Question

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

• A. $100$
• B. $1000$
• C. $10000$
• D. None of the above

Answer 1

The correct option is C

$10000$

Explanation of the correct option.

Compute the required value:

Given : $({101}^{100}–1)$

It can be written as ${(1+100)}^{100}-1$

Since, ${(1+x)}^n={}^{n}C_0{x}^0+{}^{n}C_1{x}^1+{}^{n}C_2{x}^2+........+{}^{n}C_n{x}^n$

${(1+100)}^{100}–1={}^{100}C_0{\left(100\right)}^0+{}^{100}C_1{\left(100\right)}^1+.........+{}^{100}C_{100}{\left(100\right)}^{100}-1$

$=1+{}^{100}C_1{\left(100\right)}^1+..........+{}^{100}C_{100}{\left(100\right)}^{100}-1$

$=100\times 100+{}^{100}C_2{\left(100\right)}^2+.......+{}^{100}C_{100}{\left(100\right)}^{100}$

$={\left(100\right)}^2[1+{}^{100}C_2+{}^{100}C_{3}100+.......+{}^{100}C_{100}{100}^{98}]$

This is divisible by $10000$.

Thus it will be divisible by $100$ and $1000$ also.

Therefore, the greatest integer divides the number is $10000$.

Hence option $C$ is the correct answer.