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Math Problem: You have 2 pipes...

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Originally posted by: ndee
Originally posted by: MercenaryForHire
All you said was that you wanted us to solve it. I solved it, and a lot faster than anyone using a Diophantine equation. Applicable to every single class of equations? No. A hell of a lot faster? Yes. Solved? Totally. As long as there's a method to the madness (calcuations and not just number-crunching) an elegant brute-forcing is a valid proof in every university class I've ever seen.

And it's good to see that you're finally admitting to not just "not getting it" - but still "not getting it" after many explanations have been posted. Now unless you're in front of me at waist-height, how about you shut your mouth? :roll:

- M4H
Click to expand...

After the "normal" explanations, I asked for the diophantine way, (I also said no brute-force)etc. Anyway, I'm done with you. After you had the numbers, just assume thing is alot easier then to come up with them by yourself.

Have a nice day.
Click to expand...

And I'm done with you. Now hold your head still or it'll get in your hair.

- M4H
 
Glad to see you guys finally recognized it as a Diophantine equation. All the morons on the first page did nothing but tell each other how stupid they were without solving it.

This is something I covered last quarter in my Discrete and Combinatorials class (college sophomore level class). It's part of number theory. And those who said it before are correct that you use the Euclidean Algorithm to solve it using the GCD.

More math questions that relate to what I'm studying! 😀
 
This is another simple way to solve for A and B without using Euclid algorithm and gcd.
1. 3.9A + 5.9B = 212.
2. A + B = 10N (since A and B has to be integer).

Step 3 through 13 solve for N introduced in Step 2.
3. A = 10N - B (rewrite Step 2)
4. 3.9 (10N - B) + 5.9B = 212 (plug in Step 3 to Step 1).
5. 39N + 2B = 212.
6. 39N <= 212 (from Step 5 and observation that B >= 0).
7. N <= 5.4 (solve Step 6).
8. N is even number (because of Step 5: 212 is even, 2B is even, therefore 39N is even and N is even).
9. B <= 10N (from Step 2 and observation that A >= 0).
10. 39N + 2(10N) >= 212 (plug in Step 9 to Step 5).
11. 59N >= 212 (solve Step 10).
12. N >= 3.5 (solve Step 11).
13. N = 4 (because of Step 7 and 12: 3.5 <= N <= 5.4, and Step 8: N is even).

Now, we have:
1. 3.9A + 5.9B = 212
2. A + B = 40
We can solve for A and B (2 equations, 2 unknowns: just plug in Step 2 to Step 1): A = 12 and B = 28.
 
Originally posted by: br0wn
This is another simple way to solve for A and B without using Euclid algorithm and gcd.
1. 3.9A + 5.9B = 212.
2. A + B = 10N (since A and B has to be integer).

Step 3 through 13 solve for N introduced in Step 2.
3. A = 10N - B (rewrite Step 2)
4. 3.9 (10N - B) + 5.9B = 212 (plug in Step 3 to Step 1).
5. 39N + 2B = 212.
6. 39N <= 212 (from Step 5 and observation that B >= 0).
7. N <= 5.4 (solve Step 6).
8. N is even number (because of Step 5: 212 is even, 2B is even, therefore 39N is even and N is even).
9. B <= 10N (from Step 2 and observation that A >= 0).
10. 39N + 2(10N) >= 212 (plug in Step 9 to Step 5).
11. 59N >= 212 (solve Step 10).
12. N >= 3.5 (solve Step 11).
13. N = 4 (because of Step 7 and 12: 3.5 <= N <= 5.4, and Step 8: N is even).

Now, we have:
1. 3.9A + 5.9B = 212
2. A + B = 40
We can solve for A and B (2 equations, 2 unknowns: just plug in Step 2 to Step 1): A = 12 and B = 28.
Click to expand...

niiiiiiiiiiiiiice 🙂
 
Originally posted by: br0wn
This is another simple way to solve for A and B without using Euclid algorithm and gcd.
1. 3.9A + 5.9B = 212.
2. A + B = 10N (since A and B has to be integer).

....
Click to expand...

very elegant, will work for as long as A and B have to be ints.
 
HI all,

This is a fun problem. It reminds me my elementary school Math. It was in VietNam so the system is a little bit different. They taught us to use a lot of logic without knowing much about agebra or anything.

This is how I solved it.

What we know :

3.9A + 5.9B = 212

A and B are integers.

so 3.9A + 3.9B + 2B = 212

3.9(A+B) + 2B = 212
1.95 (A+B) + B = 106

since B = interger so 1.95 (A+B) = integer

a number to multiply to 1.95 and make it an integer must be > 20

that number has to be 40.60,80 , 100 etc since 4 or 6 or 8 x5 = ..0 to make it an integer

that number can't be bigger 60 since 60 x 1.95 = 117 > 106

so 40 is the only possibility of A+B

that gives B = 28 => 106 - 78

so A = 12

Well I am not sure this is without error and can be applied to other similar situations but I think we can always break down the problem to a close solution.





 
well...I should have included 20 into my list .... but a little calculation will give B > A+B which is false.

BTW I read the solution of taking 0.1 out of each 3 x 3.9 pipes for each 2 x 5.9 pipes......very good and simple solution indeed 🙂
 
Originally posted by: bharok
why the hell is there such a long thread about an algebra problem
?
Click to expand...

Apparently you didn't read the whole thread. Thanks for your worthwhile contribution. This is not just a "normal" algebra problem.
 
Originally posted by: BigJ
Originally posted by: bharok
why the hell is there such a long thread about an algebra problem
?
Click to expand...

Apparently you didn't read the whole thread. Thanks for your worthwhile contribution. This is not just a "normal" algebra problem.
Click to expand...

i did read the thread but what i noticed is that the tread just keeps on going on even after several people have given good answers
 
Originally posted by: MercenaryForHire
My god. What are you in, elementary school?

3.9A + 5.9B = 212

Solve for A and B.

- M4H
Click to expand...

I thought the same thing. I mean...I've been out of school for a while...but this did seem like something I was doing in middle school.

Edit: After reading the thread a little longer, I did realize the problem was a little harder to solve than what I was doing in middle school.
 
Originally posted by: EyeMWing
3.9A+5.9B=212
(212-5.9B)/3.9=A

3.9((212-5.9B)/3.9)+5.9B=212
All the variables cancel. Thus, B can be any number. Any whole number in your case.

Now it's just a matter of finding out which value of B will give you a whole number value of A. How the fsck would you do that? I'm honestly not sure, but I'd just brute force it.
Click to expand...

pls don't embarrass yourself in public like this...
 
Originally posted by: PricklyPete
Originally posted by: MercenaryForHire
My god. What are you in, elementary school?

3.9A + 5.9B = 212

Solve for A and B.

- M4H
Click to expand...

I thought the same thing. I mean...I've been out of school for a while...but this did seem like something I was doing in middle school.

Edit: After reading the thread a little longer, I did realize the problem was a little harder to solve than what I was doing in middle school.
Click to expand...

You did Diophantine Equations in Middle/High School? I'm in Calc II with 2 weeks of class left and haven't seen Diophantine Equations yet.
 
Originally posted by: DrPizza
Originally posted by: br0wn
This is another simple way to solve for A and B without using Euclid algorithm and gcd.
1. 3.9A + 5.9B = 212.
2. A + B = 10N (since A and B has to be integer).

Step 3 through 13 solve for N introduced in Step 2.
3. A = 10N - B (rewrite Step 2)
4. 3.9 (10N - B) + 5.9B = 212 (plug in Step 3 to Step 1).
5. 39N + 2B = 212.
6. 39N <= 212 (from Step 5 and observation that B >= 0).
7. N <= 5.4 (solve Step 6).
8. N is even number (because of Step 5: 212 is even, 2B is even, therefore 39N is even and N is even).
9. B <= 10N (from Step 2 and observation that A >= 0).
10. 39N + 2(10N) >= 212 (plug in Step 9 to Step 5).
11. 59N >= 212 (solve Step 10).
12. N >= 3.5 (solve Step 11).
13. N = 4 (because of Step 7 and 12: 3.5 <= N <= 5.4, and Step 8: N is even).

Now, we have:
1. 3.9A + 5.9B = 212
2. A + B = 40
We can solve for A and B (2 equations, 2 unknowns: just plug in Step 2 to Step 1): A = 12 and B = 28.
Click to expand...

niiiiiiiiiiiiiice 🙂
Click to expand...

i agree. this is a great solution (best so far b/c of simplicity/cleverness).
 
People, heard of Linear Algebra? You dont' do all of algebra in middle school. This is discrete math, stuff you do after calculus generally. It isn't based on calculus but it seems that schools have you finish calc before you do any other math.
Who cares what it looks like, discrete math != easy algebra.
 
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