The total number of two digit numbers ' n ' , such that 3 ^ n + 7 ^...

∵ {7}^{n} = ( 10 - 3{)}^{n} = 10k + ( - 3{)}^{n}$</p><p>${7}^{n} + {3}^{n} = 10k + ( - 3{)}^{n} + {3}^{n}$</p><p>$10k + {\left(- 3\right)}^{n} + {\left(3\right)}^{n} = \left\{$$\begin{array}{ll}10k\text{if n=odd} \\ 10k + 2 . {3}^{n}\text{if n=even}\end{array}$$\right)

Let $n = 2t; t \in N$

$∴ {3}^{n} = {3}^{2t} = ( 10 - 1{)}^{t}$

$= 10p + ( - 1{)}^{t}$

.$= 10p \pm 1$

$∴$ if $n =$ even then ${7}^{n} + {3}^{n}$ will not be multiply of $10$

So if $n$ is odd then only ${7}^{n} + {3}^{n}$ will be multiply of $10$

$∴ n = 11, 13, 15, \ldots \ldots \ldots . , 99$

$∴$ Number of odd two digit numbers $= 45$