Archit3110
VeritasKarishma ; hello ma'am..
I am not able to understand from the highlighted part onwards..
a. how did 121 numbers come up?
b. 2^10 , 5^10 are the highest power but how did you deduce to 117 numbers
GMATinsight ; sir if possible kindly look into the question and assist.
VeritasKarishma
Noshad
Let S be the set of all positive integer divisors of 100,000. How many numbers are the product of two distinct elements of S?
A. 98
B.100
C. 117
D. 119
E. 121
100,000 = 10^5 = 2^5 * 5^5
No of distinct factors = 6*6 = 36
These are: 1, 2^1, 2^2, 5, 2^3, 2^1*5^1, 2^4, 5^2, ...
When you multiply any two of these, you will get a product with powers of 2 and/or 5.
Since 2^5 and 5^5 are the maximum powers of 2 and 5, the power of either that you can get ranges from 0 to 10.
This gives us 11*11 = 121 numbersBut no number can be multiplied by itself so the largest and smallest powers of 2 and 5 will not multiply with themselves.
Hence, you will not get the following 4 numbers:
2^10 * 5^10 (because there is only one 2^5 * 5^5)2^10 (because there is only one 2^5)
5^10 (because there is only one 5^5)
2^0*5^0 (because there is only one 2^0*5^0)
So in all, you will have 117 numbers.
Answer (C)
Calculation of Number of FactorsFor 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 oni.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
Archit3110 warrior1991