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Difficulty: Very hardOdd and Even Integers (Parity)

If xx, yy, and zz are integers, is the expression x(y+z)x(y + z) an odd integer?

(1) x2+y2+z2x^2 + y^2 + z^2 is an odd integer.
(2) xy+yz+zxxy + yz + zx is an even integer.

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. C
    BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
  4. EACH statement ALONE is sufficient.Answer
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

EACH statement ALONE is sufficient.
Each statement independently guarantees that the expression x(y+z)x(y + z) must be an even integer. Because a definitive 'NO' answer is obtained from each statement independently, each statement alone is sufficient.

Step-by-Step Solution

1
Analyze the target expression x(y+z)x(y+z).
The expression x(y+z)x(y+z) is odd if and only if both xx is odd AND (y+z)(y+z) is odd. For (y+z)(y+z) to be odd, exactly one of yy or zz must be odd and the other even.
Establishing the precise condition for the expression to be odd determines what parity combinations are needed.
2
Evaluate Statement (1): x2+y2+z2x^2 + y^2 + z^2 is an odd integer.
Since k2k^2 has the same parity as kk, x+y+zx+y+z must be odd. This occurs in two parity distributions:
- Case 1: All three of x,y,zx, y, z are odd. Here, y+z=odd+odd=eveny+z = \text{odd} + \text{odd} = \text{even}, so x(y+z)=odd×even=evenx(y+z) = \text{odd} \times \text{even} = \text{even}.
- Case 2: One variable is odd and two are even.
- Subcase 2a: xx is odd, while yy and zz are even. Then y+z=even+even=eveny+z = \text{even} + \text{even} = \text{even}, so x(y+z)=odd×even=evenx(y+z) = \text{odd} \times \text{even} = \text{even}.
- Subcase 2b: xx is even, while one of y,zy,z is odd and the other is even. Then x(y+z)=even×odd=evenx(y+z) = \text{even} \times \text{odd} = \text{even}.
In all possible cases, x(y+z)x(y+z) is even. Thus, the answer to 'Is x(y+z)x(y+z) odd?' is a definitive NO. Statement (1) is sufficient.
Testing all valid parity distributions under Statement (1) shows x(y+z)x(y+z) can never be odd.
3
Evaluate Statement (2): xy+yz+zxxy + yz + zx is an even integer.
Note that x(y+z)=xy+zxx(y+z) = xy + zx. Therefore, xy+yz+zx=x(y+z)+yz=evenxy + yz + zx = x(y+z) + yz = \text{even}.
Suppose for contradiction that x(y+z)x(y+z) were odd. Then yzyz would also have to be odd (since odd+odd=even\text{odd} + \text{odd} = \text{even}). For yzyz to be odd, both yy and zz must be odd. But if yy and zz are both odd, then y+zy+z must be even, which forces x(y+z)=x×even=evenx(y+z) = x \times \text{even} = \text{even}, contradicting our assumption that x(y+z)x(y+z) is odd.
Thus, x(y+z)x(y+z) cannot be odd under Statement (2); it must be even. The answer is a definitive NO. Statement (2) is sufficient.
Proof by contradiction demonstrates that x(y+z)x(y+z) cannot be odd under Statement (2).

Key Concept

Parity rules of sums and products, including proof by case analysis and contradiction.
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Topics

Odd and Even Integers (Parity)

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