The greatest integer which divides the number | vedclass

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Question

(C) Using the Binomial Theorem,we expand $(1 + 100)^{100}$:$(1 + 100)^{100} = 1 + \binom{100}{1}(100) + \binom{100}{2}(100)^2 + \binom{100}{3}(100)^3

Vedclass · Accepted Answer

$10000$

Vedclass · Answer

(C) Using the Binomial Theorem,we expand $(1 + 100)^{100}$:$(1 + 100)^{100} = 1 + \binom{100}{1}(100) + \binom{100}{2}(100)^2 + \binom{100}{3}(100)^3 + \dots$$(1 + 100)^{100} = 1 + 100(100) + \frac{100 	imes 99}{2}(100)^2 + \dots$$(1 + 100)^{100} = 1 + 10000 + \frac{9900}{2}(10000) + \dots$Subtracting $1$ from both sides:$101^{100} - 1 = 10000 + 4950(10000) + \dots$$101^{100} - 1 = 10000(1 + 4950 + \dots)$Thus,the number $101^{100} - 1$ is divisible by $10000$.

The greatest integer which divides the number $101^{100} - 1$ is

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