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Consider the number 21600. Find the sum of its divisors.

Answer
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Hint: Factorize the given number in its prime factor form. If a number can be written as ${p_1}^a \times {p_2}^b \times {p_3}^c....$, where ${p_1},{p_2}$ and ${p_3}$ are prime numbers, then the sum of its divisors will be $\dfrac{{{p_1}^{a + 1} - 1}}{{{p_1} - 1}} \times \dfrac{{{p_2}^{b + 1} - 1}}{{{p_2} - 1}} \times \dfrac{{{p_3}^{c + 1} - 1}}{{{p_3} - 1}} \times ....$ Use this formula to find out the sum of the divisors.

Complete step-by-step answer:
According to the question, the given number is 21600. We have to determine the sum of its divisors.
This number can be written as:
$
   \Rightarrow 21600 = 216 \times 100 \\
   \Rightarrow 21600 = {6^3} \times 100 \\
   \Rightarrow 21600 = {\left( {2 \times 3} \right)^3} \times 4 \times 25 \\
   \Rightarrow 21600 = {2^3} \times {3^3} \times {2^2} \times {5^2} \\
   \Rightarrow 21600 = {2^5} \times {3^3} \times {5^2} \\
$

Thus, the number is factorized in its prime factor form.
We know that if a number can be written as ${p_1}^a \times {p_2}^b \times {p_3}^c....$, where ${p_1},{p_2}$ and ${p_3}$ are prime numbers, then the sum of its divisors will be $\dfrac{{{p_1}^{a + 1} - 1}}{{{p_1} - 1}} \times \dfrac{{{p_2}^{b + 1} - 1}}{{{p_2} - 1}} \times \dfrac{{{p_3}^{c + 1} - 1}}{{{p_3} - 1}} \times ....$

Using above formula for $21600 = {2^5} \times {3^3} \times {5^2}$, we’ll get:
$ \Rightarrow $ Sum of divisors $ = \dfrac{{{2^{5 + 1}} - 1}}{{2 - 1}} \times \dfrac{{{3^{3 + 1}} - 1}}{{3 - 1}} \times \dfrac{{{5^{2 + 1}} - 1}}{{5 - 1}}$
$ \Rightarrow $ Sum of divisors $ = \dfrac{{{2^6} - 1}}{{2 - 1}} \times \dfrac{{{3^4} - 1}}{{3 - 1}} \times \dfrac{{{5^3} - 1}}{{5 - 1}} = \dfrac{{64 - 1}}{1} \times \dfrac{{81 - 1}}{2} \times \dfrac{{125 - 1}}{4}$
$ \Rightarrow $ Sum of divisors $ = 63 \times \dfrac{{80}}{2} \times \dfrac{{124}}{4} = 63 \times 40 \times 31$
$ \Rightarrow $ Sum of divisors $ = 78120$

Therefore, the sum of the divisors of 21600 is 78120.

Note: We can also find out the number of divisors of 21600.
We know that if a number can be written as ${p_1}^a \times {p_2}^b \times {p_3}^c....$, where ${p_1},{p_2}$ and ${p_3}$ are prime numbers, then the number of factors of this number is $\left( {a + 1} \right) \times \left( {b + 1} \right) \times \left( {c + 1} \right) \times ...$
Thus, the number of factors of $21600 = {2^5} \times {3^3} \times {5^2}$ will be:
$
   \Rightarrow {\text{ No}}{\text{. of factors }} = \left( {5 + 1} \right)\left( {3 + 1} \right)\left( {2 + 1} \right) = 6 \times 4 \times 3 \\
   \Rightarrow {\text{ No}}{\text{. of factors }} = 72 \\
$