SOLUTION: Some points were initially placed on a line. After that, a point was inserted between each pair of neighboring points. After that, such an insertion was repeated two more times. 11

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Question 1161834: Some points were initially placed on a line. After that, a point was inserted between each pair of neighboring points. After that, such an insertion was repeated two more times. 113 points are on the line as a result. How many points were placed on the line initially?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13209)   (Show Source): You can put this solution on YOUR website!


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Ignore this solution -- it performs the insertion only twice. The problem says the insertion is performed and then performed two MORE times.

Let the initial number of points be n.

There are (n-1) spaces between pairs of points; so after the first insertion the number of points is n + (n-1) = 2n-1.

For the second insertion, there are (2n-2) spaces between pairs of the (2n-1) points. So after the second insertion, the number of points is (2n-1)+(2n-2) = 4n-3.





ANSWER: The initial number of points on the line was 29.

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Corrected solution....

Let n be the number of points initially.

There are (n-1) spaces between those n points, so after the first insertion the number of points is n + (n-1) = 2n-1.

For the second insertion, there are (2n-2) spaces between the pairs of (2n-1) points, so the number of points is now (2n-1)+(2n-2) = 4n-3.

For the third insertion, there are (4n-4) spaces between the pairs of (4n-3) points, so the number of points is now (4n-3)+(4n-4) = 8n-7.

The number of points after the third insertion is 113:





ANSWER: The initial number of points was 15.
Answer by ikleyn(52879)   (Show Source): You can put this solution on YOUR website!
.

According to the problem, the insertion was repeated 3 (three) times.


After the first insertion, the number of the point is  n + (n-1) = 2n-1.


After the second insertion, the number of points is (2n-1) + (2n-2) = 4n-3.


After the third insertion, the number of points is  (4n-3) + (4n-4) = 8n - 7.


Your equation is


    8n - 7 = 113,


which implies


    8n = 113 + 7 = 120

     n           = 120/8 = 15.


ANSWER.  15 points were initially placed on the line.

Solved.


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