The greatest integer which divides the number | vedclass

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

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

Similar Questions

If the fractional part of the number $\frac{2^{403}}{15}$ is $\frac{k}{15}$,then $k$ is equal to

The numbers $a_n = 6^n - 5n$ for $n = 1, 2, 3, \ldots$ when divided by $25$ leave the remainder:

If $n \in N$,then ${7^{2n}} + {2^{3n - 3}} \cdot {3^{n - 1}}$ is always divisible by

The remainder when $10^{10} \cdot (10^{10} + 1) \cdot (10^{10} + 2)$ is divided by $6$ is

$7^{2n} + 16n - 1$ $(n \in N)$ is divisible by

Vedclass Test Series

Exam Paper Generator

Online Exam Module