Question
Answers

Find the total of all proper fractions 75600.
 $ \begin{align}
  & \text{a) 120} \\
 & \text{b) 119} \\
 & \text{c) 118} \\
 & \text{d) None of these} \\
\end{align} $

Answer Verified Verified
Hint: In order to proceed with this problem we make use of the following formulae For any number N the formula for number of proper factors of N where $ N={{p}_{1}}^{{{a}_{1}}}{{p}_{2}}^{{{a}_{2}}}...{{p}_{n}}^{{{a}_{n}}} $ and $ {{p}_{1}},{{p}_{2}},{{p}_{3}}...{{p}_{n}} $ are prime factors of N is given by $ ({{a}_{1}}+1)({{a}_{2}}+1)....({{a}_{n}}+1) $

Complete step-by-step answer:
A number is called a proper factor of N if and only if it leaves 0 as remainder when divided with N. Now to find the proper factors we will first need the prime factors of N. Prime factors are factors which cannot be further factorized.
Now let us try to find out all the prime factors of 76500.
$\begin{align}
  & 2\left| \!{\underline {\,
  75600 \,}} \right. \\
 & 2\left| \!{\underline {\,
  27800 \,}} \right. \\
 & 2\left| \!{\underline {\,
  13900 \,}} \right. \\
 & 2\left| \!{\underline {\,
  9450 \,}} \right. \\
 & 3\left| \!{\underline {\,
  4725 \,}} \right. \\
 & 3\left| \!{\underline {\,
  1575 \,}} \right. \\
 & 3\left| \!{\underline {\,
  525 \,}} \right. \\
 & 5\left| \!{\underline {\,
  175 \,}} \right. \\
 & 5\left| \!{\underline {\,
  35 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \_1 \\
\end{align}$

Hence, we can now write 75600 as $ 75600={{2}^{4}}{{3}^{3}}{{5}^{2}}{{7}^{1}}..............(1) $ .
Now we can directly apply the formula for the number of proper factors.
For any number N the formula for number of proper factors of N where $ N={{p}_{1}}^{{{a}_{1}}}{{p}_{2}}^{{{a}_{2}}}...{{p}_{n}}^{{{a}_{n}}} $ and $ {{p}_{1}},{{p}_{2}},{{p}_{3}}...{{p}_{n}} $ are prime factors of N is given by $ ({{a}_{1}}+1)({{a}_{2}}+1)....({{a}_{n}}+1) $
Comparing to the equation (1) to the formula we have N = 76500, n=4 $ {{p}_{1}}=2,{{p}_{2}}=3,{{p}_{3}}=5 $ and $ {{p}_{4}}=7 $ . Also $ {{a}_{1}}=4,{{a}_{2}}=3,{{a}_{3}}=2 $ and $ {{a}_{1}}=1 $
Hence for 76500 we get the number of factors is equal to (4+1) × (3+1) × (2+1) × (1+1).
= 5 × 4 × 3 × 2
= 5 × 4 × 6
= 5 × 24
= 120
Hence, the total number of proper factors of 76500 is 144.
So, the correct answer is “Option A”.

Note: While taking the number of factors don’t forget to add 1 to the powers.
The formula used can easily be understood though.
Here $ {{p}^{{{a}_{1}}}} $ means we have $ {{p}^{0}},{{p}^{1}},...{{p}^{{{a}_{1}}}} $ as factors hence the total options available are $ {{a}_{1}} $ +1
Hence in general $ {{p}^{n}} $ will give n + 1 options.