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A number is a product of 5 prime factors, 2 of them are same.
For explanation, I considered these to be a, b, c, d. n = a^2bcd a is reapeted twice.

To find the number of factors, say for x^a.y^b.z^c, simply multiply the powers+1, i.e. no of factors = (a+1)(b+1)(c+1)

In this case, no of factors of a^2bcd = 3.2.2.2 = 24

I learnt this formula from the forum - an old discusion between Stolyar and AkamaiBrah. That said, beware of its use, it can be used only in the simple case.

A number is a product of 5 prime factors, 2 of them are same. For explanation, I considered these to be a, b, c, d. n = a^2bcd a is reapeted twice.

To find the number of factors, say for x^a.y^b.z^c, simply multiply the powers+1, i.e. no of factors = (a+1)(b+1)(c+1)

In this case, no of factors of a^2bcd = 3.2.2.2 = 24

I learnt this formula from the forum - an old discusion between Stolyar and AkamaiBrah. That said, beware of its use, it can be used only in the simple case.

Yes, that discussion was gold. I learned it from this forum too. _________________

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...