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Questions  

 If a    b    1b    c    1c    a    1=2020 and if cacbababacbcbcbacacacbc2abaca2bcbab2=P , then the number of 

positive divisors of  P is 

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a
36
b
45
c
55
d
39

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detailed solution

Correct option is B

ab1bc1ca1=ab+bc+ca-a2-b2-c2=2020 ⇒given determinent  P =c-ac-bab-c2a-ba-cbc-a2b-cb-aca-b2 now R1→R1 +R2 +R3    ⇒  002020a-ba-cbc-a2b-cb-aca-b2=2020ab+bc+ca-a2-b2-c2     =20202=  24×52×1012   number positive divisors = 4+12+12+1=45


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