how many two digit natural numbers are there such that (n^2-9/n^2-7) is a proper fraction in its lowest form
Answers
Given : (n²-9/n²-7) is a proper fraction in its lowest form
To Find : how many two digit natural numbers are there
Solution:
n²-9 < n²-7
which is always true
and n²-9 and n²-7 are co prime
10 ≤ n ≤ 99
as there is a difference of 2 in n²-9 and n²-7 Hence
both these numbers must be odd as in even numbers both will have factor
2
Hence n² must be even
so n must be even
so possible values of n
are 10 , 12 , 14 , 16 , _________ , 98 , 100
100 = 10 + 2(n - 1)
=> n = 46
Hence 46 2 digits such numbers are possible
where (n²-9) / ( n²-7 ) is a proper fraction in its lowest form
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