@Harhsit: Normally, 0 is not considered positive.

Dave

On Aug 28, 10:45 am, harshit sethi <hshoneyma...@gmail.com> wrote:
> sorry 6 solutions y=15,12,9,6,3,0
> and x=0,4,8,12,16,20  respectively
>
> On 8/28/11, harshit sethi <hshoneyma...@gmail.com> wrote:
>
>
>
> > maximum value of y satisfying this is y=15 and for that x=0;
>
> > now decrease y by 3 and increase x by 4 ,you will have x and y
> > satisfying the equation.
>
> > keep on doing this till you reach minimum value of y i.e 0
>
> > this you can do  5 times decreasing y=15 by 3 every time
>
> > so there will be 5 solutions .
>
> > On 8/28/11, Piyush Grover <piyush4u.iit...@gmail.com> wrote:
> >> 3x+4y = 60
> >> it's a straight line equation whose x intercept is 20 and y intercept is
> >> 15.
> >> Draw it in first quadrant
> >> (as x, y are positive integers)
> >> now x = (60 - 4y)/3 = 4(15-y)/3
> >> now for y = 1, 2...15 you need to check whether (15-y) is divisible by 3
> >> or
> >> not. It's simple y = 3, 6, 9, 12
>
> >> -Piyush
>
> >> On Sun, Aug 28, 2011 at 6:38 PM, Dave <dave_and_da...@juno.com> wrote:
>
> >>> @Sivaviknesh: The smallest values of x and y are 1. The largest value
> >>> of y is (60 - 3) / 4 = 14. Solve for x: x = (60 - 4y) / 3. Since x is
> >>> an integer, 60 - 4y must be a multiple of 3. Since 60 - 3y is a
> >>> multiple of 3, (60 - 4y) - (60 - 3y) = y must be a multiple of 3.
> >>> Thus, y = 3, 6, 9, 12. Then you can solve for the corresponding values
> >>> of x.
>
> >>> Dave
>
> >>> On Aug 28, 7:46 am, sivaviknesh s <sivavikne...@gmail.com> wrote:
> >>> > *Find the number of solutions for 3x+4y=60, if x and y are positive
> >>> > integers.*
>
> >>> > Is there any standard method for solving these type of ques ..or only
> >>> trial
> >>> > and error ???
>
> >>> > --
> >>> > Regards,
> >>> > $iva
>
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