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):
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.
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.
