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02 Dec 2016, 14:23
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
77% (00:52) correct
23%
(01:31)
wrong
based on 385
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02 Dec 2016, 14:54
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SW4
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.
Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = (5)(4)(2) = 40
Now onto the question.....
100^2 = [(2)(2)(5)(5)]^2
= [(2)(2)(5)(5)][(2)(2)(5)(5)]
= (2^4)(5^4)
So, the number of positive divisors = (4+1)(4+1) =(5)(5) = 25
Answer: B
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General Discussion
02 Dec 2016, 23:21
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\(100^2 = (2^2*5^2)^2 = 2^4*5^4\)
\(N=p^a*q^b*r^c...\)
Total number of positive factors of \(N = (a+1)(b+1)(c+1)...\)
Hence: \((4+1)(4+1) = 25\)
Answer B
\(N=p^a*q^b*r^c...\)
Total number of positive factors of \(N = (a+1)(b+1)(c+1)...\)
Hence: \((4+1)(4+1) = 25\)
Answer B
