infiniti007 wrote:If 375y = x^2 and x and y are positive integers, then which of the following must be an integer?
I.) y/15
II.) y/30
III.) y^2/25
A.) I only
B.) III only
C.) I and II only
D.) I and III only
E.) I, II, and III only
Test the SMALLEST POSSIBLE CASE.
y = x²/375 = x²/(3*5*5*5).
To MINIMIZE the value of y, we must MINIMIZE the value of x.
Since y must be an integer, x² must be divisible by (3*5*5*5).
The least possible value of x such that x² is divisible by one 3 and three 5's is x=3*5*5, with the result that x² = (3*5*5)² = 3*3*5*5*5*5.
Plugging x² = 3*3*5*5*5*5 into the equation above, we get:
Least possible y = (3*3*5*5*5*5)/(3*5*5*5) = 15.
Implication:
y must be a MULTIPLE OF 15.
Test y=15 in the statements.
I: y/15 = 15/15 = 1.
II: y/30 = 15/30 = 1/2.
III: y²/25 = 15²/25 = (3*3*5*5)/25 = 9.
Statement II does not have to be an integer.
Eliminate C and E.
Since y=15 yields integer values in I and III, any greater multiple of 15 will also yield integer values in I and III.
Thus, I and III must both be integers.
The correct answer is
D.
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