Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
What is the first value of $n$ for which $n^{2}+n+41$ is not a prime?
(a)1
(b)10
(c)20
(d)40
Answer
Answer (as printed): D
Explanation
$n=1 \Rightarrow\left(n^{2}+n+41\right)=(1+1+41)=43$, which is prime. $n=10 \Rightarrow\left(n^{2}+n+41\right)=(100+10+41)=151$, which is prime. $n=20 \Rightarrow\left(n^{2}+n+41\right)=(400+20+41)=461$, which is prime. $n=40 \Rightarrow\left(n^{2}+n+41\right)=(1600+40+41)=1681$, which is divisible by 41 . Thus, 1681 is not a prime number. Hence $n=40$ for which $\left(n^{2}+n+41\right)$ is not prime.
Explanation as extracted from the printed page; notation may be imperfect.