+0

# ordered pairs

0
234
14

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

Dec 21, 2018

#1
+701
0

5 pairs.

- PM

Dec 21, 2018
edited by PartialMathematician  Dec 21, 2018
#2
+701
0

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

PartialMathematician  Dec 21, 2018
#3
+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
+102444
+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
+5225
+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
+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
+1678
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/help_77545#r5

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

The saga continues …

GA

GingerAle  Dec 21, 2018
edited by GingerAle  Dec 21, 2018
#13
+5225
+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
+102444
+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
+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
+102444
+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
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
+101798
0

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

CPhill  Dec 21, 2018
#12
+1678
0

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

GA

GingerAle  Dec 21, 2018