Consider the number 21600. Find the sum of its divisors.
Answer
361.5k+ 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 \\
$
Last updated date: 30th Sep 2023
•
Total views: 361.5k
•
Views today: 9.61k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

The poet says Beauty is heard in Can you hear beauty class 6 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
