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multiples or factors???

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by divya23 » Mon Jun 06, 2011 10:20 am
N and Y are +ve integers and 450y = n^3 which of the following is an integer?

y/3*2^2*5
y/3^2*2*5
y/3*2*5^2
y/3*2*5
y/3^3*2^2*5

[spoiler]oa = y/3*2^2*5[/spoiler]
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by Frankenstein » Mon Jun 06, 2011 10:29 am
Hi,
450=2.3^2.5^2
So, 2.3^2.5^2.y = n^3
So, y is of the form 2^2.3.5.p^3 where p is an integer
So, y/3*2^2*5 is an integer.
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by champ0007 » Mon Jun 06, 2011 10:48 am
[quote="Frankenstein"]Hi,
450=2.3^2.5^2
So, 2.3^2.5^2.y = n^3
So, y is of the form 2^2.3.5.p^3 where p is an integer
So, y/3*2^2*5 is an integer.[/quote]

Good one!!
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by iamsaurav » Mon Jun 06, 2011 12:59 pm
LHS tells us its a cube, equate both sides. As RHS is not a cube make it and find Y which comes out to be 2^2*3*5, so it must be divisible by it.

Advice- this is a symmetry based problem look for symmetry on two sides of so formed equations.
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by cans » Mon Jun 06, 2011 8:23 pm
Frankenstein wrote:Hi,
450=2.3^2.5^2
So, 2.3^2.5^2.y = n^3
So, y is of the form 2^2.3.5.p^3 where p is an integer
So, y/3*2^2*5 is an integer.
Hey i agree with IMO A
But option D is also correct
y/3*2*5
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by Frankenstein » Mon Jun 06, 2011 10:34 pm
cans wrote:
Frankenstein wrote:Hi,
450=2.3^2.5^2
So, 2.3^2.5^2.y = n^3
So, y is of the form 2^2.3.5.p^3 where p is an integer
So, y/3*2^2*5 is an integer.
Hey i agree with IMO A
But option D is also correct
y/3*2*5
Hi,
Yes.. D is correct too. But, I haven't really checked other options after getting A. Moreover, those * and ^ were so awkward, I wanted to skip this question asap!
Cheers!

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by [email protected] » Fri Jun 10, 2011 8:50 am
hey guyzzz... i did not at all understand the solution of this sum > N and Y are +ve integers and 450y = n^3 which of the following is an integer?

y/3*2^2*5
y/3^2*2*5
y/3*2*5^2
y/3*2*5
y/3^3*2^2*5

Can anybody please help me to really understand this sum.....
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by sunilrawat » Fri Jun 10, 2011 9:22 am
divya23 wrote:N and Y are +ve integers and 450y = n^3 which of the following is an integer?

y/3*2^2*5
y/3^2*2*5
y/3*2*5^2
y/3*2*5
y/3^3*2^2*5

[spoiler]oa = y/3*2^2*5[/spoiler]
3*3*5*5*2*y = n^3
to make n a perfect cube, y should be 3*5*2^2
hence option A
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by GMATGuruNY » Fri Jun 10, 2011 9:35 am
If n and y are positive integers and 450y = n³, which of the following must be an integer?

I. y/(3 x 2² x 5)

II. y/(3² x 2 x 5)

III. y/(3 x 2 x 5²)

a. None
b. I only
c. II only
d. III only
e. I, II, and III
450y is the cube of an integer.

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor:
8 is the cube of an integer because 8 = 2³ = 2*2*2.
27 is the cube of an integer because 27 = 3³ = 3*3*3.

Thus, when we prime-factorize 450y, we need to get at least 3 of every prime factor.
Here's the prime-factorization of 450y:
450y = 2 * 3² * 5² * y

Since 450 provides only one 2, two 3's, and two 5's, and we need 3 of every prime factor, the missing prime factors must be provided by y.
Thus, y must provide two more 2's, one more 3, and one more 5.
Thus, the smallest possible value of y = 2² * 3 * 5.

Onto the answer choices:

I. y/(3 x 2² x 5)
(2² * 3 * 5)/(3 x 2² x 5) = 1. The smallest possible value of y yields an integer.
Eliminate every answer choice that does not include I.
Eliminate A, C and D.

II. y/(3² x 2 x 5)
(2² * 3 * 5)/(3² x 2² x 5) = 1/3. Not an integer.
Eliminate every remaining answer choice that includes II.
Eliminate E.

The correct answer is B.
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by amit2k9 » Fri Jun 17, 2011 6:14 am
450 = 2* 3^2 * 5^2

thus 3*2^2*5 is required = 60

y/60 = integer also y/30 too.

A and D
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