If n and y are positive integers and 450y = n^3 which of the following

Question

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

Bunuel · Accepted Answer

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

"Must be an integer" means for the lowest possible value of  y .

450y=n^3  -->  2*3^2*5^2*y=n^3 . As  n  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=n^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.

unceldolan · Answer

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

KarishmaB · Answer

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.

Bunuel · Answer

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

shrivastavarohit · Answer

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

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

shrouded1 · Answer

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)

rrsnathan · Answer

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.

rajatjain14 · Answer

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

KarishmaB · Answer

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.

Bunuel · Answer

Similar questions to practice: if-n-is-a-positive-integer-and-n-2-is-divisible-by-96-then-127364.html if-n-is-a-positive-integer-and-n-2-is-divisible-by-72-then-90523.html a-certain-clock-marks-every-hour-by-striking-a-number-of-tim-91750.html if-m-and-n-are-positive-integer-and-1800m-n3-what-is-108985.html if-x-and-y-are-positive-integers-and-180x-y-100413.html n-is-a-positive-integer-and-k-is-the-product-of-all-integer-104272.html if-x-is-a-positive-integer-and-x-2-is-divisible-by-32-then-88388.html if-n-and-y-are-positive-integers-and-450y-n-92562.html if-5400mn-k-4-where-m-n-and-k-are-positive-integers-109284.html if-n-and-y-are-positive-integers-and-450y-n-92562.html Hope this helps.

jakexix · Answer

Cant we take higher values of N and Y so that it satisfies the equation,the all 3 possiblities are correct and answer is E
eg:
y= 5^4*3^4*2^5
N=5^2*3^2*2^2

Bunuel · Answer

The question asks which of the following MUST be an integer, not COULD be an integer. If y = 2^2*3*5, which is the least possible value of y, then only B is true. Please re-read the thread and follow the links to similar questions to understand the concept better.

GMATinsight · Answer

450y = n^3 
i.e.  2*3^2*5^2*y = n^3

For a number to be perfect cube the powers of all its prime factors must be multiples of 3

i.e. We need minimum  2^2*3*5  in y in order to make the number on left hand side a perfect cube

i.e. y must be a multiple of  2^2*3*5    but we can't be certain of any higher powers of respective prime factors than what is minimum hence

Answer: Option B

varshas044 · Answer

Here is what I think:

Given that 450*y = n^3 -------------- y = n^3 / 450 
450 by prime factorization = 2 * 3^2 * 5^2

Since y = integer and y > 0 then n^3 / 450 should give an integer -------- which means that n^3 should be completely divisible by 2 * 3^2 * 5^2 (=450)

This follows that the min value of n will be 2 * 3 * 5 ----------- meaning n^3 will be 2^3 * 3^3 * 5^3

Y = n^3 / 450 = (2^3 * 3^3 * 5^3) / 2 * 3^2 * 5^2
Y = 2^2 * 3 * 5

Now if you look at the options only option (i) will work.

Answer – B

achloes · Answer

For such problems, I've found it useful to remember that the only way we can get an integer from a given unknown fraction is if the denominator divides into the numerator fully. In other words, the numerator is REQUIRED to contain the ENTIRE prime factorization of the denominator.

Once this condition is met, the numerator is free to additional copies of the existing prime factors or even other new factors - more the merrier as it only increases the numerator's divisibility.

DanTheGMATMan · Answer

They love this problem. The exponent on each prime must be divisible by 3:

https://www.youtube.com/watch?v=P8vA6kfhHb0

PA0037 · Answer

Bunuel , the A option is giving us an integer - 2^2*3^1*5^1
so why is the answer B here ?

Bunuel · Answer

Option B is  I only .

orangebicycle5 · Answer

First factor 450 - 2 * 3^2 * 5^2 = n^3

In order for this to equal n^3, then the exponents need to be to a multiple of 3.

Thus, x = 2^2 * 3^1 * 5^1

Now, plug into the second part of the question. Only option 1 will create an integer as the other two options have an extra 3 and 5.

If n and y are positive integers and 450y = n^3 which of the following

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GMAT Problem Solving (PS) Questions

If n and y are positive integers and 450y = n^3 which of the following

Prep Toolkit

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