niceee 1

let m>n

LHS : 3ⁿ(3^m-n+1) = 5r

==> 3^m-n has last digit 10-1 OR 5-1;

5-1 not possibl

m-n = 4k + 2

NOW take cases........ puttin k = 0,1,2,........24...........rem. m>n

Shorter methods r welcum [1]

Possible last digits of 3^p : 1,3,9,7

Permissible last digits of 3^m & 3ⁿ for their sum to be 5r :

1,9 : 4k , 4k+2 : 25*25 ways

OR

3,7 : 4k+1 , 4k+3 : 25*25 ways

Total = 625 + 625 = 1250 ways