# Find out the number of factors of $abc$ if $\dfrac{{a + 2}}{2} = \dfrac{{b + 4}}{4} = \dfrac{{c + 6}}{6} = 12$.

Last updated date: 20th Mar 2023

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Answer

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Hint: Find out the value of $a, b$ and $c$ separately. Then calculate $abc$ and factorize it 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 number of factors of this number is $\left( {a + 1} \right) \times \left( {b + 1} \right) \times \left( {c + 1} \right) \times ...$ Use this method to find the number of factors.

Complete step-by-step answer:

According to the question, we have:

$ \Rightarrow \dfrac{{a + 2}}{2} = \dfrac{{b + 4}}{4} = \dfrac{{c + 6}}{6} = 12 .....(i)$

We will calculate the values of $a,b$ and $c$ separately.

So, from equation $(i)$, we have:

$

\Rightarrow \dfrac{{a + 2}}{2} = 12 \\

\Rightarrow a + 2 = 24 \\

\Rightarrow a = 22 .....(ii) \\

$

Again using equation $(i)$, we have:

$

\Rightarrow \dfrac{{b + 4}}{4} = 12 \\

\Rightarrow b + 4 = 48 \\

\Rightarrow b = 44 .....(iii) \\

$

Using equation $(i)$ for $c$, we have:

$

\Rightarrow \dfrac{{c + 6}}{6} = 12 \\

\Rightarrow c + 6 = 72 \\

\Rightarrow c = 66 .....(iv) \\

$

From equations $(ii)$, $(iii)$ and $(iv)$, we have:

$

\Rightarrow abc = 22 \times 44 \times 66 \\

\Rightarrow abc = 2 \times 11 \times 4 \times 11 \times 6 \times 11 \\

\Rightarrow abc = 2 \times 4 \times 6 \times {11^3} \\

\Rightarrow abc = 2 \times {2^2} \times 2 \times 3 \times {11^3} \\

$

$ \Rightarrow abc = {2^4} \times {3^1} \times {11^3}$

And 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 $\left( {a + 1} \right) \times \left( {b + 1} \right) \times \left( {c + 1} \right) \times ...$

Therefore the number of factors of $abc$ is $\left( {4 + 1} \right) \times \left( {1 + 1} \right) \times \left( {3 + 1} \right) = 5 \times 2 \times 4 = 40$

Thus, there are 40 factors of $abc$.

Note: We can also calculate the sum of factors as:

If the number is ${p_1}^a \times {p_2}^b \times {p_3}^c....$ (${p_1},{p_2}$,${p_3}$ are all prime numbers), then the sum of its factors is:

$ \Rightarrow {\text{Sum }} = \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 this for $abc = {2^4} \times {3^1} \times {11^3}$, sum of factors will be:

$ \Rightarrow {\text{Sum }} = \dfrac{{{2^5} - 1}}{{2 - 1}} \times \dfrac{{{3^2} - 1}}{{3 - 1}} \times \dfrac{{{{11}^4} - 1}}{{11 - 1}} \times ....$

Complete step-by-step answer:

According to the question, we have:

$ \Rightarrow \dfrac{{a + 2}}{2} = \dfrac{{b + 4}}{4} = \dfrac{{c + 6}}{6} = 12 .....(i)$

We will calculate the values of $a,b$ and $c$ separately.

So, from equation $(i)$, we have:

$

\Rightarrow \dfrac{{a + 2}}{2} = 12 \\

\Rightarrow a + 2 = 24 \\

\Rightarrow a = 22 .....(ii) \\

$

Again using equation $(i)$, we have:

$

\Rightarrow \dfrac{{b + 4}}{4} = 12 \\

\Rightarrow b + 4 = 48 \\

\Rightarrow b = 44 .....(iii) \\

$

Using equation $(i)$ for $c$, we have:

$

\Rightarrow \dfrac{{c + 6}}{6} = 12 \\

\Rightarrow c + 6 = 72 \\

\Rightarrow c = 66 .....(iv) \\

$

From equations $(ii)$, $(iii)$ and $(iv)$, we have:

$

\Rightarrow abc = 22 \times 44 \times 66 \\

\Rightarrow abc = 2 \times 11 \times 4 \times 11 \times 6 \times 11 \\

\Rightarrow abc = 2 \times 4 \times 6 \times {11^3} \\

\Rightarrow abc = 2 \times {2^2} \times 2 \times 3 \times {11^3} \\

$

$ \Rightarrow abc = {2^4} \times {3^1} \times {11^3}$

And 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 $\left( {a + 1} \right) \times \left( {b + 1} \right) \times \left( {c + 1} \right) \times ...$

Therefore the number of factors of $abc$ is $\left( {4 + 1} \right) \times \left( {1 + 1} \right) \times \left( {3 + 1} \right) = 5 \times 2 \times 4 = 40$

Thus, there are 40 factors of $abc$.

Note: We can also calculate the sum of factors as:

If the number is ${p_1}^a \times {p_2}^b \times {p_3}^c....$ (${p_1},{p_2}$,${p_3}$ are all prime numbers), then the sum of its factors is:

$ \Rightarrow {\text{Sum }} = \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 this for $abc = {2^4} \times {3^1} \times {11^3}$, sum of factors will be:

$ \Rightarrow {\text{Sum }} = \dfrac{{{2^5} - 1}}{{2 - 1}} \times \dfrac{{{3^2} - 1}}{{3 - 1}} \times \dfrac{{{{11}^4} - 1}}{{11 - 1}} \times ....$

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