Math Puzzles

Oct 5 05:36 UTC 2001

For math puzzles.

response 2 of 122:

Oct 5 15:09 UTC 2001

What is the operator between the fractions? It can't be multiplication,
because even the largest possible value for each (and this breaks the rules)
results in a number less that 1: 9/10^3 = 0.729

response 4 of 122:

Oct 5 19:26 UTC 2001

Algorithm:  Try each permutation of the digits 1..9, in lexicographic
order, until you find one that works.  Success guaranteed!

response 6 of 122:

Oct 5 20:53 UTC 2001

That's not "trial and error"? Well, it's not *random* trial and error. 

I found, with trials, no errors, that one can get very *close* to the
sum of unity quite easily. There should therefore be a search algorithm
in the sample space that converges on unity. That would be a little
better than either a random or exhaustive search. 

response 8 of 122:

Oct 5 23:54 UTC 2001

Ok, to find the answer, I did an exhaustive search. It took me about five
minutes to write it in VB (it was a series of 9 for..next and if..thens,
appropriately nested)and maybe 2 seconds to get an answer. That was the
exhaustive search.

Given the answer, I notice that some of my assumptions were wrong, but all
the same, I could have made an educated guess to greatly reduce John's
options. For instance, we know that the numerators will have to be largish
and the denominators smallish, since the highest POSSIBLE value is 9/12 (or
3/4), and once the 1,2, and 9 have been taken out of the pool, it's hard to
make 1/4 out of the six remaining digits (8/34, the next highest value after
those digits have been removed, is less than 1/4). 

Since I find it's sometimes easier to solve a maze by starting at the exit,
the solution is:
9/12 + 7/68 + 5/34
(5/34 = 10/68, so 5/34 + 7/68 = 17/68 = 1/4)

An exhaustive search demonstrates that this is the only answer (barring the
other five permutations of the same fractions, of course).

response 10 of 122:

Oct 7 06:29 UTC 2001

You haven't answered the question in #7. I note that my "Are you aware:"
should have been "Are you:"

If you're just guessing that an algorithm exists, then I'm not going to rack
my brain to find it. ;}

response 12 of 122:

Oct 8 02:14 UTC 2001

I rephrased the question.
Are you aware that such an algorithm exists?
If so, are you in possession of such an algorithm?

response 14 of 122:

Oct 8 14:07 UTC 2001

Well, this *is* the puzzle item; "puzzle" tends to imply that the solution
is known by the querent, or at least known to exist. ;}

Given the solution, I'm not sure if there's a precise algorithm that would
lead directly to it, or if  there are merely ways to minimizing the search
process by ruling out certain possibilities (for instance, the numerators
would have to be large-ish, but an early prediction of mine that they would
all be in (6, 7, 8, 9) turned out to be false).

response 16 of 122:

Oct 8 16:25 UTC 2001

#15 para 1> Sheesh, Rane, don't get so defensive. I was just pointing out that
my own strategies for approaching apuzzle where a solution is known to exist
is different from approaching a puzzle where that isn't known. I'm not trying
to bitch you out or nuffin. ;}

Since we've got A + B + C = 1, where A, B, and C are all fractions, it would
seem to me that the algorithm would start by identifying, first, all fractions
a/(10b+c), such that a, b, and c are all different, and a/(10b+c) can be
reduced to a simple ratio, like 1/2 or 3/4.

If that fails to produce three fractions with nine different digits, the next
step would be to identify all fraction PAIRS a/(10b+c), d/(10e+f), where a..f
are all different, and the sum a/(10b+c) + d/(10e+f) can be reduced to a
simple ratio.

In the case of this problem, the algorithm would succeed there, since 9/12
= 3/4 and 7/68 + 5/34 = 17/68 = 1/4. Something we notice that also simplifies
the algoithm is that 68 is a multiple of 34. While it's possible that D/E +
F/G = (DG + EF)/EG could be reduced even if E were not a multiple of G (or
vice versa), it's less likely (intuitively, at least) that such a fraction
pair that satisfies the solution could be found.

This isn't what I'd call an algorithm, though, as much as a way to hone wild
guesses into educated ones. For instance, I settled nearly immediately on 9/12
as a likely component, but got caught up in trying to find two other reducible
fractions using the remaining 6 digits.

response 18 of 122:

Oct 8 18:39 UTC 2001

I'm not sure my specific *stretegy* would differ, but my diligence would, and
I think that's what I meant. If I *knew* there was an answer, I'd put more
effort into finding it, assuming I were working on what is, after all, a
distraction. If I had committed to finding a solution to a puzzle of some
relevance, on the other hand, I'd want to know what solutions had already been
obtained so that I built on them instead of reconstructing them.

response 20 of 122:

Oct 8 19:41 UTC 2001

(Re #5: there are only 60,480 fraction combinations - should be some way to
take advantage of that in an exhaustive search.)

response 22 of 122:

Oct 9 18:14 UTC 2001

Re 17: I have proved that which has been said unprovable.
See http://www.geocities.com/joshudon/bignumbers/proofs.htm

response 24 of 122:

Oct 9 23:43 UTC 2001

Re  #22:"The web page you are trying to access doesn't exist on Yahoo!
GeoCities."
?????