**Choice (A):** $ab + c$:

At least one of every two consecutive positive integers $a$ and $b$ must be even.

Hence, the product $ab$ is an even number.

Now, if $c$ is odd (which happens when a is odd), $ab + c $ must be odd.

For example,

if $a = 3, b = 4$, and $c = 5,$ then $ab + c$ must equal $12 + 5 = 17$, an odd number. **Reject.**

**Choice (B):** $ab + d$:

We know that $ab$ being the product of two consecutive numbers must be even. Hence, if $d$ happens to be an odd number (it happens when $b$ is odd), then the sum $ab + d $ is also odd.

For example,

if $a = 4$, then $b = 5, c = 6$ and $ d = 7$, then $ ab + d = 3 \times 5 + 7 = 15 + 7 = 23$, an odd number. **Reject.**

**Choice (C):** $ac + d$:

Suppose $a$ is odd. Then $c$ must also be odd, being a number 2 more than $a$ .

Hence, $ac$ is the product of two odd numbers and must therefore be odd.

Now, $d$ is the integer following $c$ and must be even.

Hence, $ac + d = \text{odd + even = odd}$.

For example,

if $a = 3$, then $b = 3 + 1 = 4, c = 4 + 1 = 5$ (odd) and $d = 5 + 1 = 6$ (even) and $ac + d = 3×5 + 6 = 21$, an odd number.** Reject.**

**Choice (D):** $ac + e$:

Suppose $a$ is an odd number.

Then both $c$ and $e$ must also be odd.

Now, $ac$ is product of two odd numbers and therefore must be odd.

Summing this with another odd number $e$ yields an even number.

For example, if $ a = 1$, then $c$ must equal 3, and $e$ must equal 5 and $ac + e $ must equal $1×3 + 5 = 8$, an even number.

Now, suppose $a$ is an even number.

Then both $c$ and $e$ must also be even.

Hence, $ac + e =$

$\text{(product of two even numbers) + (an even number) =}$

$\text{(even number) + (even number) =an even number}$

For example,

if $ a = 2$, then $c$ must equal 4, and $e$ must equal 6 and the expression $ac + e $ equals 14, an even number. Hence, in any case, $ac + e$ is even. **Correct**.