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2 votes

If $a$ and $b$ are integers and $a-b$ is even, which of the following must always be even?

- $ab$
- $a^{2}+b^{2}+1$
- $a^{2}+b+1$
- $ab-b$

Migrated from GO Mechanical 3 years ago by Arjun

5 votes

Best answer

Given that $:a,b\in \mathbb{Z}$

$a-b = \text{even}$

- $\text{even} - \text{even}= \text{even}\:\color{Green}\checkmark{}$
- $\text{even} - \text{odd}= \text{odd}$
- $\text{odd} - \text{even}= \text{odd}$
- $\text{odd} - \text{odd}= \text{even}\:\color{Green}\checkmark{}$

Case$1:a=\text{even}, b = \text{even}\implies a-b=\text{even}$

- $\text{even} \times \text{even}= \text{even}$
- $\text{even} \times \text{odd}= \text{even}$
- $\text{odd} \times \text{even}= \text{even}$
- $\text{odd} \times \text{odd}= \text{odd}$

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

- $ab\:\color{Green}\checkmark{}$
- $a^{2}+b^{2}+1$
- $a^{2}+b+1$
- $ab−b\:\color{Green}\checkmark{}$

Case$2:a=\text{odd}, b = \text{odd}\implies a-b=\text{even}$

- $ab$
- $a^{2}+b^{2}+1$
- $a^{2}+b+1$
- $ab−b\:\color{Green}\checkmark{}$

So, the correct answer is $(D).$

2 votes

Since $a-b$ is even, this means

- both of them must be even or, ($a = 2n+k$ and $b=2n$ and where $k$ is even and $n$ might be even or odd so that $a-b=k$)
- both of them must be odd ($a=n$, and $b=n+k$ where $p$ is odd and $k$ is even so that $a-b=k$)

**Case I:** Both are even, say 4 and 6

A. $ab$ = 24

B. $a^2+b^2+1$ = 53

C. $a^2+b+1$ = 23

D. $ab−b$ = 18

**Case II:** Both are odd, say 3 and 7

Since options 2 and 3 are not true we will look at only options 1 and 4

A. $ab$ = 21

B. NA

C. NA

D. $ab−b$ = 14

So, $ab-b$, option $D$ is the answer.