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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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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 $$N = (2^{15})(3^{10})(5^6).$$

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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.
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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