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Date: 12/21/98 at 16:26:45
From: Katie
Subject: Math logic problems

Ol' Santa's pack held thirty toys
And though none made the same amount,
Each elf made more than two.
The elf named Cher made one more toy
Than the elf who dressed in reds,
But Cher made one less Christmas toy
Than the elf who made the sleds.
Spry Johnny Elf made racing cars.
Five toys were made by Jane.
The elf who dressed in yellow suits
The elf 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 of them all.
Another perky, smiling elf
Ole Santa's pack held 30 gifts
All tagged for girls and boys,
From the clues you've been given,
Now guess who made what toys!

Hint: there are 5 elves and you need to find out what color they are,

I tried making charts but can't figure out two colors of elves or
the toys
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Date: 12/22/98 at 11:21:58
From: Doctor Rob
Subject: Re: Math logic problems

Thanks for writing to Ask Dr. Math!

This is a cute problem.  Here are some hints that may be helpful.

Consider the numbers of toys made. There are five whole numbers, all
greater than 2, all different, which add up to 30.  They contain a run
of three consecutive ones, and a pair of which one is triple the other.
The pair must be (3,9) or (4,12), because if it were (5,15), you would
have the other three numbers adding to 10, but the smallest available
numbers are 3, 4, and 6, whose sum is 13.

If the pair is (3,9), the run cannot be (7,8,9), because the last
number would have to be another 3. The run cannot be (3,4,5), because
the last number would have to be another 9; so the run has to be
(5,6,7), and the numbers are (3,5,6,7,9). Furthermore, the green elf

If the pair is (4,12), the run cannot be (10,11,12), because the sum is
already 37 > 30.  The run cannot be (4,5,6), because then the 4 toys
would be made by an elf dressed in both red and green, which is
impossible. Thus the run must be (3,4,5), and the last number is 6.
That makes the numbers (3,4,5,6,12). Furthermore the red elf made 3,

In either case, Sue made the tops.

In the second case, we can establish that neither Cher, Johnny, Jane,
Sue, nor Marcia can be wearing yellow. That is because Cher is wearing
green, Johnny made racing cars (and the yellow elf made trains), Jane
made sleds, Sue made tops, and Marcia wears orange. This is a
contradiction, so the second case is impossible.

In the first case, you do get a solution. At this point you know:

Cher does not wear red or orange or green, does not make sleds, cars,
or tops, and makes 6 toys.

Johnny does not wear red or yellow or orange, makes cars, and doesn't
make 5, 6, 7, or 9 toys.

Jane wears red, doesn't make sleds, cars, trains, or tops, and makes
5 toys.

Sue doesn't wear red, yellow, green, or orange, makes tops, and makes
9 toys.

Marcia wears orange, doesn't make cars, trains, or tops, and doesn't
make 3, 5, 6, or 9 toys.

Now by a process of elimination, you can determine how many toys Marcia
and Johnny make, which toy Jane makes, and what color Sue wears, who
makes sleds, who makes trains, who makes 3 toys, who makes 7 toys, what
the red elf makes, how many toys the yellow elf makes, how many toys
the orange elf makes, how many cars are made, what color the elf who
makes 9 toys wears, how many trains are made, and of what toy 5 were

Now everything else is determined uniquely.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Logic
High School Puzzles
Middle School Logic
Middle School Puzzles

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