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while surfing through the net i have found the solution to my question,however i don't kno how have they reached to the answer: the question is : r,s,t are prime nos. and p,q are positive integers such that L.C.M(p,q)=r square.s to the power 4.s square,then the no. of ordered pair(p,q) is? 2) a rectangle with sides 2m-1 and 2n-1 is divided into squares of unit length by drawing parallel lines. then the no. of rectangles possible withodd side length is?

 while surfing through the net i have found the solution to my question,however i don't kno how have they reached to the answer: the question is :


r,s,t are prime nos. and p,q are positive integers such that L.C.M(p,q)=r square.s to the power 4.s square,then the no. of ordered pair(p,q) is?


                    2)          a rectangle with sides 2m-1 and 2n-1 is divided into squares of unit length by drawing parallel lines. then the no. of rectangles possible withodd side length is?

Grade:12

1 Answers

Chetan Mandayam Nayakar
312 Points
12 years ago

Solution: let p=r^as^bt^c and q=r^ds^et^f

Initially, let us ignore double counting

If LCM = r^2s^4t^2, at least one among b,e is 4 and the other can vary from 0-4

At least one among a,d is 2 and the other can vary from 0-2

At least one among c,f is 2 and the other can vary from 0-2

Thus, tentatively the number of possibilities is (4-0+1)(2-0+1)^2

=45

it is clear that if p,q are identical the fixed value of LCM implies that b=e=4,

a=d=c=f=2

so only one ordered pair is not double counted. Let n be the answer we are seeking.

2n – 1 = 45, which implies that there are 23 ordered pairs. I hope my solution is correct and comprehensible. In case it is not comprehensible please ask me whatever doubts you have.

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