Finding Perfect Squares of n Factored as a^4*b^3*c^7

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a number n when factorised can be written as a^4*b^3*c^7.find number of perfect square which are factors of n.a,b,c are prime >2.
I have no idea how to start? please help.
 
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If a, b and c are distinct primes then numbers like a^2 are perfect squares and factors of n. There are 2 possible perfect squares arising from a, only 1 from b but 3 possibilities from c so the question is how many combinations are possible with those perfect squares?
 
Tide, how dare you miss such easy things? 3 possible perfect squares from a, 2 possible perfect squares from b and 4 possiblities from c of which in total three are same. You have missed the 0th power. Now the answer is 3*2*4. However during the combination of 2, 1 and 3 we won't get 1. Now the answer 3*2*4 is because for each 3 powers of a, there are two possible powers of b and for each of b there are 4 possibilities of c when each square is taken as the product of powers of each primes.
 
vaishakh,

No, I didn't miss the zero power. I had to leave something for the student to think about!
 

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