Question

Difficulty: HardEven-Odd Properties and Sign Rules

Let xx and yy be non-zero integers that satisfy all of the following conditions:
I. (x)y+1<0(-x)^{y + 1} < 0
II. x2y+xx^2 y + x is odd
III. x<yx < y

Which of the following statements MUST be true? Select all that apply.

  1. yxy^x is an even integerAnswer
  2. xyyx^y - y is an odd integerAnswer
  3. C
    x<y-x < -y
  4. (x+y)2(x + y)^2 is an odd integerAnswer
  5. E
    (x)y<0(-x)^y < 0

Answer

The statements that must be true are 'yxy^x is an even integer', 'xyyx^y - y is an odd integer', and '(x+y)2(x + y)^2 is an odd integer'.
From the given conditions, we deduce that xx is a positive odd integer and yy is a positive even integer with x<yx < y. Therefore:
- Raising the positive even integer yy to the positive power xx yields an even integer (yxy^x is even).
- Raising the odd integer xx to the positive power yy yields an odd integer, and subtracting the even integer yy leaves an odd integer (xyyx^y - y is odd).
- The sum of odd xx and even yy is odd, and squaring an odd integer yields an odd integer ((x+y)2(x + y)^2 is odd).

Step-by-Step Solution

1
Analyze Condition I: (x)y+1<0(-x)^{y + 1} < 0
x>0x > 0 (a positive integer) and yy is an even integer.
For a number raised to an integer power to be strictly negative, the base must be negative and the exponent must be odd. Hence, x<0    x>0-x < 0 \implies x > 0, and y+1y + 1 is odd     y\implies y is even.
2
Analyze Condition II: x2y+xx^2 y + x is odd
xx is an odd integer.
Factor the expression as x(xy+1)x(xy + 1). A product of two integers is odd if and only if both factors are odd. Thus, xx must be odd. (Additionally, xy+1xy + 1 must be odd     xy\implies xy is even, which holds since yy is even).
3
Analyze Condition III: x<yx < y
yy is a positive even integer.
Since xx is a positive odd integer (x1x \ge 1) and x<yx < y, yy must also be a positive integer (y2y \ge 2).
4
Evaluate the statements based on the derived properties (xx is positive odd, yy is positive even, x<yx < y)
The statements 'yxy^x is an even integer', 'xyyx^y - y is an odd integer', and '(x+y)2(x + y)^2 is an odd integer' are guaranteed to be true.
1) evenpositive odd=even\text{even}^{\text{positive odd}} = \text{even}. 2) oddpositive eveneven=oddeven=odd\text{odd}^{\text{positive even}} - \text{even} = \text{odd} - \text{even} = \text{odd}. 3) (odd+even)2=odd2=odd(\text{odd} + \text{even})^2 = \text{odd}^2 = \text{odd}.

Key Concept

Deducing parity and signs of variables using exponent rules and arithmetic properties of even and odd numbers.
Estimated Time:2m 0s
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