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..???

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.

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TopNewestLet \(n=p_{1}^{a_{1}}p_{2}^{a_{2}}.....p_{k}^{a_{k}}\)

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.

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

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