You ain't nothin' but a hound dog

${ 6 }^{ m }+{ 9 }^{ n }\equiv { 1 }^{ m }+{ \left( -1 \right) }^{ n }\equiv 1+{ \left( -1 \right) }^{ n }\quad \left( \text{mod 5} \right)$

Since we want $1+{ \left( -1 \right) }^{ n }$ to be 0, $n$ needs to be odd, while $m$ can be anything. There are 25 odd numbers between 1 to 50 (inclusive).

Hence there are 50 possibilities of $m$ and $25$ possibilities of $n$ , making a total of $25 \cdot 50 = 1250$ possibilities.

The total number of values of x =2500 because there can be 50 values of both m and n and 50 $\times$ 50=2500.

Any power of 6 ends with the digit 6. So, all 50 values of ${ 6 }^{ m }$ end with the digit 6.

Any power of 9 end with either the digit 1 or 9. So, 25 values of ${ 9 }^{ n }$ end with the digit 1 and the other 25 end with the digit 9.

6 + 1 =7

7 is not divisible by 5. 6 + 9 = 15

We know that any number ending with the digit 5 is divisible by 5.

So, whenever ${ 9 }^{ n }$ ends with the digit 9 , x is divisible by 5. We know this occurs 25 times.

So, number of values of x divisible by 5 = 25 $\times$ 50 = 1250 (Ans)