Question

How many two digit natural numbers n
are there such that
-
n²-9
n2 - 7
is a proper
fraction in it's lowest form?​

Answer 1

Answer:

Sorry i don't know the answer to this one

Step-by-step explanation:

Sorry i don't know the answer to this one

Answer 2

Given: $\[\left( {\frac{{{n^2} - 9}}{{{n^2} - 7}}} \right)\]$ is a proper fraction in its lowest form​

To Find:  how many two-digit natural numbers are there  

Solution:

Understand that,

$\[(n^2 - 9) < (n^2 - 7)\]$  , which is always true  and $\[n^{2} - 9\]$  and $n^2-7$   are co-prime$\[10 \le n \le 99\]$ as there is  a difference of 2 in  $\[n^{2} - 9\]$  and $n^2-7$ .

Observe that , both these numbers must be odd as in even numbers both will have factor  2.

Therefore, $n^2$must be even  which implies n must be even  so possible values of n will be $10,12,14,16,-------98,100$

$\[\begin{array}{*{20}{l}}{ \Rightarrow 100 = 10 + 2\left( {n - 1} \right)}\\{ \Rightarrow n = 46}\end{array}\]$  

Hence 46 two-digits such numbers are possible  where $\[\left( {\frac{{{n^2} - 9}}{{{n^2} - 7}}} \right)\]$ is  a proper fraction in its lowest form.