Posted by Diana on Tuesday, April 26, 2011 at 8:05am.
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)

Math  drwls, Tuesday, April 26, 2011 at 9:26am
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) 
Math  Diana, Wednesday, April 27, 2011 at 8:03am
nice explanation, i understand