Question

$111.....1(91$ times$)$ is a

• A. prime number
• B. composite number
• C. not a integer
• D. integer

Answer 1

Answer: (B), (D)

composite number
integer

We have
$1111+.....1\left(91times\right)=1+10+{10}^2+.....+{10}^{90}$
$=\frac{{10}^{91}-1}{10-1}=\frac{({10}^{91}-1)}{({10}^7-1)}.\frac{\left({10}^7\right)-1}{(10-1)}=\frac{({10}^7{)}^{13}-1}{{10}^7-1}.\frac{{10}^7-1}{(10-1)}$
$=({10}^{84}+{10}^{77}+{10}^{70}+.....+{10}^7+1)({10}^6+{10}^5+{10}^4+{10}^3+{10}^2+10+1)$
$=$ product of two integers $(>2)$. so it is not a prime number.