consider the following linear equataion: 4x+11y=2011
How many solutions are there such that x and y are positive integer? (note the date 4/11/2011)
There are 182 integer values of y that make 11 y between 0 and 2002
The remainder values of 4x are:
2011, 2000, 1989, 1978, 1967, 1956, 1945, 1934, 1923, 1912... . Those 4x values resulting in an integer for x are 2000 (for y = 1), 1956 (for y = 5), 1912 (for y = 9), ... and 20 for y = 181. There are 46 such values, and therefore 46 combinations.
(x,y)
______
(500,1)
(489,5)
(478,9)
......
(5,181)
nice explanation, i understand
To determine the number of positive integer solutions for the given linear equation 4x + 11y = 2011, we need to employ a method called Diophantine Equations.
Diophantine Equations deal with finding integer solutions for equations involving two or more variables. In this case, we are looking for integer solutions (x, y) where both x and y are positive.
To solve this specific equation, we can use the fact that it is equivalent to 4x = 2011 - 11y. Since 2011 is not divisible by 4, we know that the right side of the equation cannot become divisible by 4.
Now, let's analyze the equation modulo 4:
4x ≡ 2011 - 11y (mod 4)
Since 2011 ≡ (-9) (mod 4) and -11 ≡ 1 (mod 4), we can substitute these values into the equation:
4x ≡ (-9) - y (mod 4)
Simplifying further:
4x ≡ (-9 - y) (mod 4)
Now, let's analyze the equation modulo 2:
4x ≡ (-9 - y) (mod 2)
Since -9 ≡ (-1) (mod 2), we can substitute this value into the equation:
4x ≡ (-1 - y) (mod 2)
Simplifying further:
4x ≡ (-1 - y) (mod 2)
In the equation above, we see that the right side can only take two possible values: -1 and -2. However, the left side (4x) is always divisible by 4 and hence cannot be congruent to -1 or -2 modulo 2.
Therefore, there are no positive integer solutions (x, y) for the given linear equation 4x + 11y = 2011.