21let m>n
LHS : 3n(3m-n+1) = 5r
==> 3m-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]
3m+3n = 5r
r ε N .........
find no. of ways of choosin m,n ε[1,100]
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3 Answers
21
11
Possible last digits of 3p : 1,3,9,7
Permissible last digits of 3m & 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