Question

The greater integer which divide the number ${101}^{100}-1$ is

• A. $100$
• B. $1000$
• C. $10000$
• D. $100000$

Answer 1

The correct option is C $10000$

From binomial expansion, we have ${(1+x)}^n=[1+nx+\frac{n(n-1)}{2}.x^2....x^n]$

Substitute $x=n$, we get

${(1+n)}^n=[1+nn+\frac{n(n-1)}{2}.n^2....n^n]$

or ${(1+n)}^n-1=[nn+\frac{n(n-1)}{2}.n^2....n^n]$

or ${(1+n)}^n-1=n^2[1+\frac{n(n-1)}{2}.....n^{n-2}]$

Put $n=100$

${(1+100)}^{100}-1={100}^2[1+\frac{100(100-1)}{2}.....{100}^{98}]$

${\left(101\right)}^{100}-1={100}^2[1+displaystyle\frac{100(100-1)}{2}.....{100}^{98}]$

Clearly ${\left(101\right)}^{100}-1$ is divisible by 10000.