lets say the 2 digit numbers are A and B.A+B=A+BAB-A+B=A0(this is like A x 10)-A=A x 9So, the answer will always be the multiple of 9.
Example: 757+5=1275-12=6363 is a multiple of 9
Although it seems like all multiple of nine are the same symbols, but only multiples before 90 are of the same symbols. Examples:45-9=36 32-5=27 11-2=9 81-9=72 99-18=81
Let x be one digit. and let y be the other one and z be the no. and c to be a no. varying from 1-9,z-(x+y)=9cSo whatever ans you get will always be a multiple of nine.
Lionel, you know that AB is actually A*B not A+B
My method applies to any number of digits,even 10000000000000000000 digits
How, Lionel, Prove it
Wow! Lionel, Karan & YuzheYour discussions are interesting!You are getting closer to it... Yes, it's got to do with some addition, multiplication stuff.Another hint: Think of the place value (e.g. Hundred, Tens and Ones)Once you sort this out, we will be able to see what Yuzhe said in his explanation stands for numbers greater than 90 :DOf course, the 'game' has its limitation to disappear symbols beyond 100.Look forward someone could solve this :DCheers!