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If m and n are integers, and m =
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02 Feb 2020, 04:43
02 Feb 2020, 04:43
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Question Stats:
71% (00:55) correct
28% (02:19) wrong based on 7 sessions
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If m and n are integers, and m = \(\frac{(48)(35)(51)}{(68n)}\) which of the following could be a value of n?
A. 125
B. 126
C. 127
D. 128
E. 129
A. 125
B. 126
C. 127
D. 128
E. 129
Re: If m and n are integers, and m =
[#permalink]
03 Feb 2020, 07:58
03 Feb 2020, 07:58
This question asks for prime factors: since m is an integer, n has to be a factor of m
48 = 4*12 = 2*2*2*2*3 = 2^4 *3
35 = 5*7
51 = 17 *3
68 = 2^2 *17
--> (2^4*3) * (5*7) * (17*3) / (2^2 * 17 *n) | cancel 2^2 and 17 out
= (2^2 * 3^2 * 5 * 7) / n
--> n is only allowed to consist of the primes in the numerator
A) 125 = 5*5*5 --> we only have 5^1 and not 5^3 --> not possible
B) 126 = 2 * 63 = 2 * 3 * 3 * 7 --> possible --> than m is going to be 2*5 = 10 --> we found our match --> B is the correct answer
C) 127 is a prime
D) 128 = 2* 64 = 2^6 --> not possible
E) 129 = 43 * 3 --> not possible
Hence, answer B is the correct answer
48 = 4*12 = 2*2*2*2*3 = 2^4 *3
35 = 5*7
51 = 17 *3
68 = 2^2 *17
--> (2^4*3) * (5*7) * (17*3) / (2^2 * 17 *n) | cancel 2^2 and 17 out
= (2^2 * 3^2 * 5 * 7) / n
--> n is only allowed to consist of the primes in the numerator
A) 125 = 5*5*5 --> we only have 5^1 and not 5^3 --> not possible
B) 126 = 2 * 63 = 2 * 3 * 3 * 7 --> possible --> than m is going to be 2*5 = 10 --> we found our match --> B is the correct answer
C) 127 is a prime
D) 128 = 2* 64 = 2^6 --> not possible
E) 129 = 43 * 3 --> not possible
Hence, answer B is the correct answer
Re: If m and n are integers, and m =
[#permalink]
05 Mar 2022, 01:49
05 Mar 2022, 01:49
Hello from the GRE Prep Club BumpBot!
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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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).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
