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tricky algebra problem (1 Viewer)

Maurile Tremblay

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a/(b+c) + b/(a+c) + c/(a+b) = 4

Find a solution where a, b, and c are all positive integers.

I played around with this for about 30-60 minutes (I didn't keep track) before giving up and looking at the answer. I don't think it was a waste of time. The explanation (linked below) is very good, which is why I'm posting this. It's better if, before reading it, you work on the problem long enough to convince yourself that you're not going to solve it within a reasonable amount of time (unless your math knowledge is pretty darn esoteric).

Answer: https://www.quora.com/How-do-you-find-the-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4

 
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a/(b+c) + b/(a+c) + c/(a+b) = 4

Find a solution where a, b, and c are all positive integers.

I played around with this for about 30 minutes before giving up and looking at the answer. I don't think it was a waste of time. The explanation (linked below) is very good, which is why I'm posting this. It's better if, before reading it, you work on the problem long enough to convince yourself that you're not going to solve it within a reasonable amount of time (unless your math knowledge is pretty darn esoteric).

Answer: https://www.quora.com/How-do-you-find-the-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4
Played around for 5 minutes and closest I could get was 23, 3, 3 but it came out to 4.06

 
Played around for 5 minutes and closest I could get was 23, 3, 3 but it came out to 4.06
1, 7, 30 gets to 4.00283 -- so close in value, and yet conceptually still a zillion miles away. Denominators of 37, 31, and 8 are not going to allow summing to an integer.

 
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Closest I can get is 

a3 + b3 + c3 - 3(a2b + a2c + b2a + b2c + c2a + c2b) - 5abc = 0

Not helping me at all  :loco:

 
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Closest I can get is 

a3 + b3 + c3 - 3(a2b + a2c + b2a + b2c + c2a + c2b) - 5abc = 0

Not helping me at all  :loco:
Right. I got to that point and then decided, "I can't actually solve this algebraically. I'm going to just start putting numbers into Excel..." After doing that for a while, I conceptually proved to myself that there's no way b or c could be less than 100 (starting with a=1), and that's when I looked at the explanation.

 
a/(b+c) + b/(a+c) + c/(a+b) = 4

Find a solution where a, b, and c are all positive integers.

I played around with this for about 30-60 minutes (I didn't keep track) before giving up and looking at the answer. I don't think it was a waste of time. The explanation (linked below) is very good, which is why I'm posting this. It's better if, before reading it, you work on the problem long enough to convince yourself that you're not going to solve it within a reasonable amount of time (unless your math knowledge is pretty darn esoteric).

Answer: https://www.quora.com/How-do-you-find-the-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4
One solution.

 
Right. I got to that point and then decided, "I can't actually solve this algebraically. I'm going to just start putting numbers into Excel..." After doing that for a while, I conceptually proved to myself that there's no way b or c could be less than 100 (starting with a=1), and that's when I looked at the explanation.
Got here too, started scrolling down the thread, haven't looked at the solution yet.  Dang....

 
Maurile Tremblay said:
Right. I got to that point and then decided, "I can't actually solve this algebraically. I'm going to just start putting numbers into Excel..." After doing that for a while, I conceptually proved to myself that there's no way b or c could be less than 100 (starting with a=1), and that's when I looked at the explanation.
Me too.  And then when I read the explanation, I remembered why I didn't like higher math.  Once things got too abstract I lost interest.  There's only so many puzzles one can do for the sake of doing puzzles.  At least, this one. :)   But God bless the theoretical mathematicians, where would we be without them?

 
Maurile Tremblay said:
a/(b+c) + b/(a+c) + c/(a+b) = 4

Find a solution where a, b, and c are all positive integers.

I played around with this for about 30-60 minutes (I didn't keep track) before giving up and looking at the answer. I don't think it was a waste of time. The explanation (linked below) is very good, which is why I'm posting this. It's better if, before reading it, you work on the problem long enough to convince yourself that you're not going to solve it within a reasonable amount of time (unless your math knowledge is pretty darn esoteric).

Answer: https://www.quora.com/How-do-you-find-the-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4
Took me 15 seconds on Google 

 
Maurile Tremblay said:
a/(b+c) + b/(a+c) + c/(a+b) = 4

Find a solution where a, b, and c are all positive integers.

I played around with this for about 30-60 minutes (I didn't keep track) before giving up and looking at the answer. I don't think it was a waste of time. The explanation (linked below) is very good, which is why I'm posting this. It's better if, before reading it, you work on the problem long enough to convince yourself that you're not going to solve it within a reasonable amount of time (unless your math knowledge is pretty darn esoteric).

Answer: https://www.quora.com/How-do-you-find-the-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4
Took me 15 seconds on Google 
Well that's a complete waste of 15 seconds you'll never get back.

I GAVE YOU THE LINK!!

 
Just read the answer after giving up.  Guy does a good job of explaining the answer, but I still don't really grasp it. 

 
You'd think that I'd feel better about myself after spending 15 minutes on this and then looking at the answer. But no.

 
Just read the answer after giving up.  Guy does a good job of explaining the answer, but I still don't really grasp it. 
There are very few people in the world who "really" grasp. It wouldn't surprise me if MT was one of them, but for the most part, we are all pretty dumb.

 
I was going to write a bit of optimization code to find the answer, but I am glad I read the answer first. I don't think it would have worked. I am not sure it would ever try such large numbers.I am pretty sure I would have found a local minima.

 
I haven't looked at the answer yet or anything outside of the 1st post.

First kept trying 1,2, and a different #'s then multiples  and finally tried 1 twice and finally figure out that way.

7,1,1

Never mind I'm dumb :(

 
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