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27 Oct 2020, 06:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
96% (01:05) correct
4%
(01:56)
wrong
based on 102
sessions
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If \(n=k^2pr^3\) where k, p and r are different prime numbers what is the lowest possible value of n ?
A. 180
B. 240
C. 300
D. 360
E. 720
A. 180
B. 240
C. 300
D. 360
E. 720
Updated on: 31 Oct 2020, 19:32
Originally posted by CrackverbalGMAT on 27 Oct 2020, 07:44.
Last edited by CrackverbalGMAT on 31 Oct 2020, 19:32, edited 2 times in total.
Last edited by CrackverbalGMAT on 31 Oct 2020, 19:32, edited 2 times in total.
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Bunuel
2 ways of doing this.
Method 1: If we want the lowest possible value of n, then the prime numbers have to be 2, 3 and 5.
Then we have 6 ways of putting the values of k, p and r and we get to the answer. A little application would help in the fact that the lowest value should be given to the number with the highest power. So if we take r = 2, k = 3 and p = 5, we get n = 9 * 5 * 8 = 720
Method 2:
Break each of the Options into its factors to see which of them fits our requirements.
Option A: 180 = \(2^2 * 3^2 *5\). This is not possible as we need a cubic power.
Option B: 240 = \(2^4 * 3 *5\). This is not possible.
Option C: 300 = \(2^2 * 3 *5^2\). This is not possible as we need a cubic power.
Option D: 360 = \(2^3 * 3^2 *5\). This Fits our requirement.
No need to check Option E as it is greater than 360.
Option D
Arun Kumar
27 Oct 2020, 09:01
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Asked: If \(n=k^2pr^3\) where k, p and r are different prime numbers what is the lowest possible value of n ?
n = 2^3*3^2*5 = 8*9*5 = 360
IMO D
n = 2^3*3^2*5 = 8*9*5 = 360
IMO D
