Let S be the set of all positive integer divisors of 100,000. How many

Calculation of Number of Factors
For a Number \(N=a^p*b^q*c^r*...\)where a, b, c etc are distinct prime factors of n and p, q r are their respective exponents
The number of factors of \(N = (p+1)*(q+1)*(r+1)*...\) and so on

i.e. \(100,000 = 10^5 = 2^5 * 5^5\)

hence, total factors of 100000 = (5+1)*(5+1) = 36

i.e. S = {1, \(2^1\), \(2^2\), \(5^1\), \(2^3\), \(2^1*5^1\), \(2^4\), \(2^2*5\)........100000} 36 elements in S

The biggest terms of the set are \(2^5*5^5\) and \(2^4*5^5\) and \(2^5*5^4\)

i.e. the biggest product of two terms in set S will be \(2^5*5^5\)*\(2^4*5^5\) and \(2^5*5^5\)*\(2^5*5^4\)

i.e. Biggest terms of the product will be \(2^9*5^{10}\) and \(2^{10}*5^9\)

i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^0\) i.e. 11 combinations
Similarly, i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^1\) i.e. 11 combinations
Similarly, i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^2\) i.e. 11 combinations
and so on...

i.e. \(2^0\), \(2^1\), \(2^2\)--- \(2^{10}\) etc. will have combination with \(5^{10}\) i.e. 11 combinations

i.e. Total combinations = 11*11 = 121

But as already mentioned the combinations that are not possible but included here are \(2^{10}*5^{10}\) and \(2^0*5^0\) and \(2^{10}\) and \(5^{10}\)

Total Acceptable combinations = 121 - 4 = 117

Answer: Option D

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