@A010795 Right, think of it as fraction. If we place 122 and 138 over 1, we can reduced them by 2, so there common divisor is 2. Another example is on from the addition of fractions, let x=3, then y=6/5. The only common divisor they have in common is 5.
@A010795 then their common divisor would be 2, W=122 - L=138. When the common divisor is 1, it's like asking what is the smallest sides can this rectangle be with the same proportions.
I solved this on my own! But I spent way too long stuck at the end because for some reason I kept thinking the question's wording meant the length and the width had to be prime numbers. Of course when you solve it, you make it look like a piece of cake.
@usmansharifmicky I asked the same question; but think about this. If you chose 10 and 4, the final answers would be divisible by 2, and the question asks for both numbers to have a common divisor of 1. So to answer the question properly you would still have to divide your numbers by 2 to get the same answer that you would get for 5 and 2.
x = 2/5 • y is enough of a constraint to give us a concrete solution since the greatest common divisor is 1. Had he picked any other number, he would get the same result by simplifying.
If you solve that x=2/5y, then can't you substitute 2/5y in for x in the equation 13x+7y=8x+9y? I got that x=2/5 and y=1, then subtituted both of those in each of the equations of the sides, and i got that the perimeter was 54. Did anyone else get that?
such easy questions appear in American mathematics aptitude exams? That's why the intelligence level of Indian students is far ahead as compared to American students.
Is that why USA always outperforms India at the annual IMO? Also, here are the rest of the problems on that year's AIME. Not many people can solve them. http:/www.artofproblemsolving.com/Wiki/index.php/2000_AIME_I_Problems