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If n = k^2pr^3 where k, p and r are different prime numbers what is 27 Oct 2020, 06:00
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D
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96% (01:06) correct 4% (01:56) wrong based on 109 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
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D
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If n = k^2pr^3 where k, p and r are different prime numbers what is 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.
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Bunuel
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
Show more


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

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If n = k^2pr^3 where k, p and r are different prime numbers what is 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
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If n = k^2pr^3 where k, p and r are different prime numbers what is 27 Oct 2020, 11:41
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CrackVerbalGMAT
Bunuel
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


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 E

Arun Kumar
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Hi,

I think you meant D and not E :)

Thanks,
-K
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If n = k^2pr^3 where k, p and r are different prime numbers what is 27 Oct 2020, 18:11
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[/quote]


Hi,

I think you meant D and not E :)

Thanks,
-K[/quote]

Thank you mehro023. Yes, I meant D. Rectified. Thank You.
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If n = k^2pr^3 where k, p and r are different prime numbers what is 31 Oct 2020, 13:48
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Bunuel
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
Show more

Solution:

Since we want the lowest possible value of n, we want the three smallest primes, i.e., 2, 3, and 5. Furthermore, we want the smallest prime to be raised to the largest power and the largest prime to be raised to the smallest power. Therefore, r = 2, k = 3 and p = 5 and hence,

n = 3^2 x 5 x 2^3 = 9 x 5 x 8 = 360

Answer: D
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