Granted, it has been a while since I have been involved in any math proofs, but it seems to me that there is no solution in spite of the claims of the card. The logic is as follows:
1) Note that the sum of 0 through 9 is 45.
2) In 100, the 1's place is 0.
3) Therefore, to transform the sum in 1) to a sum that adds up to 100, we need turn the 1's place into a 0, i.e., we need to move numbers that add up to 5 from the 1's place to the 10's place.
4) There are 3 ways to move a total of 5, 4&1, 2&3, and 5. In all cases, we are adding 50 (5x10) and subtracting 5 (5x1) from the original 45. So the new sum is 90.
5) We could also try moving over 15 from the 1's to the 10's place, say 7&8. Here the total would be 45 + (15x10) - (15x1) or 180.
If you still don't believe this approach, note that the second sum in the example on the card moves 1 & 3 = 4, so the sum there would be 45 + (4x10) - (4x1) = 81. The third sum in the example moves 2, 3, & 1 = 6, so the sum = 45 + (6x10) - (6x1) = 99.
I hope this is clear. Yet the card claims there is a solution. Seems like it must be a trick.
(edited to standardize notation)
This message has been edited by alanmiley on Nov 4, 2008 9:42 AM
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... but it does not have consistent internal logic.
So let's say by some strange rationale, we accept the sum 98, as an valid use of the 8 and 9 characters. If that be the case, what do we do with the 100 at the bottom? Following that logic, we used one extra 1 and two extra 0's.
A poorly constructed puzzle IMHO.
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