I tried to design this problem so that the best answer wouldn't involve simple hexagonal packing, but I'm not sure I succeeded. Obviously pennies and nickles don't really get considered. A pure hexagonal packing of dimes just won't add up to enough, some quarters have to be in the mix. The length, 4 1/4 inches, was chosen to proclude starting with a hexagonal packing of two rows of 4 quarters, but it does so barely ( 4.5 x 0.955 = 4.2974) and it's not obvious it does so when trying this with real quarters. And it's not clear to me that you can't just drop that last quarter down a little bit yet still squeeze in 6 dimes on the bottom row. Or possibly put a row of 4 quarters on top, a row of 4 on the bottom, and squeeze 6 dimes in the middle. Anyways, enclosed is my answer. Let me know if you see a more obvious way to get to \$2.60, or even better, figure out how to reach \$2.70! I was also trying to figure out how to pose a problem tiling an infinite plane with a set proportion of quarters and dimes. Depending on the ratio chosen, it looks like the densest packings could involve non-periodic pentagonal patterns suggested by my answer to the problem above. Any thoughts welcome. Brian Trial A 2 1/2 by 4 1/4 card will accomodate eight quarters and six dimes for a total of \$2.60. This is the best I could do. Bob Wainwright O O O O O o O O O O O Good one... Michael Smithson I found many ways to fit \$2.50, and one way to fit \$2.55. I've included pic of the \$2.55 solution, although it is not to scale. -aron fay I got \$2.60 on the Coin problem, but I suspect it can be improved. Seems like there ought to be a way to get another dime in there somewhere. I'll be looking forward to seeing the results in a week or so. Clint Weaver My "Coins" answer was a little off. The quarters along the top and bottom edges needed to be moved inward a little to make them correctly fit on the card. Attached is a revised version. Clinton Weaver Here's my answer for Euros, I get EUR 22.01 I would carefully check any claims that put a second cent coin, or a 2 cent coin, along with 11 2 Euro coins. There is ALMOST enough room for a second cent, but from what I could determine, not quite enough. <> Brian Trial