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07 Nov 2019, 02:16
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C
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Question Stats:
39% (02:02) correct
61%
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wrong
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If n is a positive integer, and n^2 has 25 factors, which of the following must be true.
I. n has 12 factors.
II. n > 50
III. \(\sqrt n\) is an integer.
A. I and II only
B. II only
C. III only
D. II and III
E. None
Are You Up For the Challenge: 700 Level Questions
I. n has 12 factors.
II. n > 50
III. \(\sqrt n\) is an integer.
A. I and II only
B. II only
C. III only
D. II and III
E. None
Are You Up For the Challenge: 700 Level Questions
Updated on: 13 Nov 2019, 09:50
Originally posted by TestPrepUnlimited on 11 Nov 2019, 05:58.
Last edited by TestPrepUnlimited on 13 Nov 2019, 09:50, edited 1 time in total.
Last edited by TestPrepUnlimited on 13 Nov 2019, 09:50, edited 1 time in total.
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In this case, we can create a number with 25 factors.
If the prime factorization of x is x =(2^a)*(3^b)*(5^c) .... the number of factors is (a+1)(b+1)(c+1) ...
For example, 48 = 16 * 3 = \(2^4 * 3\) has \((4+1)*(1+1) = 5 * 2 = 10\) factors.
So in this case, the number of factors, 25, must be broken down into 5 * 5 = 25 or 25 * 1 = 25. Using 5 * 5 = 25, we can have (a + 1)(b + 1) = 25, a = 4 and b = 4. This means \(x = 2^4*3^4 = n^2\). Thus \(n = 2^2*3^2\) which has 3*3 = 9 factors. We may check for the other case, \(n^2 = 2^{24}\) -> \(n = 2^{12}\) gives 13 factors). In both cases n has an odd number of factors so it is a square, hence III is true. Finally, we had \(n = 2^2*3^2 = 36\) so II is false. (Note all of these n's are just minimum values of n, in fact n can be (any prime number)^4 * (any other prime number)^4 etc)
Ans: C
If the prime factorization of x is x =(2^a)*(3^b)*(5^c) .... the number of factors is (a+1)(b+1)(c+1) ...
For example, 48 = 16 * 3 = \(2^4 * 3\) has \((4+1)*(1+1) = 5 * 2 = 10\) factors.
So in this case, the number of factors, 25, must be broken down into 5 * 5 = 25 or 25 * 1 = 25. Using 5 * 5 = 25, we can have (a + 1)(b + 1) = 25, a = 4 and b = 4. This means \(x = 2^4*3^4 = n^2\). Thus \(n = 2^2*3^2\) which has 3*3 = 9 factors. We may check for the other case, \(n^2 = 2^{24}\) -> \(n = 2^{12}\) gives 13 factors). In both cases n has an odd number of factors so it is a square, hence III is true. Finally, we had \(n = 2^2*3^2 = 36\) so II is false. (Note all of these n's are just minimum values of n, in fact n can be (any prime number)^4 * (any other prime number)^4 etc)
Ans: C
Bunuel
General Discussion
Archit3110
Major Poster
07 Nov 2019, 10:27
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for given condition
n^2 has 25 factors
IMO C III is valid
n^2 has 25 factors
IMO C III is valid
Bunuel
