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Zorluk: ZorEven-Odd Properties and Sign Rules

If aa, bb, and cc are non-zero integers such that a3b2c<0a^3 b^2 c < 0, a+ba + b is even, and b+cb + c is odd, which of the following expressions MUST be a negative even integer?

  1. ac(b2+1)a c (b^2 + 1)Cevap
  2. B
    abca b c
  3. C
    (a+c)b2(a + c) b^2
  4. D
    a2c(b+1)a^2 c (b + 1)
  5. E
    ac+ba c + b

Cevap

The expression ac(b2+1)a c (b^2 + 1) must be a negative even integer.
The expression ac(b2+1)a c (b^2 + 1) is guaranteed to be negative because ac<0a c < 0 (derived from a3b2c<0a^3 b^2 c < 0) and b2+1>0b^2 + 1 > 0 for all non-zero integers bb. It is guaranteed to be even because aa and cc have opposite parity, meaning one of them must be even, rendering aca c (and thus any integer multiple of aca c) even.

Adım Adım Çözüm

1
Analyze the sign constraint a3b2c<0a^3 b^2 c < 0.
ac<0a c < 0, which means aa and cc have opposite signs.
Since b0b \neq 0, b2b^2 is strictly positive. Dividing a3b2c<0a^3 b^2 c < 0 by b2b^2 yields a3c<0a^3 c < 0. Since a3a^3 has the same sign as aa, ac<0a c < 0.
2
Analyze the parity constraints a+ba + b is even and b+cb + c is odd.
aa and cc have opposite parity (one is even, the other is odd).
If a+ba + b is even, aa and bb share the same parity. If b+cb + c is odd, bb and cc have opposite parity. Substituting the parity of aa for bb shows that aa and cc must have opposite parity.
3
Determine the sign and parity of ac(b2+1)a c (b^2 + 1).
ac(b2+1)a c (b^2 + 1) is strictly negative and even.
Since aa and cc have opposite parity, at least one of them is even, making the product aca c an even integer. Since ac<0a c < 0 and b2+11>0b^2 + 1 \ge 1 > 0, the product of negative aca c and positive (b2+1)(b^2 + 1) is negative and even.

Anahtar Kavram

Combining sign rules (ac<0a c < 0) with even/odd parity logic across multiple variables.
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