Question

The number of different proper factors of \[3780\] is –
(A) \[45\;\]
(B) \[46\]
(C) \[47\]
(D) \[48\]

Answer 1

Hint: To solve this question, we will take prime factorization of the given number, with that we get the prime factors of it. Then we will take \[1\] (as every number is divisible by it) along with each prime factor, their exponents and make all possible combinations of it, and after counting all the combinations, we will get our required answer.

Complete step-by-step answer:
We have been given a number i.e., \[3780.\]We need to find the number of different proper factors of it.
On doing prime factorization, we get
\[3780 = {2^2} \times {3^3} \times 5 \times 7\]
Now, let’s check number of different proper factors of \[3780.\]
The first number would be, \[1\], since every number is divisible by it.
Now using prime factor and their exponents we will make all possible combinations of it.
So, prime factor \[ = {\text{ }}2\]
Prime factor \[ = {\text{ }}3\]
\[{2^2} = {\text{ }}4\]
Prime factor \[ = {\text{ }}5\]
\[2 \times 3{\text{ }} = {\text{ }}6\]
Prime factor \[ = {\text{ }}7\]
\[\begin{array}{*{20}{l}}
  {{3^2} = {\text{ }}9} \\
  {2 \times 5{\text{ }} = {\text{ }}10} \\
  {{2^2} \times 3{\text{ }} = {\text{ }}12} \\
  {2 \times 7{\text{ }} = {\text{ }}14}
\end{array}\]

\[\begin{array}{*{20}{l}}
  {3 \times 5{\text{ }} = {\text{ }}15} \\
  {2 \times {3^2} = {\text{ }}18} \\
  {{2^2} \times 5{\text{ }} = {\text{ }}20} \\
  {3 \times 7{\text{ }} = {\text{ }}21}
\end{array}\]

\[\begin{array}{*{20}{l}}
  {{3^2} = {\text{ }}27} \\
  {{2^2} \times 7{\text{ }} = {\text{ }}28} \\
  {2 \times 3 \times 5{\text{ }} = {\text{ }}30} \\
  {5 \times 7{\text{ }} = {\text{ }}35} \\
  {{2^2} \times {3^2} = {\text{ }}36}
\end{array}\]

\[\begin{gathered}
  2 \times 3 \times 7{\text{ }} = {\text{ }}42 \\
  {3^2} \times 5{\text{ }} = {\text{ }}45 \\
  2 \times {3^3} = {\text{ }}54 \\
  {2^3} \times 3 \times 5{\text{ }} = {\text{ }}60 \\
\end{gathered} \]

\[\begin{array}{*{20}{l}}
  {{3^2} \times 7{\text{ }} = {\text{ }}63} \\
  {2 \times 5 \times 7{\text{ }} = {\text{ }}70} \\
  {{2^2} \times 3 \times 7{\text{ }} = {\text{ }}84} \\
  {2 \times {3^2} \times 5{\text{ }} = {\text{ }}90} \\
  {3 \times 5 \times 7{\text{ }} = {\text{ }}105} \\
  {{2^2} \times {3^3} = {\text{ }}108} \\
  {2 \times {3^2} \times 7{\text{ }} = {\text{ }}126} \\
  {{3^3} \times 7{\text{ }} = {\text{ }}189}
\end{array}\]

\[\begin{array}{*{20}{l}}
  {2 \times 3 \times 5 \times 7{\text{ }} = {\text{ }}210} \\
  {{2^2} \times {3^2} \times {7^{}} = {\text{ }}252} \\
  {2 \times {3^3} \times {5^{}} = {\text{ }}270} \\
  {{3^2} \times 5 \times 7{\text{ }} = {\text{ }}315} \\
  {2 \times {3^3} \times {7^{}} = {\text{ }}378} \\
  {{2^2} \times 3 \times 5 \times 7{\text{ }} = {\text{ }}420}
\end{array}\]

\[\begin{array}{*{20}{l}}
  {{2^2} \times {3^3} \times 5{\text{ }} = {\text{ }}540} \\
  {2 \times {3^2} \times 5 \times 7{\text{ }} = {\text{ }}630} \\
  {{2^2} \times {3^3} \times 7{\text{ }} = {\text{ }}756} \\
  {{3^2} \times 5 \times 7{\text{ }} = {\text{ }}945} \\
  {{2^2} \times {3^2} \times 5 \times 7{\text{ }} = {\text{ }}1260} \\
  {2 \times {3^3} \times 5 \times 7{\text{ }} = {\text{ }}1890} \\
  {{2^2} \times {3^3} \times 5 \times 7{\text{ }} = {\text{ }}3780}
\end{array}\]

So, the total number of different proper factors of \[3780\] is \[48,\] out of which it has \[4\] prime factors, i.e., \[2,{\text{ }}3,{\text{ }}5{\text{ }}and{\text{ }}7.\]
So, the correct answer is “Option D”.

Note: In this question, we were asked to get the number of different proper factors of \[3780,\] Basically we were asked to get all possible numbers which is a divisor of the given number. So, for that, two numbers i.e., \[1\]and \[3780\] were compulsory, because \[1\] is a divisor of all numbers and \[3780\] because the number gets divided by itself. And for the other numbers we took the prime factors and their exponents and made all possible combinations out of it.