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

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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

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Not a prime

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(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

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