A Digital Permutation Problem (Posted on 2006-02-28)

	Submitted by K Sengupta
Rating: 3.0000 (4 votes)
Solution:	(Hide)
SOLUTION SUMMARY: The required number is 15328467. ANALYTIC SOLUTION (Submitted By Eric): So let ABCDEFG = x then 10x + H / (10^7)H + x = 3/14 or: 140x + 14H = 3x10^7H +3x or 29999986H = 137x or 218978H = x X is a seven digit positive integer and H is less than 9. H=8 fails, but H=7 gives x=1532846. ABCDEFGH = 15328467 also gives 67153284/84671532 = 23/29 ------------------- ALTERNATE SOLUTION (Submitted by tomarken): "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."

	Subject	Author	Date
	A different method	tomarken	2006-02-28 18:30:30
	re: Solution - I'll spare you	Eric	2006-02-28 17:55:03
	Solution	tomarken	2006-02-28 17:00:38