500 pennies...


New member
Apr 23, 2011
Ok, so I am not sure this is an algebra question, but it was given to me in a beginning and intermediate algebra class as a question of the week... and I am completely stumped. Any help would be appreciated.

Suppose there are 500 pennies, all facing "heads", lined up on a table. Along with the 500 pennies, are 500 people lined up.

The 1st person flips over every penny.

The 2nd person flips over every 2nd penny, starting with the second penny.

The 3rd person flips over every 3rd penny, starting with the third penny.

The 4th person flips over every 4th penny, starting with the fourth penny.

The penny flipping continues this pattern through all 500 people.

ALL the information I am given.

Wow, what a headache.

so the 250th person in line flips over the 250th penny and the 500th penny... but this does not mean this is where the penny flipping ends, because the 251st person will flip penny #251, one final time, and the 252nd person will flip penny #252 one final time... ect ect ect.


Staff member
Apr 12, 2005
Not as big a deal as it seems. Think about prime factorizations.

For example

-- 20 is flipped by 1, 2, 4, 5, 10, and 20 -- That's six flips.
-- 65 is flipped by 1, 5, 13, and 65 -- That's four flips.

There's a secret in there.

If you don't see it, try 49 and it may come to you.

Subhotosh Khan

Super Moderator
Staff member
Jun 18, 2007
Do a google search on 1000 lockers problem.