If m, n, and p are integers such that m^2 n + n^2 p + p^2 m + m n p is an odd integer, whi… — GMAT

If $m$ , $n$ , and $p$ are integers such that $m^2 n + n^2 p + p^2 m + m n p$ is an odd integer, which of the following expressions MUST be an even integer?

A
$m n + n p + p m$
B
$m + n + p + 1$
C
$m n p + 1$
D
$m^2 + n^2 + p^2 + 1$
$m + n + p$ Cevap

Cevap

The sum of the three variables,

m + n + p

, MUST be an even integer.

Analyzing the expression

m^2 n + n^2 p + p^2 m + m n p

modulo 2 reveals that

x^2

has the same parity as

x

. Substituting modulo 2 yields

m n + n p + p m + m n p

. Testing all combinations of parity for

m, n, p

shows this expression is odd if and only if exactly two of the three variables are odd and one is even. The sum of two odd integers and one even integer (

m + n + p

) is always even, so

m + n + p

must be an even integer.

Adım Adım Çözüm

Analyze the parity of the given expression modulo 2.

Since

x^2 \equiv x \pmod 2

for any integer

x

, the given expression

m^2 n + n^2 p + p^2 m + m n p \equiv m n + n p + p m + m n p \pmod 2

Squaring an integer does not change its parity.

Determine which parity combinations of

m

n

, and

p

result in an odd value.

Evaluating

m n + n p + p m + m n p \pmod 2

across all possible parity combinations shows that the expression is odd if and only if exactly two of the variables are odd and exactly one variable is even.

If all three are even or one is odd and two are even, the expression equals 0 (even). If all three are odd,

1 + 1 + 1 + 1 = 4 \equiv 0

(even). Only when exactly two variables are odd (e.g., 1, 1, 0) does

1\cdot 1 + 1\cdot 0 + 0\cdot 1 + 1\cdot 1\cdot 0 = 1

(odd).

Test the parity of

m + n + p

under the condition that exactly two variables are odd and one is even.

\text{odd} + \text{odd} + \text{even} = \text{even} + \text{even} = \text{even}

Adding two odd integers produces an even integer, and adding an even integer keeps the sum even.

Anahtar Kavram

Odd and Even Integers (Parity)

Tahmini Süre:2m 0s

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