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Q.no.-(1) The HCF of how many distinct pair of factors 18000is 75

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Quantitative Ability pankaj 1 Answer 143 views 0

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  1. Adarsh Mohan
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    July 29, 2017 at 12:20 pm

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    How to start
    18000=2^4*3^2*5^3

    *(5*3*4 = 60 Factors)*

    75=5^2*3

    Understanding the question
    So ultimately we want to know how many factors of 18000/75 are co prime
    I.e. 2^4*3*5 = 240

    Number of factors of 240= 5*2*2 = 20

    Now 1 is coprime with all factors

    60-20+1 = 41 unordered pairs
    82 ordered pairs

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