This is a famous problem from 1882, to which a prize of $1000 was awarded for the best solution.

The task is to arrange the seven numbers 4, 5, 6, 7, 8, 9, and 0, and eight dots, in such a way that an addition of two or more numbers approximates 82 as closely as possible.
Each of the numbers can be used only once.
The dots can be used in two ways: as decimal point and as symbol for a recurring decimal.

For example, the fraction 1/3 can be written as:

  . 
. 3

The dot on top of the three denotes that this digit is repeated infinitely. If a group of numbers needs to be repeated, two dots are used: one to denote the beginning of the recurring part and one to denote the end of it. For example, the fraction 1/7 can be written as:

  .         . 
. 1 4 2 8 5 7

(Note that '0.5' is written as '.5'.)

How close can you get to the number 82?

There have to be at least three numbers to be added, as each number can have at most three dots, and we need to use up 8.

Repeated single digits immediately after the decimal point are ninths (e.g., .7777... = 7/9). Repeated double digits immediately after the decimal point are 99ths, such as .67676767... = 67/99, etc.)

If they start repeating some place further on, they are divided by the appropriate power of 10, such as .00123123123123... = 123/99900, and then of course .65123123123123... = 123/99900 + 65/100, whatever that reduces to. The numbers are to be written in the dot fashion, of course, which is a little unwieldy for use here, but the point is, for example:

 .    .


.4 + .5 = 1

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Posted by Charlie on 2003-12-24 08:55:29