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Re: For a certain positive integer N, N^3 has exactly 13 unique factors. H
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18 Feb 2017, 08:19
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Bunuel wrote:
For a certain positive integer N, N³ has exactly 13 unique factors. How many unique factors does N have?
A. 1 B. 2 C. 3 D. 4 E. 5
IMPORTANT: 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------------------------------------------------
In the above rule, notice that total number of divisors is a PRODUCT [e.g., (4+1)(3+1)(1+1) =(5)(4)(2) = 40]
In this question, we're told that there are 13 divisors. There's only ONE WAY that the number 13 can be written as a product: 1 x 13
So, a number with 13 positive divisors must have a prime factorization that looks like this: prime^12
So, if N³ is equivalent to prime^12, then N must equal prime^4, since (prime^4)^3 = prime^12
For example, if N = 2^4, then N = 16 Notice that, if N = 2^4, then N³ = (2^4)^3 = 2^12 And, from the above rule, if N³ = 2^12, then the number of positive divisors of N³ = (12+1) = 13
How many unique factors does N have? If N = 16 (or any other prime^4), then the factors of 16 = {1, 2, 4, 8, 16} There are 5 factors.
For a certain positive integer N, N^3 has exactly 13 unique factors. H
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20 Mar 2017, 01:30
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Bunuel wrote:
For a certain positive integer N, \(N^3\) has exactly 13 unique factors. How many unique factors does N have?
A. 1 B. 2 C. 3 D. 4 E. 5
OFFICIAL SOLUTION
The shortest way to solve this problem is by applying the Unique Factors Trick. That trick is a tool that enables the direct calculation of the total number of factors of any number based on that number’s prime factor list. Specifically, to count the total factors of any number, prime factor the number, discard the bases, add one to the exponents, and then multiply the values obtained. The result is the total factor count.
In the case at hand, we’re told that \(N^3\) has exactly 13 factors. When we consider the Unique Factors Trick, we can see that 13 must be the result of multiplying a set of positive integers. However, 13 is prime; its only factors are 1 and 13. So our original exponents, before adding one to each, can only have been zeroes and a single 12. In other words, \(N^3\) must be the 12th power of a prime number:
\(N^3=p^{12}\)
From there, we can take the cube root and find that N is the 4th power of the same prime:
\(N=p^4\)
Using the Unique Factors Trick once more, we discard the base and add one to the exponent to determine that N has 4+1=5 total factors. The correct answer is E.
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Re: For a certain positive integer N, N^3 has exactly 13 unique factors. H
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20 Mar 2017, 04:24
Bunuel wrote:
For a certain positive integer N, N^3 has exactly 13 unique factors. How many unique factors does N have?
A. 1 B. 2 C. 3 D. 4 E. 5
since 13 is a prime number therefore 12 will be the power of a prime number and that will be N^3 therefore n will be a prime number raised to the power 4 total factor will be 4 +1 =5
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Re: For a certain positive integer N, N^3 has exactly 13 unique factors. H
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04 Apr 2019, 11:55
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