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Re: N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f [#permalink]
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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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N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f [#permalink]
so basically in this question :

we are taking the LCM of the powers because :

N/2- is a square---- but N must be a cube and a 5th power
to be a square; a cube and a fifth power we take the LCM of 5 and 3- > 15

and the same goes for the other two :

given N/3= a cube then the minimum value would be 3^4---- but when we divide N by 3 we get 3^3 which is neither a square nor a 5th power therefore when we take the LCM of 2 and 5 which is 10-----> therefore 3^10 is a square and a fifth power!
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N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f [#permalink]
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MathRevolution wrote:
[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 \(N\)?

\(A. (2^{11})(3^8)(5^8)\)

\(B. (2^{13})(3^8)(5^6)\)

\(C. (2^{15})(3^{10})(5^6)\)

\(D. (2^{17})(3^{12})(5^6)\)

\(E. (2^{19})(3^{13})(5^6)\)



I apply this shortcut while solving such questions, it's fast and easy

>For a no. to be square-- the power of 2 has to be a multiple of two
in the above case of N/2 - 2^11/2 = 2^10

>For a no. to be a cube -- The power of 3 has to be a multiple of 3
in the above case of N/3 - 3^10/3 = 3^9

and the power fifth can be narrowed by the option given..
Hence, Option C is correct

Kudos, if you find this helpful.
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N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f [#permalink]
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