SOLUTION: There is a mountain with 30 bat caves in a row that contain 340 bats in all. Any 7 caves in a row contain exactly 77 bats. Suppose the first cave has 7 times more bats than the las

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Question 1152153: There is a mountain with 30 bat caves in a row that contain 340 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. How many bats are in the 29th cave?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

There is a mountain with 30 bat caves in a row that contain 340 bats in all.
Any 7 caves in a row contain exactly 77+bats.=>each cave contains 11 bats
how many times we have 7 caves in a row, divide 30 by 7=>4
in these 4%2A7=28 caves we have 4%2A77=308 bats
means, we have 2 more caves: first and last
first+cave , then +28 caves, and last+cave =30th cave
so, the 29th cave is last in the group of +28 caves that each contains 11 bats
=>the 29th cave contains 11 bats

if the first cave c%5B1%5D has 7 times more bats than the last cavec%5B30%5D , then
c%5B1%5D=7c%5B30%5D .....eq.1
since these 28 caves contain 308 bats, the first and last cave together contain 340-308=32 bats
c%5B1%5D%2Bc%5B30%5D=32 .....eq.2....substitute c%5B1%5D from eq.1
7c%5B30%5D%2Bc%5B30%5D=32 .....eq.2
8c%5B30%5D=32
c%5B30%5D=4
go to c%5B1%5D=7c%5B30%5D .....eq.1..substitute c%5B30%5D
c%5B1%5D=7%2A4
c%5B1%5D=28

so,
1st cave:28
2nd cave: 11
3rd cave: 11
4th cave: 11
:
:
:
:
29th+cave: 11
30th cave:4


Answer by ikleyn(52802) About Me  (Show Source):
You can put this solution on YOUR website!
.

            It is a nice problem,  but formulated   INACCURATELY.

            An accurate formulation is  THIS :

              There is a mountain with 30 bat caves in a row that contain 340 bats in all. 
              Any 7 caves in a row contain exactly 77 bats. 
              Suppose the first cave has 7 times highlight%28cross%28more%29%29 as many bats highlight%28cross%28than%29%29 as the last cave. 
              How many bats are in the 29th cave?

            Why it must be so,  and why it can not be in different way,  you will see,  when you complete reading my solution.


Solution

Let me start my solution from the reference to an  Internet dictionary

            "in a row"  means  "in succession".

In this problem,  it is the only meaning of this term.


1.  Since any 7 caves in a row contain exactly 77 bats, we have, in particular

        a%5B1%5D%2Ba%5B2%5D%2Ba%5B3%5D%2Ba%5B4%5D%2Ba%5B5%5D%2Ba%5B6%5D%2Ba%5B7%5D = 77,   (1)

        a%5B2%5D%2Ba%5B3%5D%2Ba%5B4%5D%2Ba%5B5%5D%2Ba%5B6%5D%2Ba%5B7%5D%2Ba%5B8%5D = 77.   (2)


    Subtract equation (1) from equation (2).  The terms from a%5B2%5D to a%5B6%5D will cancel each other, and you will get

        a%5B8%5D - a%5B1%5D = 0,   or

        a%5B1%5D = a%5B8%5D.


    Making the same step by step, you will get

        a%5B2%5D = a%5B9%5D,
        a%5B3%5D = a%5B10%5D,
        a%5B4%5D = a%5B11%5D,
        a%5B5%5D = a%5B12%5D,
        a%5B6%5D = a%5B13%5D,
        a%5B7%5D = a%5B14%5D,
        a%5B8%5D = a%5B15%5D = a%5B1%5D

    and so on. It means that the sequence  a%5B1%5D, a%5B2%5D, a%5B3%5D, . . . , a%5B30%5D is PERIODICAL with the period 7.


    It is the FIRST IDEA of the solution.



2.  Its consequence is that  a%5B1%5D = a%5B29%5D.


    Therefore, from the condition, we have this equation

        a%5B1%5D = a%5B29%5D = 7%2Aa%5B30%5D.     (3)


    It is the SECOND IDEA of the solution.



3.  From the condition, we have 

        a%5B1%5D%2Ba%5B2%5D%2Ba%5B3%5D%2Ba%5B4%5D%2Ba%5B5%5D%2Ba%5B6%5D%2Ba%5B7%5D = 77,

        a%5B8%5D%2Ba%5B9%5D%2Ba%5B10%5D%2Ba%5B11%5D%2Ba%5B12%5D%2Ba%5B13%5D%2Ba%5B14%5D = 77,

        a%5B15%5D%2Ba%5B16%5D%2Ba%5B17%5D%2Ba%5B18%5D%2Ba%5B19%5D%2Ba%5B20%5D%2Ba%5B21%5D = 77,

        a%5B22%5D%2Ba%5B22%5D%2Ba%5B24%5D%2Ba%5B25%5D%2Ba%5B26%5D%2Ba%5B27%5D%2Ba%5B28%5D = 77.

    since any 7 caves in a row contain exactly 77 bats.


    Add these 4 equation (both sides). You will get

        a%5B1%5D + a%5B2%5D + a%5B3%5D + . . . + a%5B28%5D = 4*77 = 308.


    It means that  

        a%5B29%5D + a%5B30%5D = 340 - 308 = 32.    (4)


    It is the THIRD IDEA of the solution.



4.  Now, from equations (3) and (4) we have

        a%5B30%5D + 7%2Aa%5B30%5D = 32.


    It implies  8%2Aa%5B30%5D = 32,   or


        a%5B30%5D = 32/8 = 4.


    Hence,  a%5B29%5D = 7%2Aa%5B30%5D = 7*4 = 28.


ANSWER.  a%5B29%5D = 28.  The 29-th cave has 28 bats.

Solved.

---------------

The solution by @MathLover1 is  INCORRECT.

It is incorrect starting from its second line,  where @Mathlover1 states that it follows from the condition
that each cave has  11  bats.

This statement is  WRONG,  and  it  DOES  NOT  FOLLOW  from the condition.