When I posted this thread, I never thought it would get this much action. But I guess all your thoughts outnumbered mine. Again "thanks" for participating. Clinker
yes I disagree. If you had 4 20¢ pieces and 13 3¢ pieces that would be 17 coins and it would work. Also 39 3¢ coins and one 2¢ piece would also work. I think that would be the most coins, at 40 pieces if the goal was to get to $1.19. Like others have mentioned if the goal was to have the most value of coins and not be able to make change for $1 then the answer is only limited by the size of your pockets/purse and the strength of your belt/arms as there remains an exceptionally large number of extant 3¢ coins; from which one could never change a dollar.
Now that isn't right. Each group of five 3¢ coins would make up 15¢ toward the dollar change - for example 30 3-centers + a dime, 10 3-centers + 2 quarters and 2 dimes, etc..
You missed what he was saying. There are no quantities of ONLY 3¢ coins that will make change for a dollar. Yes, if you add other coins it can be done, but not with just 3¢ coins.
I believe the fewest # of coins you can do it with is seven. One dollar, one dime, one nickel, and four pennies.
If you include all USA coins the fewest number of coins you can do it with is five. One dollar, one dime, one nickel, and two (2) 2-cent pieces (or a 3 cent piece and a penny). I think.
p.s. I think the original post is unclear as to what's defined as "make change" and "change a dollar".
Like your first post (before you changed it), I am confused. If you are going for the fewest, why not 1? Just what do the one dime, one nickel, and four pennies add?
I guess fractional paper money isn't "today's coinage" either. Thanks for the thought provoking post.
The 15 cent fractional currency was to help leadfoot add up to $1.19 more quickly. 1 + .15 + .03 + .01 Actually, a cent which is struck on a struck 3-cent would be a 4-cent piece.... Again, thanks for the thought provoking post.