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Hello,
q15:
2x + 3y = 763
=> y = (763 - 2x)/3
It has infinite integral solutions as there are infinite integral values of 'x' for which 'y' is an integer.
But the equation has finite number of positive integral solutions and I have solved for it.
Now, for 'y' to be an integer, (763-2x) must be a multiple of 3.
A little observation tells us that it happens when
763-2x=759,753,747,741,735,729......15,9 and 3.
=> x=2,5,8,11,14,17.........371,374,377,380
For each value of 'x', y can be calculated.
The values of 'x' forms an A.P.
so,
380 = 2 + (n-1)3
=>n=127.
Hence, there are 127 solutions to the equation.
So, n-2=125
cube root of 125 =5
In REALITY:
Intuitive Conclusion : Linear Equations in 2 (or more) variables have either no integral solutions (eg 3x+6y = 4), or an infinite number of integral solutions (eg 2x+3y = 763).
Using DIOPHANTINE equation.
q15:
2x + 3y = 763
=> y = (763 - 2x)/3
It has infinite integral solutions as there are infinite integral values of 'x' for which 'y' is an integer.
But the equation has finite number of positive integral solutions and I have solved for it.
Now, for 'y' to be an integer, (763-2x) must be a multiple of 3.
A little observation tells us that it happens when
763-2x=759,753,747,741,735,729......15,9 and 3.
=> x=2,5,8,11,14,17.........371,374,377,380
For each value of 'x', y can be calculated.
The values of 'x' forms an A.P.
so,
380 = 2 + (n-1)3
=>n=127.
Hence, there are 127 solutions to the equation.
So, n-2=125
cube root of 125 =5
In REALITY:
Intuitive Conclusion : Linear Equations in 2 (or more) variables have either no integral solutions (eg 3x+6y = 4), or an infinite number of integral solutions (eg 2x+3y = 763).
Using DIOPHANTINE equation.
raghukrishnan1oy0vmz:
thanks a lot friend
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