If n and y are positive integers and 450y=n^3, which of the following must be an integer.
1) y/(3*2^2*5)
2) Y/(3^2*2*5)
3) Y/(3*2*5^2)
a) None
b) 1 only
c) 2 only
d) 3 only
e) 1, 2 and 3
integer?
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450 = 2x3²x5²vishal_2804 wrote:If n and y are positive integers and 450y=n^3, which of the following must be an integer.
1) y/(3*2^2*5)
2) Y/(3^2*2*5)
3) Y/(3*2*5^2)
As 450y is a cube of a positive integer, all the prime factors 450y must be present in triplets, i.e. powers of all the prime factors of 450y in its prime factorization must be a multiple of 3.
As, 450 contains one 2, two 3s, and two 5s, y must contain at least two 2, one 3, and one 5.
Hence, y must be a multiple of 2²x3x5.
So, y/(2²x3x5) must be an integer.
But the others may or may not be an integers.
The correct answer is B.
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If n and y are positive integers and 450y=n^3, which of the following must be an integer?
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
It almost always helps to find the prime factorization in these question types where we ask whether a certain rational expression is an integer.
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
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
It almost always helps to find the prime factorization in these question types where we ask whether a certain rational expression is an integer.
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