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How many pairs of positive integers (a,b) satisfy 2/a - 3/b = 1/5

 Dec 21, 2018
 #1
avatar+698 
0

5 pairs.

 - PM

 Dec 21, 2018
edited by PartialMathematician  Dec 21, 2018
 #2
avatar+698 
0

They are (10, 4), (15, 5), (35, 7), (60, 8), and (135, 9).

PartialMathematician  Dec 21, 2018
 #3
avatar+107 
+1

Can you walk me through the steps? I only came up with two ordered pairs (4,10) and (5,15)

MacTyBoys  Dec 21, 2018
 #4
avatar+99331 
+2

 

\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\\)

Lets look at Partial Mathemticians answers:  

They are (10, 4), (15, 5), (35, 7), (60, 8), and (135, 9).

 

(10, 4)    2/10-3/4  not equal      1/5    doesn't work,

(15, 5),      2/15 -3/5 =-7/15                 doesn't work

(35, 7),       2/35-3/7 = -13/35              doesn't work

(60, 8),        2/60 - 3/8 = -41/120          doesn't work

and (135, 9).     2/135 -3/9 =  -43/135   doesn't work

 

None of his answers work, maybe the a and b are the wrong way around. I have not looked at that.

No one should asssume that any answer is correct.   It is up to the question asker to check.

 

 

 

\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\ 5ab(\frac{2}{a} - \frac{3}{b})= 5ab(\frac{1}{5})\\ 10b-15a=ab\\ \)

.
 Dec 21, 2018
 #7
avatar+4459 
+4

flip a and b and they all work

 

but I agree a proper answer would have at least tried to show how those pairs were derived.

Rom  Dec 21, 2018
edited by Rom  Dec 21, 2018
 #8
avatar
+1

Well Rom, I agree with you and Melody, but if W/A and Mathematica 11 Home edition for that matter, don't give ANY explanations as to how they derive these numbers, then all I can think of is that they must use iteration, or trial and error essentially, to get these numbers!!. Since I have Newton-Raphson iteration and interpolation method programmed into my computer, I can essentially do the same thing they do.

Sorry, that is the only explanation I have.

Guest Dec 21, 2018
 #11
avatar+1367 
0

The new generation …

 

I suspected from the beginning that Chimp Loki used recombinant DNA from the four (4) Mr. BBs to create a mathematician.  The experiment was a partial success –Loki created a PartialMathematician

 

For those unfamiliar with the act III scenes of Web2.0calc’s Opera Camelot (Camel-Lot,)

here’s a partial synopsis of the BB’s:

https://web2.0calc.com/questions/kim-has-10-identical-lamps-and-3-identical-tables#r11

 

Partial list of posts with more detailed summaries on the pervasive phage of the BBs:

https://web2.0calc.com/questions/a-bag-contains-different-colored-beads#r10

https://web2.0calc.com/questions/help_77545#r5

https://web2.0calc.com/questions/hallppp-3-questions#r4

 

The saga continues …indecision

GA

GingerAle  Dec 21, 2018
edited by GingerAle  Dec 21, 2018
 #13
avatar+4459 
+1

I think something like this has more of a number theory type solution.

 

Chinese remainder theorem or something.

 

I will tinker with it and see if anything pops out.

Rom  Dec 21, 2018
 #14
avatar+99331 
+1

Hi Rom,

 

Thanks, but I hope you do not think it was a request from me.

This question has NOT captured my imagination.

I was just trying to make the asker and other answerers think a little more.

 

If you are looking at it because it interests you personally then I appologise for butting in :)

Melody  Dec 22, 2018
 #5
avatar
+1

There are 5 pairs of POSITIVE integers that satisfy the equation. See here:

https://www.wolframalpha.com/input/?i=2%2Fa+-+3%2Fb+%3D1%2F5,+iterate+for+a,+b

 Dec 21, 2018
 #6
avatar+99331 
+2

Yes but the asker wants YOU to explain how to find them.

I mean WITHOUT simply referring to Wolfram Alpha.

Melody  Dec 21, 2018
 #9
avatar
0

If        \(\displaystyle \frac{2}{a}-\frac{3}{b}=\frac{1}{5},\)

then

\(\displaystyle 10b-15a=ab,\)

so

\(\displaystyle b =\frac{15a}{10-a}.\)

\(\displaystyle \text{Both }a\text{ and } b\text{ are to be positive integers,}\)

so there are just nine possible values for \(\displaystyle a ,\;\;\; 1,2,\dots,9.\)

Just check each one in turn.

 Dec 21, 2018
 #10
avatar+98172 
0

Thanks, Guest.....just goes to show that there is genius in simplicity....!!!!

 

 

cool cool cool

CPhill  Dec 21, 2018
 #12
avatar+1367 
0

This is a reiteration of Melody’s post. It’s incomplete… it’s only a partial solution.   

 

 

GA

GingerAle  Dec 21, 2018

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