Question

There is a mountain with 45 bat caves in a row. Every cave has at least 2 bats and there are 490 bats in all. Any 7 caves in a row contain exactly 77 bats. Suppose the first cave has 7 times more bats than the last cave. What is the greatest possible number of bats in the 30th cave?

Answer 1

Answer:

12

Step-by-step explanation:

The first cave has 7 times more bats than the last cave.  So if the 45th cave has b bats, then the first cave has 7b bats.

There are 77 bats in every row of 7 caves.  So if there are 7b bats in the first cave, then there are 77−7b bats in caves 2 through 7.

Since there are also 77 bats in caves 2 through 8, that means cave #8 must have 7b bats.  Repeating this logic:

#1 = 7b

#2-#7 = 77−7b

#8 = 7b

#9-#14 = 77−7b

#15 = 7b

#16-21 = 77−7b

#22 = 7b

#23-28 = 77−7b

#29 = 7b

So the first 29 caves have 5(7b) + 4(77−7b) = 308 + 7b bats.

Now we do the same thing from the other end.  If cave #45 has b bats, then caves #39-#44 have 77−b bats.  And since caves #38-44 have 77 bats, then cave #38 has b bats.  Therefore:

#45 = b

#39-44 = 77−b

#38 = b

#32-37 = 77−b

#31 = b

So caves 31 through 45 have 3b + 2(77−b) = 154 + b bats.

Adding that to the first 29 caves, plus x number of bats in cave #30:

308 + 7b + x + 154 + b = 462 + 8b + x

We know this equals 490.

490 = 462 + 8b + x

28 = 8b + x

x is a maximum when b is a minimum, which is b = 2.

28 = 8(2) + x

x = 12

There are at most 12 bats in the 30th cave.