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

Last updated date: 13th Jul 2024
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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.

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:
$\Rightarrow {\text{ No}}{\text{. of factors }} = \left( {5 + 1} \right)\left( {3 + 1} \right)\left( {2 + 1} \right) = 6 \times 4 \times 3 \\ \Rightarrow {\text{ No}}{\text{. of factors }} = 72 \\$