SOLUTION: Q.1:-In a book with page numbers from 1 to 100, some pages are torn off. The sum of the numbers on the remaining pages is 4949. How many pages are torn off?
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Question 252019: Q.1:-In a book with page numbers from 1 to 100, some pages are torn off. The sum of the numbers on the remaining pages is 4949. How many pages are torn off?
Answer by kev82(151) (Show Source): You can put this solution on YOUR website!
There are many solutions to this, so I think you are missing something from the problem. The total sum of the page numbers is n(n+1)/2 = 50*101 = 5050, taking away 4949 leaves 101. So the pages that have been ripped out must total 101.
Could do 2 pages with p1 and p100
Could do 3 pages with p11, p40 and p50
Could do 4 pages with p10, p20, p31, p40
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The most number of pages you could have ripped out is 13, that is because the smallest total that can be achieved with 14 pages is the first 14 which totals 105. To find a solution for 13 take the first 13 pages which total 91, so you need another 9, replace page 13 with page 13+9=22.
If you are interested in consecutive pages then notice that any sequence of n consecutive pages will be the first n pages shifted by an integer multiple of n. So you require that n(n+1)/2 + an = 101 has an intrger solution for a. This happens when gcd(n, n(n+1)/2) divides 101. The gcd is n when n is odd, and n/2 when n is even. Because 101 is prime the only divisor is 1, which we get when n=2, therefore the only consecutive page solution is two pages, p50 and p51.
In summary there are lots of solutions, the number of pages ripped out could be any number from 2 to 13. But if you need the pages to be consecutive then is only 2 pages.
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