IF N and Y are positive integers and 450Y=N^3, which of the following is a integer
1)Y/ 3*2^2*5
2)Y/3^2*2*5
3)Y/3*2*5^2
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sumatitandon
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SOMETHING IS MISSING IN THE QUESTION.
All can be integer, if the question stands as it is!
450 = 2*3*3*5*5
450*Y = N^3
hence the value of Y can be (2^a*3^b*5^c) such that
(2*3^2*5^2)*(2^a*3^b*5^c) results in a cube of integer
that is = 1+a, 2+b, 2+c are all divisible by 3
MIN value of Y: a=2, b=1, c=1: 2^2*3*5
if question required min value of Y, then ans is 1st choice
or Y can be (a=5, 8,11.....), (b=4,7,10.......), (c=4,7,10.....)
in which case all could be integers
All can be integer, if the question stands as it is!
450 = 2*3*3*5*5
450*Y = N^3
hence the value of Y can be (2^a*3^b*5^c) such that
(2*3^2*5^2)*(2^a*3^b*5^c) results in a cube of integer
that is = 1+a, 2+b, 2+c are all divisible by 3
MIN value of Y: a=2, b=1, c=1: 2^2*3*5
if question required min value of Y, then ans is 1st choice
or Y can be (a=5, 8,11.....), (b=4,7,10.......), (c=4,7,10.....)
in which case all could be integers
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Brent@GMATPrepNow
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This one can be solved:sumatitandon wrote:IF N and Y are positive integers and 450Y=N^3, which of the following is a integer
1)Y/ 3*2^2*5
2)Y/3^2*2*5
3)Y/3*2*5^2
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 to be divisible by 3.
So, for example, 2*2*2*3*3*3*5*5*5 = (2*3*5)^3
For 2*3*3*5*5*Y to be a cube, it must be the case that Y has at least 2 additional 2's, 1 additional 3 and 1 additional 5.
So, Y = 2*2*3*5*(other possible numbers)
We see that 2*2*3*5 is the smallest possible value for Y.
Of the possible answers, only the first one (Y/ 3*2^2*5) will simplify to be an integer.
Brent Hanneson - Creator of GMATPrepNow.com
