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# Discuss this mathematical assumption

IN General N = 2^p * 3^q * 5^r * 7^s.....so on Total no of factors are = (p+1) * (q+1) * (r+1) * (s+1)...... Does anyone know about this theorem...??? If yes then explain it briefly..???

Note by Rahul Jain
3 years, 4 months ago

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Let $$n=p_{1}^{a_{1}}p_{2}^{a_{2}}.....p_{k}^{a_{k}}$$

Now, suppose you have $$k$$ boxes labelled $$1,2,...,k$$.Suppose the $$ith$$ box contains $$a_{i}$$ identical objects of $$p_{i}$$ type.

Now,you want to form all possible combinations(of any length) with those objects.

Now, consider the $$ith$$ box.You can include $$1$$ $$p_{i}$$ or $$2$$ $$p_{i}'s$$ or...or $$a_{i}$$ $$p_{i}'s$$ or no $$p_{i}$$. So there are $$a_{i}+1$$ ways of choosing the objects of $$p_{i}$$ type, $$i=1,2....,k$$

Hence, in total there will be $$(a_{1}+1)(a_{2}+1)....(a_{k}+1)$$ combinations.

Now, see that this counting is similar to counting the number of divisors.

- 3 years, 4 months ago

understood buddy...!!! Now i gotcha it completely..!! thanks @Souryajit Roy

- 3 years, 4 months ago