 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  A Digital Permutation Problem (Posted on 2006-02-28) Determine an 8-digit number of the form ABCDEFGH where each of the letters denote a distinct digit from 1 to 8 such that:

ABCDEFGH/ HABCDEFG = 3/14 , and

GHABCDEF/ EFGHABCD = 23/29.

 See The Solution Submitted by K Sengupta Rating: 3.0000 (4 votes) Comments: ( Back to comment list | You must be logged in to post comments.) A different method Comment 3 of 3 | At first I thought to go about it algebraically the same way, but that seemed like less fun so I figured I'd try a different, more intuitive approach:

I set up an Excel spreadsheet to do all the calculations as I plugged in numbers, which made this a lot easier.  First I looked for simple fractions that I could build off of.  "A" had to be 1, since any numerator starting with 2 would be too high for the first equation.  It then follows that "H" must be at least 5, since anything less would make the fraction greater than 3 / 14.  Why is that a problem?

Notice that each digit you fill in the first equation (with the exception of "H") will increase the numerator more than the denominator - for example, substituting 5 for "B" adds 5 million to the numerator but only 500,000 to the denominator.  Therefore you have to start off with a base fraction *less* than 3 / 14 and build your way up as you add digits.

Thinking that 3 / 14 can be simplified to 1.5 / 7, I started from there (which of course, in the context of the problem, was actually 15M / 71.5M).

This was close enough to the desired result to begin with that I figured "C" and "D" must be low numbers as well (to minimize the effect I mentioned earlier - this was easily confirmed as I plugged numbers into the spreadsheet).  1.532 / 7.153 worked well enough to move onto the second equation.

With half of the digits already filled in and only a 4, 6, and 8 left unplaced, it was easy enough to see that 6.7 / 8.4 was closest to 23/29, and that was basically the end of the problem.

While this was basically a guess-and-check method, I at least used very *educated* guesses as I plodded my way through.  Having somewhere likely to start from at the very least reduced the number of possible permutations to a more manageable level.

 Posted by tomarken on 2006-02-28 18:30:30 Please log in:
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