### Brain Teasers

# Menagerie

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.

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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