A book has 120 pages. A certain number of consecutive leaves are torn from the book. The sum of the page numbers on the remaining pages is 6215. The number of leaves which are torn from the book can be at most
Answers
Answer:
Step-by-step explanation:
Book has 120 pages. Pages of a book are in arithmetic progression.
Therefore, sum of pages = 120/2[1+120]
= 7260
After removal of certain pages sum = 6215
Sum of pages removed = 7260-6215
= 1045
Now, suppose 'n' leaves were removed. Each leave has 2 pages, therefore, 2n pages were removed.
We have to find n.
Let page number of first page removed = a
Number of pages removed is in AP with 1st term being 'a', total terms being 2n and common difference = 1.
Using formula for sumation of AP-
1045 = 2n/2 * [ 2a + (2n-1)d ]
=> 1045 = n[ 2(a+n) - 1 ]
Now, 1045 = 5*11*19
Let, 2(a+n)-1 = k
=> 5*11*19 = n*k
Clearly, k is odd and k>n
For highest value of n,
If n=19, k=55 (valid)
=> n cannot be 55 ( as then k<n)
Therefore, n cannot be 5 or 11.
=> n = 19
Therefore, at most 19 leaves were removed.