Question

The last two digits of the number $3^{400}$ is

• A. $81$
• B. $43$
• C. $29$
• D. $01$

Answer 1

The correct option is D $01$

$3^{400}=(3^4{)}^{100}={81}^{100}=(1+80{)}^{100}=1+{}^{100}{C}_1\cdot 80+\dots +{}^{100}{C}_{100}{80}^{100}=1+100k,k\in N$

Hence, the last two digits is $01$