Question

The number $101$ ${}^{100}-1$ is divisible by:

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

Answer 1

Answer: (A), (B), (C)

The correct options are
A $100$
B $1000$
C $10000$

${101}^{100}-1$
$=(1+100{)}^{100}-1$
$=1+{}^{100}{C}_{1}100+{}^{100}{C}_2{100}^2+{}^{100}{C}_3{100}^3...{}^{100}{C}_{100}{100}^{100}-1$
$={}^{100}{C}_{1}100+{}^{100}{C}_2{100}^2+{}^{100}{C}_3{100}^3...{}^{100}{C}_{100}{100}^{100}$
$=\left(100\right)\left(100\right)+{}^{100}{C}_2{100}^2+{}^{100}{C}_3{100}^3...{}^{100}{C}_{100}{100}^{100}$
$={10}^{4}k$
Hence the following binomial expansion is divisible by $100$, $1000$, $10000$