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Question

Answers

$ \begin{align}

& \text{a) 120} \\

& \text{b) 119} \\

& \text{c) 118} \\

& \text{d) None of these} \\

\end{align} $

Answer
Verified

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.

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.