# The REAL-WORLD answer for 1000 \$1s in 10 Envelopes --- it ain't a power of two, geeks!

Guys, I did not answer your puzzler with \$1000 and 10 envelopes because I thought the solution was too easy. But now I realize you do not know it. Yeah, you found your solution with power of two, but that just proves you are geeks from MIT.

OK, have you heard of 200 bills? The Euro has them, and before the Euro many countries had theirs. Here is the REAL-WORLD solution that we have used since elementary school. Six bills: 500, 200, 200, 50, 20, 20; and four coins: 5, 2, 2, 1. Here you go, you can make these into your 10 envelopes. We do this with the cents as well: 50, 20, 20, 5, 2, 2, 1 cents. This is the REAL WORLD outside the USA. We think, dudes! :-)))

Sincerely
Rudy (prof. of Math…)
Czech Republic

In the REAL WORLD, we read the question. (Does that make me a geek?) The question specifically says 1,000 \$1 bills, not tens, fifties or anything else. Make sure you are correct before slamming a whole country. In the USA the Math Profs also learn to read!

David L

@RudyB - Yep, you flunked the test, you answered the wrong question. You might be right if the question was “What is the minimum number of bills needed to pay all amounts from \$1 to \$1,000.”

F

Show me how to make \$11 with your solution? Or \$101?

In the USA, our math has to add up… Prof.

@RubyB - Here in the unreal world, don’t need no stinkin’ degree in math to deal with money. We learn that around second grade. Joke’s on you, Hoss…

Getting back to the actual problem, I think everyone agrees that Ray is correct that that distribution he gave will work, but I wonder it it is unique? Are there are other ways of distributing the 1000 bills into 10 envelopes that will meet the objective?

I think if there were 1023 bills, there is no other way. But there may be with 1000 bills.

489, 256, 128, 64, 32, 16, 8, 4, 2, and 1,

500, 250, 125, 62, 32, 16, 8, 4, 2, and 1?

The smallest six are still powers of two, allowing any value from \$0 to \$63, adding the next envelope of \$62 gives \$0 to \$125, then adding \$125 gives \$0 to \$250, adding \$250 and then \$500 extends the range to \$1000.

The second solution has more “round” numbers while still, I think, meeting the requirements.

@AETA - I think you’re correct, there are other combinations that would fit in the 10 evelopes that would allow counting to \$1,000. But 10 envelopes is the minimun number needed.

I’ll just bet that RudyB is a really “wild and crazy guy.”

I bet so too.
After all, who in his right mind in this day and age of debit cards and electronic banking transactions, would walk around with several envelopes with large amounts of cash in them?

@asecular…And you know this HOW???

@AETA, good post. You are right, the answer proffered-up by theCar Talk brothers is not the only answer. Good on you for discovering a second one.