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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?
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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