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If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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22 Oct 2008, 11:12
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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
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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22 Oct 2008, 11:36
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rishi2377 wrote: If n and y are positive integers and 450y = n^3, which of the following must be an interger
I. y/ (3 * 2^2 * 5)
II. y/ (3^2 * 2 * 5)
III. y/ 3 * 2 * 5^2
A. None. B. I only. C. II only. D. III only. E. I, II, and III
OA to follow 450y = n^3 > y = n3/450 > n3 is divisible by 450 & n is +ve integer. look for values of n^3 such that n is integer and n is divisible by 450. Clearly we have to look in the order of 1000's to find a value such that n3 is divisible by 450 and n is integer. n3=1000 n =10 not divisible by 450 n3=8000 n =20 not divisible by 450 n3=27000 n =30 divisible by 450 y = 2700/45 = 60 Look for I, II , III which have denominators 60,90,150. only I does and hence B



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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22 Oct 2008, 11:39
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rishi2377 wrote: If n and y are positive integers and 450y = n^3, which of the following must be an interger
I. y/ (3 * 2^2 * 5)
II. y/ (3^2 * 2 * 5)
III. y/ 3 * 2 * 5^2
A. None. B. I only. C. II only. D. III only. E. I, II, and III
OA to follow 450y = n^3 means 450y is a cube. 450 = 3² * 2 * 5² for a cube, the powers of the prime factors have to be multiples of 3 so if 450y is to be a cube then y has to be 3 * 2² * 5 ( to make the three prime factors have powers with multiples of 3) so y/ 3 * 2² * 5 has to be an integer
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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24 Jan 2010, 00:25
I am not sure I fully understand the explanation.
Method1 asks for a technique to find a value of n divisible by 450.This seems to be a time consuming method The other method I dont seem to understand.450=3^2*2*5^2 I also understand the statement "for a cube, the powers of the prime factors have to be multiples of 3"
What I dont understand is how be 3 * 2² * 5 makes each of the prime factors have powers with multiples of 3
Could someone help me on this please.



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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24 Jan 2010, 02:34
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gautamsubrahmanyam wrote: I am not sure I fully understand the explanation.
Method1 asks for a technique to find a value of n divisible by 450.This seems to be a time consuming method The other method I dont seem to understand.450=3^2*2*5^2 I also understand the statement "for a cube, the powers of the prime factors have to be multiples of 3"
What I dont understand is how be 3 * 2² * 5 makes each of the prime factors have powers with multiples of 3
Could someone help me on this please. We have \(450y=2*3^2*5^2*y=n^3\), as \(y\) and \(n\) are positive integers, \(y\) must complete the powers of 2, 3 and 5 so that these powers will be the multiples of 3. Hence the least value of \(y\) is \(2^2*3*5\). In this case \(2*3^2*5^2*y=(2*3^2*5^2)*(2^2*3*5)=2^3*3^3*5^3=(2*3*5)^3=n^3\). As the least value of \(y\) must be \(2^2*3*5\), then \(\frac{y}{2^2*3*5}\), will equal to 1. Choice B. Well generally speaking (and little complicating), \(y\) must be of a form \(y=2^{2+3p}*3^{1+3q}*5^{1+3r}*x^3\), where \(p\), \(q\) and \(r\) are the integers \(\geq0\) and \(x\) (\(x\geq{0}\)), is some other multiple of \(y\) which also has the power of \(3\). For example \(y\) can be \(y=2^5*3^7*5^{16}*91^3\), in this case \((2*3^2*5^2)*(2^5*3^7*5^{16}*91^3)=(2^2*3^3*5^6*91)^3=n^3\).
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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14 Mar 2010, 15:10
great explanation! thanks! Bunuel wrote: gautamsubrahmanyam wrote: I am not sure I fully understand the explanation.
Method1 asks for a technique to find a value of n divisible by 450.This seems to be a time consuming method The other method I dont seem to understand.450=3^2*2*5^2 I also understand the statement "for a cube, the powers of the prime factors have to be multiples of 3"
What I dont understand is how be 3 * 2² * 5 makes each of the prime factors have powers with multiples of 3
Could someone help me on this please. We have \(450y=2*3^2*5^2*y=n^3\), as \(y\) and \(n\) are positive integers, \(y\) must complete the powers of 2, 3 and 5 so that these powers will be the multiples of 3. Hence the least value of \(y\) is \(2^2*3*5\). In this case \(2*3^2*5^2*y=(2*3^2*5^2)*(2^2*3*5)=2^3*3^3*5^3=(2*3*5)^3=n^3\). As the least value of \(y\) must be \(2^2*3*5\), then \(\frac{y}{2^2*3*5}\), will equal to 1. Choice B. Well generally speaking (and little complicating), \(y\) must be of a form \(y=2^{2+3p}*3^{1+3q}*5^{1+3r}*x^3\), where \(p\), \(q\) and \(r\) are the integers \(\geq0\) and \(x\) (\(x\geq{0}\)), is some other multiple of \(y\) which also has the power of \(3\). For example \(y\) can be \(y=2^5*3^7*5^{16}*91^3\), in this case \((2*3^2*5^2)*(2^5*3^7*5^{16}*91^3)=(2^2*3^3*5^6*91)^3=n^3\).



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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18 Dec 2010, 04:00
viperm5 wrote: great explanation! thanks! Bunuel wrote: gautamsubrahmanyam wrote: I am not sure I fully understand the explanation.
Method1 asks for a technique to find a value of n divisible by 450.This seems to be a time consuming method The other method I dont seem to understand.450=3^2*2*5^2 I also understand the statement "for a cube, the powers of the prime factors have to be multiples of 3"
What I dont understand is how be 3 * 2² * 5 makes each of the prime factors have powers with multiples of 3
Could someone help me on this please. We have \(450y=2*3^2*5^2*y=n^3\), as \(y\) and \(n\) are positive integers, \(y\) must complete the powers of 2, 3 and 5 so that these powers will be the multiples of 3. Hence the least value of \(y\) is \(2^2*3*5\). In this case \(2*3^2*5^2*y=(2*3^2*5^2)*(2^2*3*5)=2^3*3^3*5^3=(2*3*5)^3=n^3\). As the least value of \(y\) must be \(2^2*3*5\), then \(\frac{y}{2^2*3*5}\), will equal to 1. Choice B. Well generally speaking (and little complicating), \(y\) must be of a form \(y=2^{2+3p}*3^{1+3q}*5^{1+3r}*x^3\), where \(p\), \(q\) and \(r\) are the integers \(\geq0\) and \(x\) (\(x\geq{0}\)), is some other multiple of \(y\) which also has the power of \(3\). For example \(y\) can be \(y=2^5*3^7*5^{16}*91^3\), in this case \((2*3^2*5^2)*(2^5*3^7*5^{16}*91^3)=(2^2*3^3*5^6*91)^3=n^3\). As y=2*2*3*5=60,so B as well as C is also correct because II. y/ (3^2 * 2 * 5) is equal to 60/60=1 and that is an integer. The same way III. y/ 3 * 2 * 5^2 can also be the integer because when we multiply 5*5*5 to y=2*2*3*5 ie 60*125 it provides the III. y/ 3 * 2 * 5^2 with the integer value. So the answer should be E. Let me have ur thoughts on the same.



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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18 Dec 2010, 04:19



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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18 Dec 2010, 04:27
Yeah I got that...Thanks for your quick response.



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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14 Apr 2012, 09:01
Bunuel wrote: gautamsubrahmanyam wrote: I am not sure I fully understand the explanation.
Method1 asks for a technique to find a value of n divisible by 450.This seems to be a time consuming method The other method I dont seem to understand.450=3^2*2*5^2 I also understand the statement "for a cube, the powers of the prime factors have to be multiples of 3"
What I dont understand is how be 3 * 2² * 5 makes each of the prime factors have powers with multiples of 3
Could someone help me on this please. We have \(450y=2*3^2*5^2*y=n^3\), as \(y\) and \(n\) are positive integers, \(y\) must complete the powers of 2, 3 and 5 so that these powers will be the multiples of 3. Hence the least value of \(y\) is \(2^2*3*5\). In this case \(2*3^2*5^2*y=(2*3^2*5^2)*(2^2*3*5)=2^3*3^3*5^3=(2*3*5)^3=n^3\). As the least value of \(y\) must be \(2^2*3*5\), then \(\frac{y}{2^2*3*5}\), will equal to 1. Choice B. Well generally speaking (and little complicating), \(y\) must be of a form \(y=2^{2+3p}*3^{1+3q}*5^{1+3r}*x^3\), where \(p\), \(q\) and \(r\) are the integers \(\geq0\) and \(x\) (\(x\geq{0}\)), is some other multiple of \(y\) which also has the power of \(3\). For example \(y\) can be \(y=2^5*3^7*5^{16}*91^3\), in this case \((2*3^2*5^2)*(2^5*3^7*5^{16}*91^3)=(2^2*3^3*5^6*91)^3=n^3\). Thanks. Now is clear because the question that I had, it was in the form n3 and NOT n^3 so I didn't think how to solve this problem because the to quantities did not balance each other
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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12 May 2012, 08:00
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rishi2377 wrote: If n and y are positive integers and 450y = n^3, which of the following must be an interger
I. y/ (3 * 2^2 * 5)
II. y/ (3^2 * 2 * 5)
III. y/ (3 * 2 * 5^2)
A. None. B. I only. C. II only. D. III only. E. I, II, and III
OA to follow A difficult problem which can be easy if we understand the question states that 450y=n^3 that means 450y must be to the third power of n ,so if a number raised to the power of cube then all numbers must be in 3 no's for example if we take 216 then if we factorize 216 we get 2*2*2*3*3*3 that means in cube power we have each number 3 times so let's go to our problem here we have 3 2's and 3 3's so 450y must have all numbers in 3 no's so if we factorize 450 we get 450=2*3*3*5*5 as we need all numbers in 3 no's that means we should have 3 2's,3 3's,3 5's so we need extra 2 2's,a 3 and a 5 to make 450y perfect cube so y must be 2^2*3*5=>4*3*5 =>60 so y must be 60 to make 450y a perfect cube so if we look at given options 1. y/ (3 * 2^2 * 5) => y/3*4*5 =>y/60 as we know that y is 60 so 60/60=1 which is integer 2. y/ (3^2 * 2 * 5) => y/9*2*5 =>y/90 as we know that y is 60 so 60/90 =>2/3 which is not integer 3. y/ (3 * 2 * 5^2) => y/3*2*25 =>y/150 as we know that y is 60 so 60/150 =>2/5 which is not integer so we get only 1 as integer so our answer is b Hope it's helpful.



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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11 Jul 2012, 12:41
Can someone please explain to me why y cannot simply be 2.
If 450y = n^3, then cannot it be 3x3x5x5x2 and y=2 so that would make 3x3x5x5x2x2 so it ends up 30x30x30 = nxnxn. Then if y=x and the equation works, then y/I, II, or III would not work.
In this case, the answer should be A, None. Please help, I've been racking my brain over why y cannot be 2, but it makes perfect sense.



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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12 Jul 2012, 02:09
mthasan1 wrote: Can someone please explain to me why y cannot simply be 2.
If 450y = n^3, then cannot it be 3x3x5x5x2 and y=2 so that would make 3x3x5x5x2x2 so it ends up 30x30x30 = nxnxn. Then if y=x and the equation works, then y/I, II, or III would not work.
In this case, the answer should be A, None. Please help, I've been racking my brain over why y cannot be 2, but it makes perfect sense. If \(y=2\) then \(450y=2^2*3^2*5^2=30^2=900\) not \(30^3\) as you've written.
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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19 Oct 2014, 19:23
rishi2377 wrote: If n and y are positive integers and 450y = n^3, which of the following must be an interger
I. y/ (3 * 2^2 * 5)
II. y/ (3^2 * 2 * 5)
III. y/ 3 * 2 * 5^2
A. None. B. I only. C. II only. D. III only. E. I, II, and III Hello There, 450 X y = n^3 Factors of 450 = (3^2) X (5^2) X 2 Since, there is a cube on RHS, lets try to make LHS also a cube. To do this, we need one 3, one 5 and two 2s which will result in 3^3 X 5^3 X 2^3 Now, y = 3 X 5 X 2^2 = 60. Now verifying with options, only I results in an integer. Answer  B
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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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26 Dec 2014, 12:01
450y = n^3 means 450y is a cube.
450 = 3² * 2 * 5²
for a cube, the prime factors should be in a set of 3
so Y/3*2²*5



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Re: If n and y are positive integers and 450y = n^3 which of the following [#permalink]
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07 Mar 2015, 10:33
Could someone please explain why 2,3,and 5 have to be in powers that is a multiple of 3? Recognizing that seems to be the crux of the problem. I am not sure why if n^3 that means the prime factors have to be in powers of 3 also. If anyone could provide explanations or links for this rule it would be much appreciated! Thanks!



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