Question 1152153
.


            It is a nice problem,  but formulated   INACCURATELY.


            An accurate formulation is  THIS :


<pre>
              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(cross(more))}}} <U>as many</U> bats {{{highlight(cross(than))}}} <U>as</U> the last cave. 
              How many bats are in the 29th cave?
</pre>

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



<U>Solution</U>


Let me start my solution from the reference to an &nbsp;<U>Internet dictionary</U>


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;"in a row"  &nbsp;means  &nbsp;"in succession".


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



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

        {{{a[1]+a[2]+a[3]+a[4]+a[5]+a[6]+a[7]}}} = 77,   (1)

        {{{a[2]+a[3]+a[4]+a[5]+a[6]+a[7]+a[8]}}} = 77.   (2)


    Subtract equation (1) from equation (2).  The terms from {{{a[2]}}} to {{{a[6]}}} will cancel each other, and you will get

        {{{a[8]}}} - {{{a[1]}}} = 0,   or

        {{{a[1]}}} = {{{a[8]}}}.


    Making the same step by step, you will get

        {{{a[2]}}} = {{{a[9]}}},
        {{{a[3]}}} = {{{a[10]}}},
        {{{a[4]}}} = {{{a[11]}}},
        {{{a[5]}}} = {{{a[12]}}},
        {{{a[6]}}} = {{{a[13]}}},
        {{{a[7]}}} = {{{a[14]}}},
        {{{a[8]}}} = {{{a[15]}}} = {{{a[1]}}}

    and so on. It means that the sequence  {{{a[1]}}}, {{{a[2]}}}, {{{a[3]}}}, . . . , {{{a[30]}}} is <U>PERIODICAL</U> with the period 7.


    It is the <U>FIRST IDEA</U> of the solution.



2.  Its consequence is that  {{{a[1]}}} = {{{a[29]}}}.


    Therefore, from the condition, we have this equation

        {{{a[1]}}} = {{{a[29]}}} = {{{7*a[30]}}}.     (3)


    It is the <U>SECOND IDEA</U> of the solution.



3.  From the condition, we have 

        {{{a[1]+a[2]+a[3]+a[4]+a[5]+a[6]+a[7]}}} = 77,

        {{{a[8]+a[9]+a[10]+a[11]+a[12]+a[13]+a[14]}}} = 77,

        {{{a[15]+a[16]+a[17]+a[18]+a[19]+a[20]+a[21]}}} = 77,

        {{{a[22]+a[22]+a[24]+a[25]+a[26]+a[27]+a[28]}}} = 77.

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


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

        {{{a[1]}}} + {{{a[2]}}} + {{{a[3]}}} + . . . + {{{a[28]}}} = 4*77 = 308.


    It means that  

        {{{a[29]}}} + {{{a[30]}}} = 340 - 308 = 32.    (4)


    It is the <U>THIRD IDEA</U> of the solution.



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

        {{{a[30]}}} + {{{7*a[30]}}} = 32.


    It implies  {{{8*a[30]}}} = 32,   or


        {{{a[30]}}} = 32/8 = 4.


    Hence,  {{{a[29]}}} = {{{7*a[30]}}} = 7*4 = 28.


<U>ANSWER</U>.  {{{a[29]}}} = 28.  The 29-th cave has 28 bats.
</pre>

Solved.


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


The solution by @MathLover1 is &nbsp;INCORRECT.


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


This statement is &nbsp;<U>WRONG</U>, &nbsp;and &nbsp;<U>it &nbsp;DOES &nbsp;NOT &nbsp;FOLLOW</U> &nbsp;from the condition.