If n and y are positive integers and 450y=n raised to 3, which of the following must be an integer?
I.y/3*2raised to 2*5
II.y/3raised to 2*2*5
III.y/3*2*5raised2
A.None
B I only
C II only
D III only
E I,II, and III
GMATPrep Test 2 Q15
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Try to prove that I, II and III DON'T have to be integers.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
To this end, plug in the MINIMUM POSSIBLE VALUE for y.
450y = n³ implies that 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 at least 3 of every prime factor, the missing prime factors must be provided by y.
Thus, y must provide at at least two more 2's, one more 3, and one more 5.
Thus, the MINIMUM possible value of y = 2² * 3 * 5.
Plug y = 2² * 3 * 5 into 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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It often helps to find the prime factorization in these question types where we ask whether a certain rational expression is an integer.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 = n^3
2*3*3*5*5*y = n^3
For 2*3*3*5*5*y to be a cube, we need the number of 2's, 3's and 5's in the prime factorization to each be divisible by 3.
So, for example, 2*2*2*2*2*2*3*3*3*5*5*5 = (2*2*3*5)^3
For 2*3*3*5*5*y to be a cube, it must be the case that the prime factorization of y includes at least two additional 2's, one additional 3 and one additional 5.
So, y = 2*2*3*5*(other possible numbers)
Now check the option.
I. Must y/(3 * 2^2 * 5) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3 * 2^2 * 5)
= some integer
Since this must be an integer, we can eliminate A, C and D, which leaves us with B or E.
II. Must y/(3^2 * 2 * 5) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3^2 * 2 * 5)
= 2*(other possible numbers)/3
Not necessarily an integer
Since this need not be an integer, we can eliminate E, which leaves us with B.
NOTE: At this point we have the correct answer. But let's check III for "fun"
III. Must y/(3 * 2 * 5^2) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3 * 2 * 5^2)
= 2*(other possible numbers)/5
Not necessarily an integer
Answer: B
Cheers,
Brent