Friday Puzzler -- four fractions

Re: Friday Puzzler -- four fractions

Henry Higginbotham

I've found three, one being 1/2+2/3+6/7+41/42. It appears any solution would have to lean heavily to the smaller fractions (1/2, 2/3 ...) and that the denominators of the three smallest values have to be factors of the denominator of the largest.

It seems there should be a way of predicting which combinations yield an integer (always 3 in this case?), but I'm drawing a blank on that.

Re: Friday Puzzler -- four fractions

Larry Barrett

Here is one:

1/2 + 3/4 + 5/6 + 11/12 = 3

Good observations

Alex Y

The one you identified is, I believe, the one with the largest possible denominator. (Would be nice if someone skilled at programming could whip up a program to confirm, though.)

I believe you are right about heavily depending on the smaller fractions, and I THINK that 1/2 has to be one of the fractions.

It took me a while to come up with the observation you made so quickly about the smaller denominators all having to be divisors of the largest one. But I haven't figured out how to use that observation to aid in the hunt for solutions.

What are the other two you have found?

Good one

Re: Good observations

Henry Higginbotham

The other two were 1/2+2/3+7/8+23/24 and 1/2+3/4+4/5+19/20.

1/2+2/3+8/9+17/18 also works. With Larry's answer, that's five.

I looked at possible combinations through around 79/80 without finding anything else, but if 41/42 is in fact the end of the line, there must be some way of proving that. Sometimes the three smallest fractions that satisfy the factor requirement will run the sum past 3.000, and you can just stop there (although maybe that's obvious and useless).

Re: Good observations

Alex Y

I have been looking at this one with the aid of Excel -- definitely not the best tool for the job, but one has to use the tools that he knows how to operate. :-( I came up with 14 answers, but discovered that I had double counted, and not only that, one of my (double-counted) solutions had the same fraction occurring twice. As I said, not the best tool--it doesn't prevent operator error. ;-)

Anyway, I am now down to six solutions, which I feel pretty sure is all.

I found it helpful to look at the smallest pair of fractions combined with the larger pair. The smallest pair must add to less than 1 1/2, which limits it quickly to pairs that contain either a 1/2 or 2/3 (or both). Then a stronger (but less readily apparent to me) test is that the smaller two plus the lowest two that are larger than them must add to three or less. E.g., 1/2 and 9/10 adds to 1 4/10, which is < 1 1/2. But the smallest candidates for the larger pair (without even considering the divisbility issue you mentioned) is 10/11 and 11/12. those two +1 4/10 >3, so 1/2, 9/10 is not a smallest pair.

This got me to (1/2,2/3), (1/2,3/4) and (1/2,4/5) as the only possible lowest pairs.

Re: Good observations

Henry Higginbotham

Hmmm. My "factors" idea needs a little tweaking. I assume the sixth solution you mentioned is 1/2+2/3+9/10+14/15, which doesn't work quite the way I had guessed.

A couple more observations:

1. For any three smaller fractions, there is only one larger one that will work. I noticed the 14/15 solution by throwing out the combinations already used and working with the narrow choices that were left.

2. I don't believe there are any solutions involving 1/2 and 4/5.

Re: Good observations

Larry Barrett

I have been following your and Alex's observations and thinking about this. I have not opened Excel so far, but have some questions.

1. Henry, you said that you think there are no solutions involving 1/2, 4/5. But you just listed one - 1/2+3/4+4/5+19/20. Or perhaps you meant that there are no solutions where 1/2 and 4/5 are the lowest pairs.

2. The observation that the smaller denominators must all be factors of the largest denominator seems like a logical place to start. It at least makes it easy to convert all the fractions to the same common denominator and add the numerators. But is there a logical way to eliminate a solution where the denominatiors are not all factors of the largest denominator?

3. Alex started this question with an example of a 3 fraction solution - 1/2+2/3+5/6=2. Are there other solutions to the 3 fraction problem? Can this be extended to a 2 fraction problem, adding to 1? (I don't think so.) Can this be extended to a 5 fraction problem, adding to 4?

Good catch, Larry!

Henry Higginbotham

Yes, I was commenting on Alex's remark that (1/2, 4/5) is one of the "only possible lowest pairs." I should have read it more critically before posting.

Note, too, that I had to revise my factors idea after noticing that (1/2+1/3+1/10=1/15) didn't quite work out. But the denominators are all factors of 30, which still allows for an integer sum. Seems more obvious with hindsight.

As for a 2-fraction problem: The integer sum must be 1 (since the sum of any two fractions must be less than two), and the two smallest (1/2 and 2/3) are already a bust.

I think you can eliminate other 3-fraction solutions quickly by just trying a few. Once your largest fraction takes you above 2.000, going further only makes it worse. But I'm not the one to attempt any sort of logical proof.

This does sound very much like other puzzles I've seen wherein some smart guy (read: not me) comes up with a brilliant, structured method to solve and prove everything. Your turn--get on it!

Re: Good observations

Alex Y

Hmmm. My "factors" idea needs a little tweaking. I assume the sixth solution you mentioned is 1/2+2/3+9/10+14/15, which doesn't work quite the way I had guessed.

Yes, that is the sixth one I had found. And I totally missed the fact that it violated the "lower denominators a divisor of highest denominator" rule.

1. For any three smaller fractions, there is only one larger one that will work.

Specifically, 3 minus the sum of the other three

2. I don't believe there are any solutions involving 1/2 and 4/5.

Correct. Using my spreadsheet, I found the "solution" of 1/2 + 4/5 +4/5 + 9/10. Obviously missed the "distinct fractions" part of the problem!

Your new queston

Alex Y

Can this be extended to a 5 fraction problem, adding to 4?

My quick answer is "yes". On the other hand, whether there are any solutions will take more cogitation. ;-)

Re: Your new queston

Alex Y

Here's one:

1/2 + 3/4 + 7/8 + 11/12 + 23/24 = 4