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# If n and y are positive integers and 450y=n^3

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If n and y are positive integers and 450y=n^3  [#permalink]

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Updated on: 11 Apr 2010, 23:58
13
74
00:00

Difficulty:

45% (medium)

Question Stats:

62% (01:41) correct 38% (01:11) wrong based on 2158 sessions

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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

Originally posted by abhi758 on 11 Apr 2010, 22:56.
Last edited by abhi758 on 11 Apr 2010, 23:58, edited 1 time in total.
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Re: Number Properties  [#permalink]

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12 Apr 2010, 04:54
40
36
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

"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: $$\frac{y}{{3*2^2*5}}=\frac{2^2*3*5}{{3*2^2*5}}=1$$.

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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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31 Oct 2013, 04:49
15
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 non-math-overbrains.

Greets
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Re: Number Properties  [#permalink]

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12 Apr 2010, 22:15
1
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, 01:58
4
6
shrivastavarohit wrote:
Moderator,

Can you elaborate more on the solution, seems unfathomable for simple minds like mine .

How did we deduce that "Minimum value of Y"?

$$x$$ and $$y$$ are integers and $$450y=x^3$$ --> $$450y$$ equals to cube of an integer. $$450y=2*3^2*5^2*y=x^3$$. The smallest value of $$y$$ for which $$2*3^2*5^2*y$$ is a cube of an integer is when $$y=2^2*3*5$$. In this case $$450y=(2*3^2*5^2)*(2^2*3*5)=(2*3*5)^3$$. Of course $$y$$ can take another values as well, for example $$y=2^5*3^4*5^7$$ and in this case $$450y=(2*3^2*5^2)*(2^5*3^4*5^7)=(2^3*3^2*5^3)^3$$, but the smallest value of $$y$$ is when $$y=2^2*3*5$$.

You can check the similar problems at:
og-quantitative-91750.html#p704028
division-factor-88388.html#p666722
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Re: Number Properties  [#permalink]

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13 Apr 2010, 02:28
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, 02:32
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, 22:00
1
Thanks Bunnel! your first explanation makes it crystal clear..
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Re: problem solving  [#permalink]

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Updated on: 27 Jun 2010, 20:35
1
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.

Originally posted by jakolik on 27 Jun 2010, 05:37.
Last edited by jakolik on 27 Jun 2010, 20:35, edited 1 time in total.
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Re: Tough integer properties question!  [#permalink]

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11 Oct 2010, 23:14
2
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

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08 Jan 2012, 04:29
kotela wrote:

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, 21:45
11
3
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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19 Apr 2014, 21:11
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, 10:59
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).

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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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22 May 2014, 12:11
2
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, 19:25
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, 09:53
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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16 Jul 2015, 08:39
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  [#permalink]

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29 Nov 2016, 19:13
Our goal is to get figure out why so that x is a perfect square:

450 = (2)(3^2)(5^2)

To make a perfect square we are missing the following: (2^2)(3)(5) --> This is the minimum value y must take on in order for us to get the cube root of x to be an integer.

Option I is the only choice given that will give us an integer value for x. Thus, B is the correct answer.
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Re: If n and y are positive integers and 450y=n^3  [#permalink]

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28 Dec 2016, 10:08
Great Official Question.
Here is what i did in this one =>

x is an integer => x^3 must be a perfect cube.

Hence the general expression for y => 5*3*2^2*prime^3

Hence least value of y will be 5*3*2^2

Only 1 is in integer

Hence B

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Re: If n and y are positive integers and 450y=n^3 &nbs [#permalink] 28 Dec 2016, 10:08

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# If n and y are positive integers and 450y=n^3

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