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# If x, y and z are integers, is x + y + 2z even?

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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]

basically x+y+z+z = even + even = even

i. not suff as x + z = even = not sure about any of x and y being even or odd

ii. same as above..
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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]
If x, y and z are integers, is x + y + 2z even?
(1) x + z is even
(2) y + z is even
x+y+2z------ 2Z is automatically even, because any number odd or even multiplied by even gives us even.
statement 1: x+z is even even+even=even , odd+odd= even so we are not sure about x, because for z it is not important to know even or odd.
Statement 2: y+z is even either both are even or odd, so we are not sure about y here.
Combining the 2 statements we still do not know if x and y are odd or even.
so x + y + 2z let's take case by case
E+O+E=O
O+E+E=O
E+E+E=E
O+O+E=E
Please if i am wrong correct me . thank you
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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]
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wardalaknaoui1 wrote:
If x, y and z are integers, is x + y + 2z even?
(1) x + z is even
(2) y + z is even
x+y+2z------ 2Z is automatically even, because any number odd or even multiplied by even gives us even.
statement 1: x+z is even even+even=even , odd+odd= even so we are not sure about x, because for z it is not important to know even or odd.
Statement 2: y+z is even either both are even or odd, so we are not sure about y here.
Combining the 2 statements we still do not know if x and y are odd or even.
so x + y + 2z let's take case by case
E+O+E=O
O+E+E=O
E+E+E=E
O+O+E=E
Please if i am wrong correct me . thank you

Hello

I think if you take the statements together properly, you will see that x/z are either both even or both odd. Similarly y/z are both even or both odd (you yourself have explained this)
Now if z is even, it means both x/y are also even, then x+y+2z will be E+E+E = even
And if z is odd, it means both x/y are also odd, then x+y+2z will be O+O+E = even

So it has to be even.

Or you could simply add the two: x+z is even and y+z is also even; Add the two to get x+z + y+z = E + E
Or x+y+2z = Even
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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]
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Hello wardalaknaoui1,

When we combine both statements, we will get two even numbers. When two even numbers are added we always get another even number.
So, (x+z)+(y+z) = even + even = even = x+y+2z

Question is asking whether x+y+2z is even or not and we got our answer when we combine both the statements.

Hope it helps.
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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]
Bunuel wrote:
If x, y and z are integers, is x + y + 2z even?

(1) x + z is even

(2) y + z is even

2z will always be even, so the question basically asks us whether x+y is even.
Now I does not deal in y and II does not deal in x, so don't waste time looking for A and B as answer by taking each statement individually.

Combined..
x+z and y+z, both are even, so their SUM should also be even..
So, x+z+y+z or x+y+2z is even
Suff

C
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If x, y and z are integers, is x + y + 2z even? [#permalink]
Question: Is x + y + 2z = even?

For the above equation to hold true, x+y would have to be even. x+y can only be even if both x and y are even or both are odd. Notice that z itself can be either even or odd, we just know that the 2z term will always be even.

Stmt 1: x + z = even. 2 cases: either x and z are both even or both are odd. In either case, we don't know what y will be so we can't figure out if x + y is even or odd. Not sufficient.

Stmt 2: y + z = even. Using the same logic as stmt 1, we don't know anything about x so we can't figure out if x+y is even or odd. Not sufficient

Together: add both statements: x + z + y + z = even because (x+z) is even and (y+z) is even. So together they will be even.
Adding it yields: x + y + 2z = even. This directly answers the question in the stem.
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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]
Step 1: Analyse Question Stem

Any number of the form 2n, where n is an integer, is always even.

The question clearly says that x, y and z are integers. Therefore, the term 2z is definitely even. This can be used to break down the question stem and rephrase it.

Is x + y + 2z = even?

Since 2z is even, question can be rephrased as, Is x + y + even = even?

By moving the even term from the LHS to the RHS, this can further be rephrased as, Is x + y = even – even?

Since even – even = even, the question now becomes, Is x + y = even?

Essentially, we are trying to find out if x and y are both odd OR both even.

Step 2: Analyse Statements Independently (And eliminate options) – AD / BCE

Statement 1: x + z is even

This means that x and z are both odd OR both even. Statement 1 does not provide any information about y.

The data in statement 1 is insufficient to find if x and y are of the same nature.
Statement 1 alone is insufficient. Answer options A and D can be eliminated.

Statement 2: y + z is even

This means that y and z are both odd OR both even. Statement 2 does not provide any information about x.

The data in statement 2 is insufficient to find if x and y are of the same nature.
Statement 2 alone is insufficient. Answer option B can be eliminated.

Step 3: Analyse Statements by combining

From statement 1: Both x and z are of the same nature.
From statement 2: Both y and z are of the same nature.
From this, it is clear that both x and y are of the same nature i.e. they are both odd or both even.

The combination of statements is sufficient to answer the question, Is x + y = even, with a definite YES.
Statements 1 and 2 together are sufficient. Answer option E can be eliminated.

The correct answer option is C.
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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]
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Re: If x, y and z are integers, is x + y + 2z even? [#permalink]
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