Write last two digits of the number

$3^{400}=(9)^{200}$'

$=(10-1)^{200}$

$={ }^{200} C_{0}(10)^{200}+{ }^{200} C_{1}(10)^{199}(-1)^{1}+\ldots .+{ }^{200} C_{198}(10)^{2}(-1)^{198}+{ }^{200} C_{199}(10)^{1}(-1)^{199}+{ }^{200} C_{200}(-1)^{200}$

$=100\left[(10)^{198}+{ }^{200} C_{1}(10)^{197}(-1)^{1}+\ldots . .+{ }^{200} C_{198}(-1)^{198}\right]+200(10)^{1}(-1)^{199}+(-1)^{200}$

$=100\left[(10)^{198}-{ }^{200} C_{1}(10)^{197}+\ldots \ldots+{ }^{200} C_{198}-2(10)\right]+1$

$=100($ a natural number $)+1$

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