Learner
1057 Points
Joined May 2010
Hi Renjith, here is my solution. Hope its correct & u sleep well 2day

Step 1:Select the highest integer in LHS..lets assume it as n(for convenience sake). Split it as 1+(n-1). For example, in the eq. 4+5=9, 5 is the highest integer in LHS. 5 will be written as 1+(5-1) or 1+4. So the eq. is rewritten as 4+1+4=9. Similarly eq. 5+4=21 is rewritten as 1+4+4=21.
2)Step 2:Multiply the 2nd number in the LHS of eq. with the 3rd number. In eq. 4+1+4=9, multiply 1(2nd no) with 4(3rd no). Add the resultant product,i.e.4, to 1st number & 2nd number. Thus we get 4+4+1 which is the same as RHS,i.e.9. Similarly in eq.1+4+4=21, we multiply 4 with 4. The resultant product is 16 to which we add 1 & 4. The sum=16+1+4=21=RHS.
Step 3:Now lets apply the above steps to find 6+9. We write 6+9 as 6+1+8. Multiply 1 with 8. 8 the resultant product is added to 6 & 1. 8+6+1=15. So 6+9=15. Now letz check the reverse sum, 9+6. 9+6 is written as 1+8+6. Multiply 8 with 6. The resultant product 48 is added to 1st number & 2nd number. 48+1+8=57. Therefore 9+6=57.