Suppose you had $1000 that you split into 10 separate envelopes so that you could give any amount of money from $1 to $1000 when asked. How much money would you put in each envelope?

Congratulations to Matt Ginsky for being the first to submit the correct answer. The correct distribution of the $1000 into 10 envelopes would be: 1, 2, 4, 8, 16, 32, 64, 128, 256, and 489. (Do you notice a trend in the first 9 numbers?)

## Wednesday, September 10, 2008

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## 2 comments:

The conditions of this scenario preclude a particularly specific and definite answer. If for example, you could only use one envelope, you would put all $1,000 in one envelope in varying denominations and that would be the end of it. The problem becomes more interesting if you are allowed to use more than one envelope but cannot leave any money in that envelope when you’ve finished. If that’s the case, the scenario really becomes intriguing; you would essentially need any envelopes as you have bills, which, in my calculation, with only dollar integers would be 18 envelopes. You would need to ascertain 1 one dollar bill, 2 two dollar bills, 1 five dollar bill, 1 ten dollar bill, 4 twenty dollar bills, and 9 one hundred dollar bills. This combination results in the ability to dispense any integer amount between $1 and $1000 (no fractional amounts of a dollar) at will but by carrying the fewest number of bills (only 18 total bills). If cents were taken into account, you would want to have 3 quarters, 1 dime, 2 nickels, 5 pennies, 3 one dollar bill, 1 two dollar bills, 1 five dollar bill, 1 ten dollar bill, 4 twenty dollar bills, and 9 one hundred dollar bills. If you could have ONLY ten envelopes, you could place the bills and change into the envelopes in any combination as long as you could remove only part of the currency and yet keep the rest in the envelope (e.g. you have a ten and a five in an envelope and you need only the ten).

The ten envelopes would contain the following: 1,2,4,8,16,32,64,128,256,489.

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