Question

The last two digits of the number $3^{400}$ are $01$. State true or false.

Answer 1

We can write,

$3^{400}=(3^2{)}^{200}$

$3^{400}=9^{200}=(-1+10{)}^{200}$

Using $(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x+.....{+}^n{C}_n{x}^n$

$\Rightarrow 3^{400}=(-1+10{)}^{200}$

$\Rightarrow 3^{400}{=}^{200}{C}_0{+}^{200}{C}_1(-1)10{+}^{200}{C}_2{10}^2-\dots {+}^{200}{C}_{200}{10}^{200}$

$\Rightarrow 3^{400}=1-200\left(10\right){+}^{200}{C}_2\left(100\right)+{10}^2[{-}^{200}{C}_3\times 10+\dots {+}^{200}{C}_{200}{10}^{198}]$

$\Rightarrow 3^{400}=1-2000+\frac{(200\times 199)}{2}+100k,k\in N$

$\Rightarrow 3^{400}=1-2000+19900+100k=17901+100k$

Hence, the last two digits are $01$

Hence, it is a True statement.