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

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(C) Let $S$ be the number consisting of $91$ ones. This can be written as a geometric series: $S = 1 + 10 + 10^2 + \dots + 10^{90}$.Using the sum form

Vedclass · Accepted Answer

Not prime

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(C) Let $S$ be the number consisting of $91$ ones. This can be written as a geometric series: $S = 1 + 10 + 10^2 + \dots + 10^{90}$.Using the sum formula for a geometric progression,$S = \frac{10^{91} - 1}{10 - 1} = \frac{10^{91} - 1}{9}$.Since $91 = 7 	imes 13$,we can write $10^{91} - 1 = (10^{13})^7 - 1$.Using the algebraic identity $x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + \dots + 1)$,we have $(10^{13})^7 - 1 = (10^{13} - 1)( (10^{13})^6 + (10^{13})^5 + \dots + 1)$.Thus,$S = \frac{10^{13} - 1}{9} 	imes (10^{78} + 10^{65} + \dots + 1)$.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:

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