Register

ISI2015-DCG-13

• recategorized by
363 views

1 Answer

0 votes
0 votes

if we take $n=4$ then $n^2+1=17$ which is not divisible by $3$ and $9$.

So option a and c are eliminated.

Option B. is correct option.

For proof refer this

 

User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

1 votes
1 votes
1 answer
1
gatecse asked Sep 18, 2019
403 views
ISI2015-DCG-6
The coefficient of $x^2$ in the product $(1+x)(1+2x)(1+3x) \dots (1+10x)$ is $1320$ $1420$ $1120$ None of these
The coefficient of $x^2$ in the product $(1+x)(1+2x)(1+3x) \dots (1+10x)$ is$1320$$1420$$1120$None of these
User Avatar
1 votes
1 votes
2 answers
2
gatecse asked Sep 18, 2019
534 views
ISI2015-DCG-8
Let $S=\{0, 1, 2, \cdots 25\}$ and $T=\{n \in S: n^2+3n+2$ is divisible by $6\}$. Then the number of elements in the set $T$ is $16$ $17$ $18$ $10$
Let $S=\{0, 1, 2, \cdots 25\}$ and $T=\{n \in S: n^2+3n+2$ is divisible by $6\}$. Then the number of elements in the set $T$ is$16$$17$$18$$10$
User Avatar
2 votes
2 votes
1 answer
3
gatecse asked Sep 18, 2019
505 views
ISI2015-DCG-9
Let $a$ be the $81$ – digit number of which all the digits are equal to $1$. Then the number $a$ is, divisible by $9$ but not divisible by $27$ divisible by $27$ but not divisible by $81$ divisible by $81$ None of the above
Let $a$ be the $81$ – digit number of which all the digits are equal to $1$. Then the number $a$ is,divisible by $9$ but not divisible by $27$divisible by $27$ but not ...
User Avatar
1 votes
1 votes
1 answer
4
gatecse asked Sep 18, 2019
512 views
ISI2015-DCG-12
The highest power of $3$ contained in $1000!$ is $198$ $891$ $498$ $292$
The highest power of $3$ contained in $1000!$ is$198$$891$$498$$292$
User Avatar
Link Copied!