A) 19 |
B) 20 |
C) 3 |
D) 1 |
D) 1 |
We have = \( \Large \frac{19^{100}}{20} \)
=> \( \Large \frac{(20-1)^{100}}{20} \)
=> \( \Large \frac{(-1)^100}{20} \)=>\( \Large \frac{1}{20} \)
Remainder = 1
Required remainder = \( \Large 19^{100} \)
= \( \Large (-1)^{100} \)=1
[ 20 = 19 + 1 so 19 replace by (-1)]