jimmyrep: No way! :-)

This 'problem' appears in fairy tales some times.  A character promises to do something only he can do for a king and before hand asks for payment of one grain of wheat on one square, two on the second, 4 on the third, etc.  The king agrees and blithely promises forfeiture of the kingdom for the non-payement, because he doesn't know how to add up a geometric series. 

The task is done.  Then the reckoning is made.

turingtest · Answer

ah...
The smart way.

anonymous · Answer

yes its 2^64 - 1  - a Geometric seRIES right?

turingtest · Answer

I was just about to review series...

anonymous · Answer

hardly worth subtracting the 1!!!!