
Consider the number 21600. Find the sum of its divisors.
Answer
513.3k+ views
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 \\
$
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 \\
$
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
