Dragon's TailDate: 04/01/2003 at 05:20:29 From: Peggy Subject: Dragons Dear Dr. Math, This is my question: An n-dragon is a set of n consecutive positive integers. The first two-thirds of them is called the tail, the remaining one-third the head, and the sum off the numbers in the tail is equal to the sum of the numbers in the head. E.g., the 9 consecutive integers 2, 3, 4, 5, 6, 7, 8, 9, and 10 form a 9-dragon. Its tail is 2, 3, 4, 5, 6, 7 six numbers with sum 27 Its head is 8, 9, 10 three numbers with sum 27 Find the sum of the tail of a 99,999-dragon. Date: 04/01/2003 at 06:01:37 From: Doctor Jeremiah Subject: Re: Dragons Hi Peggy, If k is the first integer, then the sum of n integers starting with k is is k + k+1 + k+2 + ... + k+(n-2) + k+(n-1). In that sum there are n of the letter k and also the sum from 1 to n-1. The sum from 1 to n is n(n+1)/2, so the sum from 1 to n-1 will be (n-1)n/2 and the sum of the first n integers is nk + (n-1)n/2. In a 99,999 dragon there will be 66,666 integers in the tail and 33,333 in the head. The tail starts at k so the 66,666 integers in the tail can be calculated as above. The head starts at k+66,666 so the sum of the 33,333 integers in the head can be calculated as well. The two sums must be the same. So solve for k and then you can get the sum of the tail easily. Let's try it with the 9 dragon. In a 9 dragon there will be 6 integers in the tail and 3 in the head. The tail starts at k so the 6 integers in the tail can be calculated like this: The first integer is k and there are 6 integers (n=6) so Sum = nk + (n-1)n/2 becomes Sum = 6k + (6-1)6/2 which is: Sum = 6k + 15 The head starts at k+6, so the sum of the 3 integers in the head can be calculated like this: The first integer is k+6 and there are 3 integers (n=3) so Sum = n(k+6) + (n-1)n/2 becomes Sum = 3k+18 + (3-1)3/2 which is Sum = 3k + 21 The two sums must be the same. So solve for k: Sum = 6k + 15 = 3k + 21 6k - 3k = 21 - 15 3k = 6 k = 2 Now we said k was the first integer in the tail, so that means the dragon starts at 2: 2,3,4,5,6,7,8,9,10 And it's almost exactly the same with any other dragon. You can simplify all that into one simple equation. Start from this equation: Sum = nk + (n-1)n/2 For the tail n is 2n/3 which gives: Sum_of_Tail = (2n/3)k + ((2n/3)-1)(2n/3)/2 For the head n is n/3 and k is k+2n/3 which gives: Sum_of_Head = (n/3)(k+(2n/3)) + ((n/3)-1)(n/3)/2 Which produces: (2n/3)k + ((2n/3)-1)(2n/3)/2 = (n/3)(k+(2n/3)) + ((n/3)-1)(n/3)/2 And after some algebra: k = (n+3)/6 Which implies that some dragons do not exist (for example, a 12-dragon). - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 04/01/2003 at 20:13:03 From: Peggy Subject: Thank you (Dragons) Thanks for your prompt reply! I get what you mean now. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/