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25 Apr 2024, 16:18
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D
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Difficulty:
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Question Stats:
78% (01:10) correct
22%
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A number n can be written as the product of four prime numbers, exactly two of which are the same. How many different positive divisors does n have, including 1 and n?
A. 5
B. 8
C. 9
D. 12
E. 16
A. 5
B. 8
C. 9
D. 12
E. 16
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25 Apr 2024, 23:57
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delphsan
This implies that \(n = a^2bc\), where a, b, and c are prime numbers. The number of different positive divisors of n is thus (2 + 1)(1 + 1)(1 + 1) = 12.
Answer: D.
Finding the Number of Factors of an Integer
First, make the prime factorization of an integer \(n = a^p * b^q * c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\), and \(p\), \(q\), and \(r\) are their respective powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and \(n\) itself.
Example: Finding the number of all factors of 450: \(450 = 2^1 * 3^2 * 5^2\)
The total number of factors of 450, including 1 and 450 itself, is \((1+1)(2+1)(2+1) = 2*3*3 = 18\) factors.
28 Apr 2024, 05:45
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delphsan
\(n = a^2*b*c\) where a, b and c are distinct prime number
CONCEPT: If \(N = a^p*b^q*c^r...\)
where a, b, c... are distinct primes
Number of factors of \(N = (p+1)*(q+1)*(r+1)*...\)
where a, b, c... are distinct primes
Number of factors of \(N = (p+1)*(q+1)*(r+1)*...\)
i.e. Factors of \(n = a^2*b*c\) will be (2+1)*(1+1)*(1+1) = 12
Answer: Option D
