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