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N is a positive integer. N/2, N/3 and N/5 are a square, a cu

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N is a positive integer. N/2, N/3 and N/5 are a square, a cu

by Max@Math Revolution » Wed Jul 31, 2019 10:52 pm

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[GMAT math practice question]

N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a fifth power, respectively. What is the minimum possible value of N?

A. (2^{11})(3^8)(5^8)
B. (2^{13})(3^8)(5^6)
C. (2^{15})(3^{10})(5^6)
D. (2^{17})(3^{12})(5^6)
E. (2^{19})(3^{13})(5^6)
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edited:

by deloitte247 » Sat Aug 03, 2019 10:54 pm
N is divisible by 2, 3 and 5. i.e N/2: N/3; N/5.
$$N\ can\ be\ written\ as\ 2^x\cdot3^y\cdot5^z$$
A perfect square will always have an even power for all its prime factors.
A perfect cube will always have a multiple of 3 as the power for all its prime factors.
In the same vein, the fifth power should have the power of its prime factor as multiple of 5.
$$So,\ for\ option\ A,\ N=2^{11}\cdot3^8\cdot5^8;\ \frac{N}{2}=2^{10}\cdot3^8\cdot5^8;\ which\ is\ a\ perfect\ square,$$
$$While\ \frac{N}{3}=2^{11}\cdot3^7\cdot5^8\ is\ not\ a\ perfect\ cube.\ $$
Hence, option A cannot be the correct answer.
$$For\ option\ B,\ N=2^{13}\cdot3^8\cdot5^6;\ \frac{N}{2}=2^{12}\cdot3^8\cdot5^6;\ is\ a\ perfect\ square,$$
$$While\ \frac{N}{3}=2^{13}\cdot3^7\cdot5^6\ is\ not\ a\ perfect\ cube.\ $$
Therefore, option B cannot be the correct answer.
$$For\ option\ C,\ N=2^{15}\cdot3^10\cdot5^6;\ \frac{N}{2}=2^{14}\cdot3^10\cdot5^6;\ is\ a\ perfect\ square,$$
$$While\ \frac{N}{3}=2^{15}\cdot3^9\cdot5^6\ is\ also\ a\ perfect\ cube.\ $$
$$Thus,\ \frac{N}{5}=2^{15}\cdot3^{10}\cdot5^5\ is\ a\ fifth\ power;\ Option\ C\ is\ therefore\ the\ CORRECT\ answer.$$

Thanks. I hope this helps!<i class="em em---1"></i>
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by Max@Math Revolution » Sun Aug 04, 2019 5:36 pm
=>

Since N/2, N/3 and N/5 are integers, N should have prime factors of 2, 3, and 5. Thus, the minimum value of N has the form (2^a)(3^b)(5^c) for some positive integers, a, b and c.
Since N/2 = (2^{a-1})(3^b)(5^c) is a square, b and c are even numbers and a is an odd number.
Since N/3 = (2^a)(3^{b-1})(5^c) is a cube, a and c are multiples of 3 and b has remainder 1 when it is divided by 3.
Since N/5 = (2^a)(3^b)(5^{c-1}) is a fifth power, a and b are multiples of 5 and c has remainder 1 when it is divided by 5.

Since a is a multiple of 3 and 5 and an odd number, the smallest positive value of a is 15.
Since b is a multiple of 2 and 5 and has remainder 1 when it is divided by 3, the smallest positive value of b is 10.
Since c is a multiple of 2 and 3 and has remainder 1 when it is divided by 5, the smallest positive value of c is 6.

Thus, the minimum value of N is N = (2^{15})(3^{10})(5^6).

Therefore, C is the answer.
Answer: C
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