



 
That kind of puzzles me too! I suppose if you wrote the numbers in base 13, it works:
Steve In January 2010 we received the following from Andrew. I'm not asking a question, but rather providing an answer (what I believe to be the answer) to a previous question. The question was about a math riddle that a parents child had received. The question was SEVEN+EIGHT=TWELVE. Steve who answered the question had said that if you use base 13, it works. The problem was that this question was given to my 13 year old niece and nephew as well. I came up with an answer based upon the multiples of each number. 7 is prime and thus can only have 1 and itself as a multiple, so 2 multiples. 8 has 4 multiples. 1, 2, 4, and 8. Finally, 12 has 6 multiples. 1, 2, 3, 4, 6, and 12. So when you look at the equation or the riddle, it'll read 2+4=6. That's the best I could come up with but I wanted to convey that answer to Steve, just in case he was still curious about it. Thanks Andrew, In January 2012 we received the following from Graham and his son. You have a couple different answers, but my son just brought it home as well and what was needed for him is to replace the SEVEN and EIGHT with the numbers (cannot use a number twice) that when you apply the same to TWELVE you will have an actual addition problem. Solve for what you know. T has to be 1 because if S and E were 9 and 8, you would have 17. Likewise, S+E must be greater than 9. I must be zero because E+I=E. That also means that V+G is less than 10 or there would be carryover. After that we sorted the numbers as best we could highest to lowest and then plugged We got:
There are other combos but this is the one came to. Thanks to both of you. We really appreciate it.  


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