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

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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

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