Menagerie
Logic puzzles require you to think. You will have to be logical in your reasoning.
Zachary has a private zoo. He has five groups of animals in his zoo: snakes, birds, mammals, insects, and spiders. Assume that, typically: animals have 1 head, snakes have 0 legs, birds have 2 legs, mammals have 4 legs, insects have 6 legs, and spiders have 8 legs. Zachary has some unusual animals in his zoo. He has: a snake with 3 heads, a bird with 2 heads, a mammal with 3 legs, an insect with 4 legs, and a spider with 7 legs. From the following information, determine how many of each group of animals that Zachary has in his menagerie.
1) There are a total of 100 heads and 376 legs.
2) Each group has a different quantity of animals.
3) The most populous group has 10 more members than the least populous group.
4) There are twice as many insect legs as there are bird legs.
5) There are as many snake heads as there are spider heads.
Answer
A) Since there are 3 extra heads (2 snake and 1 bird), there are only 97 animals (clue 1). Find a set of five different numbers whose sum is 97 (clue 2) and whose size difference between the least populous and most populous group is 10 (clue 3). There are 17 sets that meet these requirements.
13, 18, 21, 22, 23
13, 19, 20, 22, 23
14, 15, 21, 23, 24
14, 16, 20, 23, 24
14, 16, 21, 22, 24
14, 17, 19, 23, 24
14, 17, 20, 22, 24
14, 18, 19, 22, 24
14, 18, 20, 21, 24
15, 16, 17, 24, 25
15, 16, 18, 23, 25
15, 16, 19, 22, 25
15, 16, 20, 21, 25
15, 17, 18, 22, 25
15, 17, 19, 21, 25
15, 18, 19, 20, 25
16, 17, 18, 20, 26
B) There are twice as many insect legs as there are bird legs (clue 4). To find the possible bird and insect group sizes, multiply the quantity of insects by 6 (legs per insect). Subtract 2 (an insect is missing 2 legs). Divide by 2 (twice as many insect legs as bird legs). Divide by 2 (legs per bird) to get the quantity of birds. The formula is: B = ((I * 6)  2) / 4. There are three combinations that work: 13 and 19, 15 and 22, 17 and 25.
((13 insects * 6 legs/insect)  2 missing legs) / 4 = 19 birds
((15 insects * 6 legs/insect)  2 missing legs) / 4 = 22 birds
((17 insects * 6 legs/insect)  2 missing legs) / 4 = 25 birds
Of the 17 sets (from step A), only 5 contain these sizes.
13, 19, 20, 22, 23 (13 insects & 19 birds)
15, 16, 17, 24, 25 (17 insects & 25 birds)
15, 16, 19, 22, 25 (15 insects & 22 birds)
15, 17, 18, 22, 25 (15 insects & 22 birds or 17 insects & 25 birds)
15, 17, 19, 21, 25 (17 insects & 25 birds)
C) There are as many snake heads as spider heads (clue 5). Since there is a snake with two extra heads, there are actually 2 fewer snakes compared to spiders. Only 2 combinations (from step B) yield the required head count.
13 insects, 19 birds, 20 snakes, 22 spiders, 23 mammals
15 mammals, 17 insects, 19 snakes, 21 spiders, 25 birds
D) The only group combination (from step C) that produces 376 legs (clue 1) is: 15, 17, 19, 21, 25. The other combination yields 380 legs.
(15 mammals * 4 legs/mammal)  1 missing leg = 59 legs
(17 insects * 6 legs/insect)  2 missing legs = 100 legs
19 snakes * 0 legs/snake = 0 legs
(21 spiders * 8 legs/spider)  1 missing leg = 167 legs
25 birds * 2 legs/birds = 50 legs
Thus, there are: 19 snakes (21 heads and 0 legs), 25 birds (26 heads and 50 legs), 15 mammals (15 heads and 59 legs), 17 insects (17 heads and 100 legs), and 21 spiders (21 heads and 167 legs). This produces 97 animals with 100 heads and 376 legs.
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Comments
scallio
Sep 03, 2006
 Owww... my brain hurts. That took way too long! Great teaser!

jazzmusician46
Sep 06, 2006
 Wow! 
rachayl
Sep 14, 2006
 Wow, that was algebralicious 
Cindy650
Nov 07, 2006
 5599277Wow 
javaguru
Mar 20, 2009
 The difficulty rating on this one seems kind of high. The problem solves much more easily and elegantly than the given solution.
As pointed out, there are 97 animals, or an average of between 19 and 20 of each. Since the difference between the fewest and most is 10, the fewest is around 14 to 15 and the most is around 24 to 25.
First give the unfortunate animals back their missing legs, making 380 legs. So where A = arachnids (spiders), I = insects, M = mammals and B = birds, the formula for the number of legs is:
8A + 6I + 4M + 2B = 380
4A + 3I + 2M + B = 190
So I and B must either both be even or both be odd. The relationship between birds and insects is:
4B = 6I  2
B = (3I  1)/2
This forces I to be odd, so I and B are both odd. Plugging in odd values in the range for I gives:
I = 13, B = 19
I = 15, B = 22
I = 17, B = 25
I = 19, B = 28
Given these choices, it's probably I = 17, B = 25. It could be I = 13, B = 19, but that means that the most would be 23 and the other two need to total 97  55 = 42, constraining them to (19,23), (20,22) and (21,21), and since there are two more snakes than spiders, (19,23) won't work. So try I = 17, B = 25.
4A + 51 + 2M + 25 = 190
4A + 2M = 114
2A + M = 57
So M has to be odd, and the least has to be 15. Since there are two more snakes than spiders, there have to be either 15 mammals or 15 snakes. Try 15 for M first since you already know M has to be odd:
2A + 15 = 57
2A = 42
A = 21
S = A  2 = 19 (snakes)
Check it:
A + I + M + B + S = 97
21 + 17 + 15 + 25 + 19 = 97
Took a couple minutes, but it solved easily with very little trial and no error.

javaguru
Mar 20, 2009
 I got my spiders and snakes mixed up twice above...there are more spiders than snake, not snakes than spiders. 
grilled_cheese3
Jun 04, 2009
 that was the easiest teaser ive solved. by the way i never got a lower than an a on my report card and got a letter from the president saying im the smartest kid in my school district. elementary school that is. 
javaguru
Jun 06, 2009
 What does being smart have to do with this being "easiest". You must not have done many of the teasers here as there are some really stupid simple teasers here.
Oh, and I'd really be surprised if the school district sent a letter proclaiming someone as the smartest kid in the district. 
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