One linked question #1

Step 1: Prime factorize $32400$ .

$32400=2^{4} \times 3^{4} \times 5^{2}$

Step 2: Since the sum of the sum of a number $n= p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times \ldots p_{n}^{a_{n}}$ is $\dfrac{p_{1}^{a_{1}+1} - 1}{p_{1} - 1} \times \dfrac{p_{2}^{a_{2}+1} - 1}{p_{2} - 1} \times \ldots \dfrac{p_{n}^{a_{n}+1} - 1}{p_{n} - 1}$ , where $p_{1}$ , $p_{2}$ $\ldots$ $p_{n}$ are distinct prime factors of $n$ and $a_{1}$ , $a_{2}$ $\ldots$ $a_{n}$ are highest powers of $p_{1}$ , $p_{2}$ $\ldots$ $p_{n}$ in $n$ respectively.

So, using this formula we get sum of the divisors of $32400$ as

$\dfrac{2^{4+1} - 1}{2 - 1} \times \dfrac{3^{4+1} - 1}{3- 1} \times \ldots \dfrac{5^{2+1} - 1}{5 - 1} = 31 \times 121 \times 31$

So, taking it's square root we get, the answer as $\boxed{341}$ .

Easiest way is prime factorization. Here are the tricks which I followed -

$32400=324\cdot { 10 }^{ 2 }=4\cdot 81\cdot { 10 }^{ 2 }\\ ={ 2 }^{ 2 }\cdot { 3 }^{ 4 }\cdot { 2 }^{ 2 }\cdot { 5 }^{ 2 }\\ ={ 2 }^{ 4 }\cdot { 3 }^{ 4 }\cdot { 5 }^{ 2 }$

Using sum of divisors formula, the sum is -

$\left( 2^{ 4 }+2^{ 3 }+2^{ 2 }+2+1 \right) \left( 3^{ 4 }+3^{ 3 }+3^{ 2 }+3+1 \right) \left( 5^{ 2 }+5+1 \right) =31\cdot 121\cdot 31\\ ={ 11 }^{ 2 }\cdot { 31 }^{ 2 }$

Square root -

$\sqrt { { 11 }^{ 2 }\cdot { 31 }^{ 2 } } =11\cdot 31=341$

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Further Readings: Sum of Factors

nice problem