Question 251525: Santa has 30 toys, none of his elves made the same amount and each elf made more than 2. Cher, the elf made one more toy then the elf who was dressed in red, but Cher made one less toy then the elf who made each sled. Spry Johnny Elf made racing cars and 5 toys were made by Jane. The elf who dressed in yellow made each and every train. The who who always dressed in green, made one-third as many as Sue. Cute Marcia Elf was dressed in orange and one elf dressed in blue. The elf who made the spinning tops made the most toys out of everyone. Another perky elf made each and every ball. Which elf dressed in which color and makes which toy? Also, how many toys did each elf make?
Found 2 solutions by richwmiller, kev82:
Answer by richwmiller(17219)   (Show Source):
Answer by kev82(151)  (Show Source):
You can put this solution on YOUR website! This becomes a standard logic puzzle once you figure out what the number of toys are.
We know that we are looking for 5 numbers which add to 30, all different, all greater than 2. If we take the 4 smallest numbers, then 30 minus that is our largest possible number, 30-(3+4+5+6)=12. So we must choose from the numbers 3-12. We know we need a 5, and we need one number that is 3 times the other. This means we must either take 3,5,9 or 4,5,12. If we take 3,5,9 then we need two more numbers that sum to 13. the only choice is 6,7. If we take 4,5,12 then the remaining two numbers must add to 9, the only choices are 3,6.
So the numbers must be 3,5,6,7,9 or 3,4,5,6,12.
I can't see any way to eliminate either of these choices from the constraints, but every time I try to solve 3,4,5,6,12 it leads to a contradiction. The solution to 3,5,6,7,9 is:
Cher,yellow,train,6
Johnny,green,car,3
Jane ,red,ball,5
Sue,blue,spin,9
Marcia,orange,sled,7
Thanks for this, it was fun. I'd like to see how to eliminate the other set of numbers other than trial and error though. Where did you find it?
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