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N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f
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01 Aug 2019, 00:08
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Difficulty:
45% (medium)
Question Stats:
62% (01:47) correct 38% (02:07) wrong based on 45 sessions
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[GMAT math practice question]
\(N\) is a positive integer. \(\frac{N}{2}, \frac{N}{3}\) and \(\frac{N}{5}\) are a square, a cube and a fifth power, respectively. What is the minimum possible value of \(N\)?
Re: N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f
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01 Aug 2019, 04:38
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This is a slightly different question compared to the usual ones on the concept of squares and cubes.
A perfect square is the square of an integer. A perfect square always has an even power for all its prime factors.
A perfect cube is the cube of an integer. A perfect cube always has a multiple of 3 as the power for all its prime factors.
Similarly, the fifth power should have the powers of its prime factors as multiples of 5.
With this information, let’s look at the question. The best approach to solve this question in the shortest time is to eliminate options, taking them one by one.
For option A, N = \(2^{11} * 3^8 * 5^8\). So, \(\frac{N}{2}\) = \(2^{10} * 3^8 * 5^8\) which is a perfect square. \(\frac{N}{3}\) = \(2^{11} * 3^7 * 5^8\) which is not a perfect cube. Option A can be eliminated.
Following the same procedure for option B, we see that although \(\frac{N}{2}\) is a perfect square, \(\frac{N}{3}\) is not a perfect cube and so option B can also be eliminated.
When we come to option C, we see that \(\frac{N}{2}\) is a perfect square, \(\frac{N}{3}\) is a perfect cube and \(\frac{N}{5}\) is a fifth power. Option C has to be the answer.
Re: N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f
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04 Aug 2019, 18:57
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Since \(\frac{N}{2}, \frac{N}{3}\) and \(\frac{N}{5}\) 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 \(\frac{N}{2} = (2^{a-1})(3^b)(5^c)\) is a square, \(b\) and \(c\) are even numbers and \(a\) is an odd number.
Since \(\frac{N}{3} = (2^a)(3^{b-1})(5^c)\) is a cube, \(a\) and \(c\) are multiples of \(3\) and \(b\) has remainder \(1\) when it is divided by \(3.\)
Since \(\frac{N}{5} = (2^a)(3^b)(5^{c-1})\) 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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62% (01:47) correct 
