1 1 vote The symbol $\mid$ reads as "divides", and $\nmid$ as "does not divide". For instance, $2 \: \mid \:6$ and $2 \: \nmid \: 5$ are both true. Consider the following statements. There exists a positive integer $a$ such that $(2 \mid (a^3 -1))$ and $( 2 \mid a)$. There exists a positive integer $b$ such that $6 \nmid (b^3 -b)$. What can you say about these statements? Only i is true Only ii is true Both i and ii are true Neither i nor ii is true Quantitative Aptitude cmi2016 quantitative-aptitude number-system + – go_editor 947 views answer comment Share Follow Print See 1 comment 1 1 comment reply chirudeepnamini commented Oct 19, 2019 reply Follow flag @Arjun sir, the question is mistyped. Statement 1 is "There exists a positive integer a such that (2 | ($a^{3}$ − 1)) and (2 | a)." https://www.cmi.ac.in/admissions/sample-qp/pgcs2016.pdf 0 0 replyShare Please log in or register to add a comment.
3 3 votes i. is False because 2 does not divides 1 ("and" is used so no need to check for 2 divides $(a^3-1)$) ii. $(b^3-b) = b(b^2-1) = b(b-1)(b+1) = (b-1)b(b+1)$ multiplication of three consecutive numbers means one of them must be even making whole multiplication even and also one of them must be divisible by 3 an even number divisible by 3 is also divisible by 6 which makes ii statement False so ans is option D Lokesh . answered Dec 30, 2016 Lokesh . comment Share Follow See 1 comment 1 1 comment reply DIBAKAR MAJEE commented Apr 18, 2020 reply Follow flag Option B true 0 0 replyShare Please log in or register to add a comment.
0 0 votes Answer: soujanyareddy13 answered May 6, 2021 soujanyareddy13 comment Share Follow 0 reply Please log in or register to add a comment.