Question

Last three digits in ${19}^{100}$ is

• A. $001$
• B. $010$
• C. $100$
• D. $999$

Answer 1

The correct option is A $001$

$(19{)}^{100}$

$=(20-1{)}^{100}$

$=20k+{}^{100}{C}_{98}\cdot {20}^2-{}^{100}{C}_{99}20+1$

$=20k+1980000-2000+1$

$=20k+1978001$

since the last tree digits of $20k$ will be zeros so the last three digits of $(19{)}^{100}$ will be same as $1978001$

So,he last three digits of $(19{)}^{100}$ are $001$