Question

What is the sum of the factors of ${2^4} \times {3^3}?$

Answer 1

Hint:In this question we have to find the sum of the factors, so here we have numbers in the exponential form. The number of the form ${a^m}$ is of the exponential form, here $a$ is the base and $m$ is the exponent or power. If we have expressions of the form ${X^a} \times {Y^b} \times {Z^c}$,where $X,Y,Z$ are prime numbers and $a,b,c$ are their respective powers. We will apply the formula of sum of factors to solve this question: $N = \dfrac{{{X^{a + 1 - 1}} - 1}}{{X - 1}} \times \dfrac{{{Y^{b + 1 - 1}} - 1}}{{Y - 1}} \times \dfrac{{{Z^{c + 1 - 1}} - 1}}{{Z - 1}}$ .

Complete step by step answer:
Let us first understand the definition of a factor. We know that the factors of a number are defined as the numbers that divide the original number evenly or exactly leaving no remainder. For example, factors of $10$ are $1,2,5,10$.As per the question we have ${2^4} \times {3^3}$ .By comparing with the formula, we have:
$X = 2,Y = 3,a = 4,b = 3$

So we will now put the values in the formula and we can write:
$\dfrac{{{2^{4 + 1 - 1}} - 1}}{{2 - 1}} \times \dfrac{{{3^{3 + 1 - 1}}}}{{3 - 1}}$
We can simplify the value:
$\dfrac{{{2^5} - 1}}{{2 - 1}} \times \dfrac{{{3^4} - 1}}{{3 - 1}}$
We can write the value of ${2^5}$ as
 $2 \times 2 \times 2 \times 2 \times 2 = 32$
Similarly, the value of ${3^4}$ can be written as
$3 \times 3 \times 3 \times 3 = 81$
By substituting the values back in the expression, we have:
$\dfrac{{32 - 1}}{1} \times \dfrac{{81 - 1}}{2}$
On simplifying we have:
$31 \times \dfrac{{80}}{2}$
It gives us:$31 \times 40 = 1240$

Hence the required answer is $1240$.

Note: We should note that there are basically three types of formulas considered for factors. So we can also calculate the total number of factors with the formula $N = (a + 1)(b + 1)(c + 1)$.Similarly we can calculate the product of factors also with formula i.e. $N = {N^{\dfrac{{Total\,\,No.\,of\,factors}}{2}\,}}$.