The number of two digit numbers n for which  

 $7^n+3^n$  is divisible by 10 is 

We find that 

$\begin{array}{l}{7}^n=(10-3{)}^n\\ \Rightarrow 7^n{=}^n{C}_0{10}^n{-}^n{C}_1{10}^{n-1}\times 3+{\dots }^n{C}_{n-1}10\times (-3{)}^{n-1}\\ {+}^n{C}_n(-1{)}^n{3}^n\end{array}$

$\Rightarrow 7^n=10k+(-1{)}^n{3}^n$

$\Rightarrow 7^n=10k-3^n$ if  $n$  is  odd

$\Rightarrow 7^n+3^n=10k$  if  $n$  is  odd

$\Rightarrow 7^n+3^n$ 1 is divisible by 10

Thus, $7^n+3^n$  is divisible by 10 for all odd values of  $n$. 

If $n$  is even, 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{7}^n=10k+3^n \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[\text{ From (i) }\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{7}^n+3^n=10k+2\times 3^n\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$

Let $n=2m.$ Then 

$\begin{array}{l}{3}^n=3^{2m}=(10-1{)}^m\\ \Rightarrow 3^n{=}^m{C}_0{10}^m{-}^m{C}_1{10}^{m-1}+\dots \\ {+}^m{C}_{m-1}\times 10\times (-1{)}^{m-1}{+}^m{C}_m(-1{)}^m\\ \Rightarrow 3^n=10p\pm 1\\ \Rightarrow 2\times 3^n=2\times 10p\pm 2\end{array}$

$\Rightarrow 2\times 3^n$  is not divisible by 10. 

$\therefore 7^n+3^n$ is not divisible by 10, if  is even. 

Thus , is divisible by 10 for all odd natural numbers. 

Two digit odd natural numbers are:   $11,13,15,\dots ,99.$

Let their number be k. Then 

$99=11+(k-1)\times 2\Rightarrow 2k=90\Rightarrow k=45$ 

AUTER  We observe that   $7^n+3^n$ is a multiple of 10 only for odd

 values. 

Two digit odd natural numbers are:   $11,13,15,\dots ,99$

Clearly, these are 45 numbers.