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If n and y are positive integers and 450y=n^3 [#permalink]
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11 Apr 2010, 23:56
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If x and y are positive integers and 450y=x^3, which of the following must be an integer? i) \(\frac{y}{{3*2^2*5}}\) ii) \(\frac{y}{{3^2*2*5}}\) iii) \(\frac{y}{{3*2*5^2}}\) a. None b. i only c. ii only d. iii only e. i, ii and iii Please explain your answers..
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Last edited by abhi758 on 12 Apr 2010, 00:58, edited 1 time in total.



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Re: Number Properties [#permalink]
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abhi758 wrote: If x and y are positive integers and \(450y=x^3\), which of the following must be an integer? i) \(\frac{y}{{3*2^2*5}}\) ii) \(\frac{y}{{3^2*2*5}}\) iii) \(\frac{y}{{3*2*5^2}}\) a. None b. i only c. ii only d. iii only e. i, ii and iii Please explain your answers.. "Must be an integer" means for the lowest possible value of \(y\). \(450y=x^3\) > \(2*3^2*5^2*y=x^3\). As \(x\) and \(y\) are integers, \(y\) must complete the powers of 2, 3, and 5 to cubes (generally to the multiple of 3). Thus \(y_{min}=2^2*3*5\), in this case \(2*3^2*5^2*y=(2*3*5)^3=x^3\). Notice that for this value of \(y\) only the first option is an integer: \(\frac{y}{{3*2^2*5}}=\frac{2^2*3*5}{{3*2^2*5}}=1\). Answer: B.
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Re: Number Properties [#permalink]
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12 Apr 2010, 23:15
Moderator, Can you elaborate more on the solution, seems unfathomable for simple minds like mine . How did we deduce that "Minimum value of Y"?
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Re: Number Properties [#permalink]
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13 Apr 2010, 02:58
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Re: Number Properties [#permalink]
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13 Apr 2010, 03:28
Thanks for the quick response it makes more sense now I will check out the problems (similar) on the threads mentioned by you. Is this really a GMAT question from the calculation it looks solving this should take more than 2 mins. Posted from my mobile device
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Re: Number Properties [#permalink]
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13 Apr 2010, 03:32
Oops !!! One more important thing why is plugging in values applicable to these kind of problems? It seemed pretty simple from the question to plug in 2 integer values in the final equation and check for the integer output btw which miserably failed on all three equations. Posted from my mobile device
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Re: Number Properties [#permalink]
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14 Apr 2010, 23:00
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Thanks Bunnel! your first explanation makes it crystal clear..



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Re: problem solving [#permalink]
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27 Jun 2010, 06:37
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I am not sure if I got the question right. The way I understand this Q isthat 450y = n*n*n and the question is whether the given options are integers.
The answer of above Q is as follows: We know that: n = cubic root (450y) = cubic root (2X3X3X5X5) For n to be an integer, y should be factor of 2X2X3X5.
So y=(2X2X3X5)k where k is any natural number. if k = 1, y=2X2X3X5 so the answer is b.
I hope this explanation meets your satisfaction.
Last edited by jakolik on 27 Jun 2010, 21:35, edited 1 time in total.



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Re: Tough integer properties question! [#permalink]
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atomjuggler wrote: Raising the white flag on this one. I really hope I'm not missing something obvious Source: GMATPrep Test #1 If n and y are positive integers and 450y=n^3, which of the following must be an integer? I. \(\frac{y}{3*2^2*5}\) II. \(\frac{y}{3^2*2*5}\) III. \(\frac{y}{3*2*5^2}\) A. None B. I only C. II only D. III only E. I, II, and III \(450*y=n^3\) Since 450y is a whole cube, in the prime factorization of 450y, all the exponents on the prime factors must be multiples of 3. \(450*y = y * 3^2 * 5^2 * 2\) So y must have atleast \(3, 5, 2^2\) as prime factors. y must be of the form form 3x5x4xA where A is also another perfect cube. By this logic the answer is (I) only or B, as in all cases y/(3x5x4) has to be an integer. (ii) or (iii) cannot be true if A=1 Answer is (b)
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Re: Algebra [#permalink]
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08 Jan 2012, 05:29
kotela wrote: Can anyone please help me in solving this problem....... lets find prime factors of 450 = 2* 3* 3* 5* 5 for 450* y to be equal to n^3 450y needs to have 2,3and 5 three times ( as n is an integer ) i.e. 450 y = (2* 3* 3* 5* 5) (2*2*3*5) * x = n^3 now y = (2*2*3*5) * x only option 1 will then give an integer



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Re: 450y=n^3, finding integer [#permalink]
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19 Jun 2012, 22:45
enigma123 wrote: If n and y are positive integers and 450y = n^3, which of the following must be an integer?
1. \(\frac{Y}{3 * 2^2 * 5}\)
2. \(\frac{Y}{3^2 * 2 * 5}\)
3. \(\frac{Y}{3 * 2 * 5^2}\)
A) None
B) I only
C) II Only
D) III only
E) I, II and III onmly. The critical point in this question is that n and y are integers. Since n is an integer, n^3 must be the cube of an integer. So in n^3, all the prime factors of n must be cubed (or have higher powers which are multiples of 3) Now consider 450y = n^3 450 is not a perfect cube. So whatever is missing in 450, must be provided by y to make a perfect cube. \(450 = 45*10 = 2 * 3^2 * 5^2\) To make a perfect cube, we need at least two 2s, a 3 and a 5. These missing factors must be provided by y. Therefore, \(2^2*3*5\) must be a factor of y. This means \(y/2^2*3*5\) must be an integer.
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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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28 Oct 2013, 13:36
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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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31 Oct 2013, 05:49
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I finally get it. If you want the cuberoot of 450y to be an integer, you need the primefactors to be cubed.
So the primefactors we have for 450 are 2, 3², 5². To be perfect cubes, we need to multiply by 2², 3, 5 which in this case equals y.
so we get that (2*3²*5²) * (2²*3*5) = 2³*3³*5³. If we'd cuberoot x now, we'd get an integer.
This leads to the conclusion that y / 2²*3*5 = 1 = integer. So answer B.
I tried explaining in simpler words so I'd get it myself, I hope it helps the you other nonmathoverbrains.
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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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19 Apr 2014, 22:11
Concept Tested: For a perfect square the powers should occur in pairs. For a perfect cube powers should occur in cubes. 450y = x^3 5^2 * 3^2 * 2 * y = x^3 For LHS to become a cube Y must be y= 5*3*2^2 Only (1) matches it.
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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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22 May 2014, 11:59
I could understand the options
But tell y my alternate approach is wrong 450y = x^3
X and Y are integer 450 = 2*3^2*5^2
X=((2*3^2*5^2)y)/x^2 Note: as x is an integer if x = 2*3^2*5^2
X = y/ (2*3^2*5^2)
And its option c).
Please correct me.
Thanks in advance.



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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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Hi rrsnathan, Your approach is wrong because you are assuming 1 fixed value of \(x = 2*3^2*5^2\) But question explicitly asks which option MUST be an integer always. In your assumption, Option C comes out to be an integer but that may not be true in other cases.So, Right method to solve it is: \(450y = x^3\) \((2*3^2*5^2) * y = x^3\) For LHS to be a cube (As R.H.S is a cube), \(y = 2^2*3*5*k^3\) (where k is a positive integer) Hence, Option (b) is always correct Rgds, Rajat
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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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22 May 2014, 20:25
rrsnathan wrote: I could understand the options
But tell y my alternate approach is wrong 450y = x^3
X and Y are integer 450 = 2*3^2*5^2
X=((2*3^2*5^2)y)/x^2 Note: as x is an integer if x = 2*3^2*5^2
X = y/ (2*3^2*5^2)
This is incorrect. Note that in the denominator, you have x^2, not x. You are assuming \(x = 2*3^2*5^2\) and putting it as it is in the denominator. Also note that since you have the same x on the left hand side, if you put \(x = 2*3^2*5^2\), you get \(2*3^2*5^2 = (2*3^2*5^2)*y/(2*3^2*5^2)^2\) i.e. \(y = (2*3^2*5^2)^2\) Basically, you have assumed a value of x and got a value of y. There is no reason for you to assume the value of x as \(2*3^2*5^2\). The point here is that y must have some values such that when it multiplies 450, it gives a perfect cube. This is the method discussed above in my post.
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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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23 May 2014, 10:53



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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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Re: If n and y are positive integers and 450y=n^3 [#permalink]
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16 Jul 2015, 09:39
450 = (5^2) * (3^2) * (2) so, in order for 450y to be a cube of some number, y must be equal to 5*3*(2^2) = 60. Only option (i) gives an integer (60/60). So, Ans(B).
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Re: If n and y are positive integers and 450y=n^3
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