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If n and y are positive integers and 450y=n^3 [#permalink]
 11 Apr 2010, 23:56
Question Stats:
59% (02:02) correct
40% (01:39) wrong based on 5 sessions
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..
Last edited by abhi758 on 12 Apr 2010, 00:58, edited 1 time in total.
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Re: Number Properties [#permalink]
12 Apr 2010, 05:54
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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]
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]
13 Apr 2010, 02:58
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Re: Number Properties [#permalink]
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]
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]
14 Apr 2010, 23:00
Thanks Bunnel! your first explanation makes it crystal clear..
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Re: problem solving [#permalink]
27 Jun 2010, 06:37
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]
12 Oct 2010, 00:14
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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 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 * 2So 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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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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Must be an Interger [#permalink]
26 Feb 2012, 09:56
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 Hi, I couldnt figure how divisibility rule applies here? I tried to find y but post that it just looked like it could be anything
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If n and y are positive integers, and 450y= n^3, which of the following must be an integer ?
I. y / 3 x 2^2 x 5
II. y / 3^2 x 2 x 5
III. y / 3 x 2 x 5^2
A- None
B- I only
C- II only
D- III only
E- I,II, and III
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450y=n^3, finding integer [#permalink]
19 Jun 2012, 13:38
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.
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Re: integers n and y [#permalink]
19 Jun 2012, 13:42
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Re: 450y=n^3, finding integer [#permalink]
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 * 5^2To 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: 450y=n^3, finding integer
[#permalink]
19 Jun 2012, 22:45
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