6 6 votes If $a$ and $b$ are integers and $a-b$ is even, which of the following must always be even? $ab$ $a^{2}+b^{2}+1$ $a^{2}+b+1$ $ab-b$ Quantitative Aptitude gate2017-me-2 general-aptitude quantitative-aptitude number-theory + – ♦Arjun 547 views answer comment Share Follow Add Sync Questions 0 reply Please log in or register to add a comment.
Best answer 10 10 votes 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 $(D).$ Lakshman Bhaiya answered Feb 5, 2020 • moved May 18 by Arjun Lakshman Bhaiya comment Share Follow 0 reply Please log in or register to add a comment.
3 3 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. m2n037 answered Feb 14, 2018 • moved May 18 by Arjun m2n037 comment Share Follow 0 reply Please log in or register to add a comment.
3 3 votes The possible cases of a-b even is when:- a and b is odd a and b is even so, substituting values of a=5 and b=3 or a=8 and b=4 and checking options we get answer as (d) option. Setika Mehra answered Nov 4, 2020 • moved May 18 by Arjun Setika Mehra comment Share Follow 0 reply Please log in or register to add a comment.
1 1 vote and following above procedure we can get answer as option D. aashish1406 answered Sep 13, 2023 • moved May 18 by Arjun aashish1406 comment Share Follow 0 reply Please log in or register to add a comment.