If a, b, and c are integers such that a^2b + b^2c + c^2a is an odd integer, then the produ… — GMAT

If $a$ , $b$ , and $c$ are integers such that $a^2b + b^2c + c^2a$ is an odd integer, then the product $(a+b)(b+c)(c+a)$ must be an even integer.

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The statement is true because the sum

(a+b) + (b+c) + (c+a) = 2(a+b+c)

is always an even integer, which mathematically prevents all three factors from being odd simultaneously.

The statement is true. The sum of the three factors

(a+b) + (b+c) + (c+a) = 2(a+b+c)

is guaranteed to be an even integer. If the product

(a+b)(b+c)(c+a)

were odd, each individual factor would have to be odd. But the sum of three odd integers is always odd, contradicting the fact that

2(a+b+c)

is even.

Adım Adım Çözüm

Analyze the parity of

a^2b + b^2c + c^2a

using parity rules for exponents.

Since

x^2

has the same parity as

x

for any integer

x

a^2b + b^2c + c^2a

has the same parity as

ab + bc + ca

. Thus,

ab + bc + ca

is an odd integer.

Squaring an integer preserves its parity.

Determine possible parity combinations for

a

b

, and

c

For

ab + bc + ca

to be odd, either exactly two of

\{a, b, c\}

are odd, or all three are odd.

If 0 or 1 variable is odd, every product term (

ab, bc, ca

) is even, making the total sum even.

Evaluate the parity of the factors

(a+b)

(b+c)

, and

(c+a)

If two variables are odd and one is even, the sum of the two odd variables produces an even factor. If all three variables are odd, the sum of any two of them produces an even factor (all three factors become even).

The sum of two odd integers is always even.

Determine the parity of the product

(a+b)(b+c)(c+a)

Because at least one factor in the product is even under all valid scenarios, the product

(a+b)(b+c)(c+a)

must be an even integer.

Any product of integers containing at least one even factor is even.

Anahtar Kavram

Parity preservation under exponents and algebraic parity constraints of sums and products of integers.

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