Question

Write last two digits of the number 3⁴⁰⁰.

Answer 1

$$\begin{gathered} 3^{400}={\left(9\right)}^{200} \\ ={\left(10-1\right)}^{200} \\ {=}^{200}{C}_0{\left(10\right)}^{200}{+}^{200}{C}_1{\left(10\right)}^{199}{\left(-1\right)}^1+.....{+}^{200}{C}_{198}{\left(10\right)}^2{\left(-1\right)}^{198}{+}^{200}{C}_{199}{\left(10\right)}^1{\left(-1\right)}^{199}{+}^{200}{C}_{200}{\left(-1\right)}^{200} \\ =100\left[{\left(10\right)}^{198}{+}^{200}{C}_1{\left(10\right)}^{197}{\left(-1\right)}^1+.....{+}^{200}{C}_{198}{\left(-1\right)}^{198}\right]+200{\left(10\right)}^1{\left(-1\right)}^{199}+{\left(-1\right)}^{200} \\ =100\left[{\left(10\right)}^{198}{-}^{200}{C}_1{\left(10\right)}^{197}+.....{+}^{200}{C}_{198}-2\left(10\right)\right]+1 \\ =100(\mathrm{a}\mathrm{natural}\mathrm{number})+1 \end{gathered}$$

Hence, last two digits of the number 3⁴⁰⁰ is 01.