SAT Numbers

Correct Answer: B

Solution 1:

Tip: Replace variables with numbers.
(A) If $x=1$ and $y=2$ , $x+y = 3 =\text{odd}$ , and $(x+y)^{2}+xy =(1+2)^{2}+1\cdot 2 = 3^{2} + 2 = 9+2=11$ , which is odd. Eliminate (A).

(C) If $x=2$ and $y=1$ , $x+y = 3=\text{odd}$ ,and $(x+y)^{2}+xy =(2+1)^{2}+2\cdot 1 = 3^{2} + 2 = 9+2=11$ , which is odd. Eliminate (C).

(D) If $x=1$ and $y=2$ , $x+y =3= \text{odd}$ , and $(x+y)^{2}+xy =(1+2)^{2}+1\cdot 2 = 3^{2} + 2 = 9+2=11$ , which is odd. Eliminate (D).

(E) If $x=-2$ and $y=1, xy = -2.$ Then, $x+y = -2 + 1 = -1 = \text{odd},$ and $(x+y)^{2} + xy = (-1)^{2} + (-2) = 1-2=-1,$ which is odd. Eliminate choice E.

We eliminated all choices except (B), which is the correct answer.

Solution 2:

Tip: Know the properties of even and odd numbers.
If $x+y$ is odd, then $(x+y)^{2} = \underbrace{(x+y)}_{\text{odd}} \times \underbrace{(x+y)}_{\text{odd}} = \text{odd} \times \text{odd} = \text{odd}$ .

$x+y$ is odd only if: $x$ is even and $y$ is odd, or if $x$ is odd and $y$ is even. In either case, $xy$ will be even, because even $\times$ odd $=$ even.

So, $\underbrace{(x+y)^{2}}_{\text{odd}}+\underbrace{xy}_{\text{even}}=$ odd $+$ even $=$ odd.

Therefore, answer choice (B) is correct.

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Incorrect Choices:

(A) , (C) , (D) , and (E)
See Solution 1 for how we can eliminate these choices by replacing the variable with a number.

Let $E$ be a even number and $O$ be a odd number. We know that:

$E + E = E$

$O + O = E$

$E + O = O$

$O + E = O$

Then:

$(x + y)^{2} = O$

And:

$xy = E$

Finally we have that:

$O + E = O$

The answer is $B$

x+y is only odd if and only if x or y is odd (not both). Square of an odd number is odd while product of even and odd number is even. Furthermore, sum of an odd and even number is odd. So, the answer must be an odd number.

If $x + y$ is an odd number, then it means one of them is odd and one of them is even. Now let's see the equation $(x+y)^{2} + xy$ ... $(x+y)^2$ means that an odd number is multiplied by itself, which means an odd number is multiplied by odd. An odd number multiplied by another odd number gives a product of an odd number. $xy$ would say that an odd number is multiplied by an even number, which would give a product of an even number.

Overall, if you add an odd number with an even number, you would get an odd number, so the answer is $\boxed{B}$ .

since the sum of the numbers is odd that means both of them must have different parities. Since they have different parities their product must be even. The final number formed is the sum of an odd(square of an odd number is odd) and an even number therefore it must be an odd number hence B

Assume x and y are integers and x+y is odd. Then by definition: (x+y) = 2n+1 for some n where n is an integer. Then (x+y)^2 = (2n+1)(2n+1)=4n^2+4n+1

Now with a quick little move we get: 2(2n^2+2n)+1 which is odd, since it is a number of the form 2m+1, (m = 2n^2+2n where m is an integer). Now, I don't even have to worry about xy being even or odd, because for (x+y)^2 + xy to be even, then it must be divisible by 2.

In other words: (x+y)^2+xy = 2c where c is an integer, which implies (x+y)^2 and xy must both be divisible by 2.

But we have already shown that (x+y)^2 is odd, therefore not divisible by 2.

QED

Note: xy has to be even, and in fact, one of x or y has to be odd. Just run through the three options:

1) x and y could both be even. (Contradicts x+y is odd.) 2) x and y could both be odd. (Contradicts x+y is odd.) 3) One of x or y is odd. Let's say x is even, y is odd. Doesn't contradict x+y odd. And since x =2n where n is an integer, then xy = 2ny which is a number of the form 2m where m is an integer. (Here: m=ny).