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Joined: 20 Feb 2017
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If n > 6, and if n is a multiple of 6, which of the followin
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19 Mar 2020, 21:32
19 Mar 2020, 21:32
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If n > 6, and if n is a multiple of 6, which of the following is always a factor of n?
(A) \(n - 6\)
(B) \(n + 6\)
(C) \(\frac{n}{3}\)
(D) \(\frac{n}{2} + 3\)
(E) \(\frac{n}{2} + 6\)
(A) \(n - 6\)
(B) \(n + 6\)
(C) \(\frac{n}{3}\)
(D) \(\frac{n}{2} + 3\)
(E) \(\frac{n}{2} + 6\)
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WE:Education (Education)
If n > 6, and if n is a multiple of 6, which of the followin
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12 Feb 2021, 12:02
12 Feb 2021, 12:02
n > 6 and is a multiple of 6
Therefore, n can be 12, 18, 24, 30, ..............
A. n - 6 is a factor of n
Not necessarily;
when n = 12
n - 6 = 6: It is a factor
but, when n = 18
n - 6 = 12: not a factor
B. n + 6 is a factor of n
NO;
when n = 12
n + 6 = 18: Not a factor
C. \(\frac{n}{3}\) is a factor of n: YES
when n = 12
\(\frac{n}{3}\) = 4: It is a factor
when n = 18
\(\frac{n}{3}\) = 6: It is a factor
when n = 24
\(\frac{n}{3}\) = 8: It is a factor, and so on ..
D. \(\frac{n}{2} + 3\) is a factor of n
No;
when n = 12
\(\frac{n}{2} + 3\) = 9: Not a factor
E. \(\frac{n}{2} + 6\) is a factor of n
Not necessarily;
when n = 12
\(\frac{n}{2} + 6\) = 12: It is a factor
but, when n = 18
\(\frac{n}{2} + 6\) = 15: Not a factor
Hence, option C
Therefore, n can be 12, 18, 24, 30, ..............
A. n - 6 is a factor of n
Not necessarily;
when n = 12
n - 6 = 6: It is a factor
but, when n = 18
n - 6 = 12: not a factor
B. n + 6 is a factor of n
NO;
when n = 12
n + 6 = 18: Not a factor
C. \(\frac{n}{3}\) is a factor of n: YES
when n = 12
\(\frac{n}{3}\) = 4: It is a factor
when n = 18
\(\frac{n}{3}\) = 6: It is a factor
when n = 24
\(\frac{n}{3}\) = 8: It is a factor, and so on ..
D. \(\frac{n}{2} + 3\) is a factor of n
No;
when n = 12
\(\frac{n}{2} + 3\) = 9: Not a factor
E. \(\frac{n}{2} + 6\) is a factor of n
Not necessarily;
when n = 12
\(\frac{n}{2} + 6\) = 12: It is a factor
but, when n = 18
\(\frac{n}{2} + 6\) = 15: Not a factor
Hence, option C
Re: If n > 6, and if n is a multiple of 6, which of the followin
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07 Mar 2024, 06:30
07 Mar 2024, 06:30
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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