IIT JEE , AIEEE Forum - Mathematics , Algebra

karthik2007

Are you sure the sum is right? Anyway, this is how you proceed :

By prime factorization, you get 45321 = 15107 x 3 (Let me know if there are any prime factors of 15107 itself)

Now, abcde can take any of these values each.. that is the solution can be:

15107x3x1x1x1, 15107x1x1x1x3, etc etc...

So this is a permutation problem now. The number of ways to arrange the 2 numbers 15107 and 3 among the five variables, abcde.
This is 5!/3! = 20 ways.

But the solution can also be like (15107x3), 1, 1, 1, 1.

So total no of ways are 20+5! = 20+120 = 140 ways, ie, 140 positive integral solutions.

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