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The number of ways in which 45000 can be expressed as product of 3 Co- primes ?

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    Brute force will do the job. Try to look at the prime factorization and combine the primes into three groups, then see when co-primes appear... – 2017-02-19

3 Answers 3

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That slightly depends on whether you regard $1$ as coprime with itself, which would allow $(1,1,45000)$. Otherwise the distinct prime factors of $45000=2^3\cdot 3^2\cdot 5^4$ are $2,3,5$ and none of these can occur in more than one factor if those factors are to be coprime. So we can have each prime represented in separate factors, $(8,9,625)$, or we can include two of the prime-power factors in one factor and take $1$ as a factor: $(1,72,625)$, and two more.

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    The ambiguity you're describing has more to do with whether using $1$ twice counts as supplying $3$ coprimes. I don't think you'll find much difference of (informed) opinion on whether $1$ is coprime with itself :). – 2017-02-19
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Hint:

Not many, as $45,000$ has only $3$ prime divisors.

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$45000=2^3\cdot 3^2\cdot 5^4$

Case 1

$1, \text{(factor consisting one of } 2^3,3^2,5^4), \text{(last factor)}$

Count: $\dbinom{3}{1}$

Case 2

$1, \text{(factor consisting two of } 2^3,3^2,5^4), \text{(last factor)}$

Count: $\dbinom{3}{2}$

Case 3

$1, \text{(factor consisting all the three of } 2^3,3^2,5^4), \text{(last factor)}$

Count: $\dbinom{3}{3}$

Case 4

$\text{(factor consisting one of } 2^3,3^2,5^4), \text{(factor consisting one of } 2^3,3^2,5^4), \text{(last factor)}$

Count: $\dbinom{3}{2}$

Required count: $10$