answer of the crackthe cat blog “Algebra 5” should be 24C2*2.
Arun, the product of the two numbers needs to be 600. So for each value for x there will be only 1 value for y and not 23.
thank you sir
one more doubt : Algebra 6 , 1st question if (60-x)=2400 and (y+40)=1 then x will be negative thus contradicting the situation.
Right you are Arun, I have edited it. Thanks for pointing out and sorry for the delayed reply!
In second ques also (0,0) is a solution which must be subtracted.. isn’t it?
Yes, that’s why there are 71 and not 72 solutions.
In the first question too (0,0) can be a solution when 600 is written as 20*30. But since positive integer solutions are being asked for, I think this case should be neglected and the number of solution needs to be 23.
Kindly let me know your thoughts.
No? If x – 20 = 20, x is 40 and not 0. Only when x – 20 is -20 will the issue arise….
I’m having a pretty big doubt regarding this method of finding number of Integer solutions. 1st question the factors are 1,2,3,…,200,300,600. And they can be (x,y) = (21,630), (22,330),…,(40,60), (45,54).
But if I see from other angle then (x,y) can also be (620,31), (320,32), (220,33),…,(60,45), (50,50), (54,55).
That makes a total of 48 positive integer solutions. Am I right, Sir?
And if that’s right, then no. of integer solutions will be 96 solutions including (0,0). That means 95 integer solutions are possible right, Sir?
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