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09 May 2005, 10:06
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I have a generic question on how to do something faster.
Say we have a eqn :
Y = (-3/5) X + 30
where 0<= X <=50 and 0<= Y <=30.......Both X,Y are integers only.
Ques is to find out how many solutions are there for this eqn i.e. how many integer (X,Y) pairs are there. I know one can enumerate and then count the number of solutions. I wud be interested to know if someone knows some trick that can make this process of counting etc go quickly.
Say we have a eqn :
Y = (-3/5) X + 30
where 0<= X <=50 and 0<= Y <=30.......Both X,Y are integers only.
Ques is to find out how many solutions are there for this eqn i.e. how many integer (X,Y) pairs are there. I know one can enumerate and then count the number of solutions. I wud be interested to know if someone knows some trick that can make this process of counting etc go quickly.
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09 May 2005, 18:11
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since value of y can be only integer, there are 11 solutions because x can only be ultiples of 5 so the values of y are also 11 integers.
value of x: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
value of y: 30, 27,........................................., 0.
i know i am going to enumerate, but it's not time consuming. but about advance mathematics, i have no idea however i love it, i love it, i love it. if i get a chance to take advance mathmatics/econometrics, i will definitely take.
value of x: 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
value of y: 30, 27,........................................., 0.
i know i am going to enumerate, but it's not time consuming. but about advance mathematics, i have no idea however i love it, i love it, i love it. if i get a chance to take advance mathmatics/econometrics, i will definitely take.
