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### Dragon's Tail

```Date: 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

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:

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)

```
Associated Topics:
High School Number Theory

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