Question 11 Mark
The last two digits of the numbers $3^{400}$ are $01.$
Answer
View full question & answer→True.
Given that $3^{400}=(9)^{200}=(10-1)^{200}$
$\therefore(10-1)^{200}=\ ^{200}\text{C}_0(10)^{200}-\ ^{200}\text{C}_1(10)^{199}+...$
$-\ ^{200}\text{C}_{199}(10)^1+\ ^{200}\text{C}_{200}(1)^{200}$
$=10^{200}-200\times10^{199}+...-10\times200+1$
So$,$ it is clear that last two digits are $01.$
Hence$,$ the given statement is True.
Given that $3^{400}=(9)^{200}=(10-1)^{200}$
$\therefore(10-1)^{200}=\ ^{200}\text{C}_0(10)^{200}-\ ^{200}\text{C}_1(10)^{199}+...$
$-\ ^{200}\text{C}_{199}(10)^1+\ ^{200}\text{C}_{200}(1)^{200}$
$=10^{200}-200\times10^{199}+...-10\times200+1$
So$,$ it is clear that last two digits are $01.$
Hence$,$ the given statement is True.