# Difficult Number System Solved QuestionAptitude Discussion

 Q. $a, b, c, d$ and $e$ are five consecutive integers in increasing order of size. Which one of the following expressions is not odd?
 ✖ A. $ab + c$ ✖ B. $ab + d$ ✖ C. $ac + d$ ✔ D. $ac + e$

Solution:
Option(D) is correct

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.