The number 111...1 (91 times) is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $N = 111...1$ ($91$ times).$N = \sum_{k=0}^{90} 10^k = \frac{10^{91}-1}{10-1} = \frac{10^{91}-1}{9}$.Since $91 = 7 	imes 13$,we can write $10

Vedclass · Accepted Answer

Not a prime

Vedclass · Answer

(A) Let $N = 111...1$ ($91$ times).$N = \sum_{k=0}^{90} 10^k = \frac{10^{91}-1}{10-1} = \frac{10^{91}-1}{9}$.Since $91 = 7 	imes 13$,we can write $10^{91}-1 = (10^7)^{13}-1$.Using the algebraic identity $x^n-1 = (x-1)(x^{n-1} + x^{n-2} + ... + 1)$,we have:$10^{91}-1 = (10^7-1)((10^7)^{12} + (10^7)^{11} + ... + 1)$.Thus,$N = \frac{10^7-1}{9} 	imes ((10^7)^{12} + (10^7)^{11} + ... + 1)$.Since $\frac{10^7-1}{9} = 1111111$ and the second factor is clearly greater than $1$,$N$ is a product of two integers greater than $1$.Therefore,$N$ is a composite number,which means it is not a prime number.

The number $111...1$ ($91$ times) is

Similar Questions

The coefficient of $x^{10}$ in the expansion of $(1+x^2-x^3)^8$ is

The number of terms in the expansion of $(x+y+z)^{10}$ is

The coefficient of $a^{3} b^{4} c^{5}$ in the expansion of $(bc + ca + ab)^{6}$ is

Among the positive divisors of the number $12600$,if $n_1$ is the number of divisors which are multiples of $3$ and $n_2$ is the number of divisors which are multiples of $14$,then $n_1 + n_2 =$

The number of sequences of ten terms,whose terms are either $0$,$1$,or $2$,that contain exactly five $1$s,exactly three $2$s,and two $0$s,is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module