Quote:
Note: a perfect square, is an integer that can be written as the square of some other integer. For example 16=4216=42, is a perfect square. Similarly, a perfect cube is an integer that can be written as the cube of some other integer. For example, 27=3327=33 is a perfect cube.
Make prime factorization of 2,700: xyz=2,700=22∗33∗52xyz=2,700=22∗33∗52.
(1) yy is an even perfect square and zz is an odd perfect cube. If yy is either 2222 or 22∗5222∗52 and z=33=odd perfect squarez=33=odd perfect square then xx must be a perfect square which makes x√x an integer: x=52x=52 or x=1x=1. But if z=13=odd perfect cubez=13=odd perfect cube then xx could be 3333 which makes x√x not an integer. Not sufficient.
(2) z√z is not an integer. Clearly insufficient.
(1)+(2) As from (1) z√≠integerz≠integer then z≠1z≠1, therefore it must be 3333 (from 1), so xx must be a perfect square which makes x√x an integer: x=52x=52 or x=1x=1. Sufficient.
Answer: C
Hi Bunuel,
I need your help on this one please.
in S1: it says
y = perfect sq, therefore it can have values such as 1, 4, 25 or 100
z = odd perfect cube, therefore it can have values such as 1 or 27.
and therefore S1 is NS.
in S2: it says
Under the root z is not an integer. Therefore it gives us that z is not equal to 1 and it is equal to 27.
NS.
Combining S1 and S2, we get z = 27 but for y still the values are 1, 4, 25 or 100.
how can we say that combining the equation gives us the answer?
Thank you.