Question

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

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

Answer 1

Answer: (C)

By Binomial theorem
$(1+x{)}^n=\left[1+nx+\frac{n(n-1)}{2}\cdot x^2\dots +x^n\right]$
or $(1+x{)}^n-1=nx+\frac{n(n-1)}{2}{x}^2\dots +x^n$
If $x=n,(1+n{)}^n-1=n^2+\frac{n(n-1)}{2}{n}^2\dots n^n$
$(1+n{)}^n-1=n^2\left[1+\frac{n(n-1)}{2}\dots +n^{n-2}\right]$
Put $n=100$,
$(1+100{)}^{100}-1=(100{)}^2\left[1+\frac{100(100-1)}{2}\dots +100^{98}\right]$
$(101{)}^{100}-1=(100{)}^2\left[1+\frac{100\times 99}{2}\dots +100^{98}\right]$
Clearly $(101{)}^{100}-1$ is divisible by
$(100{)}^2=10000$