Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
What is the remainder when $2^{31}$ is divided by 5?
(a)1
(b)2
(c)3
(d)4
Answer
Answer (as printed): C
Explanation
$2^{31}=2\times2^{30}=2\times(2^{2})^{15}=2\times4^{15}$. When $n$ is odd, $(x^{n}+a^{n})$ is divisible by $(x+a)$. $\therefore (4^{15}+1^{15})$ is divisible by $(4+1)$ $\Rightarrow (4^{15}+1)$ is divisible by 5 $\Rightarrow (2^{30}+1)$ is divisible by 5 $\Rightarrow$ On dividing $2^{30}$ by 5, we get $(5-1)$ i.e. 4 as remainder. $\therefore$ Remainder obtained on dividing $2^{31}$ by 5 = Remainder obtained on dividing $(2\times4)$ i.e. 8 by 5 $=3$.
Explanation as extracted from the printed page; notation may be imperfect.