Thread: John,Mike,and Charles

John,Mike,and Charles divide a pile of pennies. If the number of pennies in the pile is even, Mike will get half of the pile. If the number of pennies in the pile is odd, one penny will be given to Charles and John will get half of the pennies remaining in the pile. This process is repeated until the pile is gone. How many pennies will Mike receive if the original pile contains 2005 pennies?

Originally Posted by aznmartinjai

John,Mike,and Charles divide a pile of pennies. If the number of pennies in the pile is even, Mike will get half of the pile. If the number of pennies in the pile is odd, one penny will be given to Charles and John will get half of the pennies remaining in the pile. This process is repeated until the pile is gone. How many pennies will Mike receive if the original pile contains 2005 pennies?

Step through the algorithm:

stage 1 2005 pennies 1 to C, 1002 to J, leaves 1002
stage 2 1002 pennies 501 to M, leaves 501
stage 3 501 pennies 1 to C, 250 to J, leaves 250
stage 4 250 pennies 125 to M, leaves 125
stage 5 125 pennies 1 to C, 62 to J, leaves 62
stage 6 62 pennies 31 to M, leaves 31
stage 7 31 pennies 1 to C, 15 to J, leaves 15
stage 8 15 pennies 1 to C, 7 to J, leaves 7
stage 9 7 pennies 1 to C, 3 to J, leaves 3
stage 10 3 pennies 1 to C, 1 to J, leaves 1
stage 11 1 pennies 1 to C, 0 to J, leaves 0

Now do some adding:

C gets: 8 pennies
M gets: 657 pennies
J gets: 1340 pennies

I'm pretty sure that there is a neat direct way of doing this, but the question looks to me as though it is meant to introduce you to the idea of an algorithm, so I have solved it by tracing the process

RonL

thanks you so much captainblack =)