Expert: Ahmed Salami - 12/10/2008

Question
Asked the question with one other expert, and this was my answer:I have already answered this question for you once before? Not so? Can't ever recall asking the question on this site.? Maybe you can help me?

I'm Rick and I'd like to resolve a math riddle that conventionally is
very easy to solve, but in the "EXACT" way it's worded it becomes an aberration, I'd like to know why?

In taking individual numbers apart in all ways, and then putting back the
together in all ways  the result should always = the original sum . The
following does not. A very bright teacher gave this riddle to me.



Take 3x10=30. We give the 30 to and individual. The individual decides to give
back 5 of the 30. He takes five from the 30but then decides to only give back 3 and keep 2 for himself.
Common sense tells us in giving back  3 from the original sum
of 30, that original sum of 30 now becomes 27, but how can this be when we
consider the individual kept 2 and when added to the now su of 27, we can now only
come up with a sum of 29, not the original sum of 30? Why is this so? Again I
know how to get the original 30 in all ways except this one?

Thanks

Rick  

Answer
Hi Rick,
Not a very clear question the way you've put it. The statement "that original sum of 30 now becomes 27" is not understandable because that situation doesnt happen. If you're saying the sum remains 27 with the individual, all you need do is add the 3 you got back to it.
Lastly, why add 2 to 27 when its already part of 27 (i.e 2 + 25)???

Regards