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Zorluk: ZorOdd and Even Integers (Parity)

If xx, yy, and zz are integers such that (x+1)(y+2)(x + 1)(y + 2) is an odd integer and y(z+3)y(z + 3) is an even integer, which of the following expressions MUST be an even integer?

  1. x+y+zx + y + zCevap
  2. B
    xy+zxy + z
  3. C
    (y+z)2+1(y + z)^2 + 1
  4. D
    xz+yxz + y
  5. E
    (x+1)yz(x + 1)yz

Cevap

The expression x+y+zx + y + z MUST be an even integer.
The condition (x+1)(y+2)=odd(x + 1)(y + 2) = \text{odd} implies both (x+1)(x + 1) and (y+2)(y + 2) are odd, so xx is even and yy is odd. Then y(z+3)=eveny(z + 3) = \text{even} with yy being odd requires (z+3)(z + 3) to be even, so zz is odd. Summing an even integer xx and two odd integers yy and zz yields even+odd+odd=even\text{even} + \text{odd} + \text{odd} = \text{even}. Thus, the expression representing the sum of xx, yy, and zz must always be even.

Adım Adım Çözüm

1
Determine the parity of xx and yy from the first given condition.
xx is even and yy is odd.
The product (x+1)(y+2)(x + 1)(y + 2) is odd, which means both factors (x+1)(x + 1) and (y+2)(y + 2) must be odd. If x+1x + 1 is odd, then xx is even. If y+2y + 2 is odd, then yy is odd.
2
Determine the parity of zz using the parity of yy and the second given condition.
zz is odd.
The product y(z+3)y(z + 3) is even. Since yy is odd, (z+3)(z + 3) must be even for the product to be even. If z+3z + 3 is even, then zz must be odd.
3
Evaluate the parity of x+y+zx + y + z.
x+y+zx + y + z is even.
Substituting the parities gives even+odd+odd=even+even=even\text{even} + \text{odd} + \text{odd} = \text{even} + \text{even} = \text{even}.

Anahtar Kavram

Parity rules for addition and multiplication of integers
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