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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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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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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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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 $$N$$?

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

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

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

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

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

_________________
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 31 May 2018 Posts: 185 Location: United States Concentration: Finance, Marketing Re: N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f [#permalink] ### Show Tags 01 Aug 2019, 01:35 N is a positive integer and is divisible by 2,3 and 5 so N is of form = $$2^a*3^b*5^c$$ now given $$\frac{N}{2}$$ is a square (that means powers of N are multiple of 2 $$\frac{N}{2}$$ = 2^(a-1)*3^b*5^c (when a-1 ,b and c divided by 2 result is integer) a = 2I+1, b = 2I, c = 2I a= 3,5,7...., b = 2,4,6...., c = 2,4,6..... similarly, $$\frac{N}{3}$$ is a cube (that means powers of N are multiple of 3) $$\frac{N}{3}$$ = 2^a*3^(b-1)*5^c (when a ,b-1 and c divided by 3 result is integer) a = 3I, b = 3I+1 , c = 3I a = 3,6,9..., b = 4,7,10..., c = 3,6,9... $$\frac{N}{5}$$ is a fifth power (that means powers of N are multiple of 5) $$\frac{N}{5}$$ = 2^a*3^b*5^(c-1) (when a ,b and c-1 divided by 5 result is integer) a = 5I, b = 5I+1 , c = 5I a= 5,10,15.. b = 6,11,16... c= 5,10,15 taking the LCM of all the 3 cases of value of a, b and c (to satisfy $$\frac{N}{2}$$ square, $$\frac{N}{3}$$ cube,$$\frac{N}{5}$$ fifth power) a = 15,30,45...., b= 10,20,30...., c = 6,12,18 minimum values of a,b, and c = 15,10, and 6 so minimum value of N = 2^15*3^10*5^6 CrackVerbal Quant Expert Joined: 12 Apr 2019 Posts: 218 Re: N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f [#permalink] ### Show Tags 01 Aug 2019, 04:38 1 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. Hope this helps! _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7741 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f [#permalink] ### Show Tags 04 Aug 2019, 18:57 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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] 04 Aug 2019, 18:57
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