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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