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

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

Question

(C) The number $S$ consists of $91$ ones,which can be written as a geometric series: $S = 1 + 10 + 10^2 + \dots + 10^{90} = \frac{10^{91} - 1}{10 - 1}

Vedclass · Accepted Answer

Not prime

Vedclass · Answer

(C) The number $S$ consists of $91$ ones,which can be written as a geometric series: $S = 1 + 10 + 10^2 + \dots + 10^{90} = \frac{10^{91} - 1}{10 - 1} = \frac{10^{91} - 1}{9}$. Since $91 = 7 	imes 13$,we can use the identity $x^n - 1 = (x^k - 1)(x^{n-k} + x^{n-2k} + \dots + 1)$ where $k$ is a divisor of $n$. Let $x = 10^{13}$,then $10^{91} - 1 = (10^{13})^7 - 1 = (10^{13} - 1)((10^{13})^6 + (10^{13})^5 + \dots + 1)$. Thus,$S = \frac{10^{13} - 1}{9} 	imes ((10^{13})^6 + (10^{13})^5 + \dots + 1)$. Since $S$ is the product of two integers greater than $1$,it is a composite number and therefore not prime.

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

Similar Questions

Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2\sum_{n=1}^{40}(P_{n})^{\frac{1}{n}}=\frac{3^{\alpha}-1}{3^{\beta}}$ and $\gcd(\alpha,\beta)=1$, then $\alpha+\beta$ is equal to

If $a_1, a_2, a_3, \ldots, a_{10}$ is a geometric progression and $\frac{a_3}{a_1}=25$,then $\frac{a_9}{a_5}$ equals

If the roots of the equation $x^3 - ax^2 + bx - c = 0$ are in $GP$,then $\frac{b^3}{a^3}$ is equal to:

The sum of the first two terms of a $G.P.$ is $1$ and every term of this series is twice its previous term. Then,the first term will be:

If the roots of the equation $8x^3 - 14x^2 + 7x - 1 = 0$ are in $G.P.$,then the roots are

Vedclass Test Series

Exam Paper Generator

Online Exam Module