Is x an even integer? (1) x is the square of an integer. (2) x is the

Statement 1) and 2) by themselves are clearly insufficient, since we can find a host of numbers which can satisfy each of the 2 conditions.

Combining 1) and 2) , we know that x is both a square of an integer and cube of an integer.

Let us take x = 64 = 8^2 = 4^3 [Here x is even]

Let us take x = 1 = 1^2 = 1^3 [Here x is odd]

Clearly insufficient. Hence E) should be the answer.

i used trial error first x= 1 both square and cube of 1 in this case x is odd
x=64 square of 8 even cube of 4 in this case x is even so E.

A - being as square x can be either even or odd - 2^2 or 3^2 - not A

B - Being a cube x can be either even or odd - 2^3 or 3^3 - so not B

lets see if we can find using both - if x is both cube and square of a number - it should 6th power or some integer - can be either odd or even - ex - 2^6 or 3^6

finally we cannot determine whether x is even or odd

So answer is E­

Analyzing the question:
One property of powers, from observation, is that squaring or cubing an integer doesn't change its parity (whether it is odd or even). This is because an integer needs a factor of 2 to be even. An odd integer does not contain a factor of 2, so even if you multiply it to itself many times, it will never contain that 2. Therefore if x is even, x to any positive integer exponent will be even, and vice versa.

Then (1) and (2) are individually insufficient because the statements themselves do not allow you to determine whether x is odd or even. Combining them us x is some integer to the power of 6 but again we are left with the same problem. The answer is E.­

Do not forget the three Nos : -1,0,1

St 1- square of integer - possible values: 0,1 ( Yes and No ans)
St -2 -Cube of an integer - Possible values: 0,1 ( Yes and No ans)
1+2 Possible values : 0,1 ( Yes and No ans)

So E­

Statements alone are not sufficient if we combine just x^2=x^3
x=0 and x=1 that means we have two options.
Answer E

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