Let S= {(m, n) :m, n $$\epsilon$$ {1, 2, 3, .... , 50}}. lf the number of elements (m, n) in S such that $$6^m+9^n$$ is a multiple of 5 is p and the number of elements (m, n) in S such that m + n is a square of a prime number is q, then p +q is equal to ________.

$$p$$ is the number of elements $$(m\ ,\ n)$$ such that $$6^m+9^n$$ is a multiple of 5.

$$6^m+9^n=0$$ (mod 5) 

6 = 1 (mod 5) 

$$6^m=1^m=1$$ (mod 5) 

9 = 4 (mod 5) 

$$9^n=4^n$$ (mod 5) 

$$1+4^n=0$$ (mod 5) 

Now, $$4^1=4$$ (mod 5)

$$4^2=1$$ (mod 5)

$$4^3=4$$ (mod 5)

$$4^4=1$$ (mod 5)

Now, we will get a remainder of 1 if we divide 1 by 5 (for any value of $$m$$) and a remainder of either 1 or 4 from $$4^n$$ if $$n$$ is even or odd, respectively. 

It is given that $$6^m+9^n$$ is a multiple of 5. 

It means that the remainder that we get from $$6^m+9^n$$ must be a multiple of 5. 

So, we need the remainder of 4 from $$4^n$$. 

Hence, $$n$$ will be odd. 

$$m$$ can take any value from 1 to 50 and $$n$$ can only take odd numbers. 

So, the number of elements of $$(m\ ,\ n)$$ for which $$6^m+9^n$$ must be a multiple of 5 is $$\left(50\times25\right)=1250$$

$$p$$ = 1250 

Now, $$q$$ is the number of elements $$(m\ ,\ n)$$ such that $$(m+n)$$ is a square of a prime number. 

$$1\le\left(m\ ,\ n\right)\le50$$

$$2\le\left(m\ +\ n\right)\le100$$

Square of prime numbers which is less than 100 = $$2^2,\ 3^2,\ 5^2,\ 7^2$$

So, $$(m+n)$$ = 4, 9, 25 and 49

For $$1\le\left(m\ ,\ n\right)\le50$$, if $$(m+n)\le51$$, there will be a total of $$(m+n-1)$$ elements.  

For $$(m+n)$$ = 4, $$(m\ ,\ n)$$ = (1,3) , (2,2) , (3,1) i.e. 3 combinations. 

For $$(m+n)$$ = 9, $$(m\ ,\ n)$$ will have 8 combinations.

For $$(m+n)$$ = 25, $$(m\ ,\ n)$$ will have 24 combinations.

For $$(m+n)$$ = 49, $$(m\ ,\ n)$$ will have 48 combinations.

Total combinations of $$(m+n)$$ = $$(3+8+24+48)$$ = 83

$$q$$ = 83

So, $$(p+q)$$ = $$1250+83$$ = 1333

Hence, the sum of the values of $$p$$ and $$q$$ is 1333.