2017th edition

Number Theory Level 4

We have a peculiar series in which a lead number $(say \space a_{1})$ is written first and then the number of consecutive numbers $(say \space a_{2}, a_{3}, \dots)$ that follow the lead number (including the lead number) is equal to the lead number itself i.e. $\underbrace{a_{1}, a_{2}, a_{3}, \dots, a_{a-2}, a_{a-1}, a_{a}}$ . And then we continue the series with the number $a_{2}$ and follow the same procedure.

If the series starts from 1 as $\{$ 1, $\,$ 2, 3, $\,$ 3, 4, 5, $\,$ 4, 5, 6, 7, ... $\}$ , find the smallest $n$ , where the $n$ th term $a_n=2017$ .

$\underbrace{\color{#3D99F6}{1}}_\color{#3D99F6}{1}, \underbrace{\color{#3D99F6}{2}, 3}_\color{#3D99F6}{2}, \underbrace{\color{#3D99F6}{3}, 4,5}_\color{#3D99F6}{3}, \underbrace{\color{#3D99F6}{4}, 5,6,7}_\color{#3D99F6}{4},...$

The series above comprises segments of a leader integer (blue) followed by consecutive follower integers such that, if a segment starts with a leader integer $\color{#3D99F6}{k}$ , it is followed by $k-1$ follower integers: $k+1$ , $k+2$ , $k+3$ ,... $k+k-1$ , before the next segment, which starts with leader integer $\color{#3D99F6}{k+1}$ .

Find the smallest $n$ such that the $n$ th term of the series is 2017.

The answer is 509545.

1 solution

Ashish Menon
Jun 6, 2016

Lets break up the sequence to $\color{#3D99F6}{1} \ | \ 2,\color{#3D99F6}{3} \ | \ 3,4,\color{#3D99F6}{5} \ | \ 4,5,6,\color{#3D99F6}{7} \cdots$ .
So, we see that the last term of these segments follow the AP $1,3,5,7,\cdots$ . Firstly, we have to find which term of this AP is $2017$ .
$2017 = 1 + (x - 1)×2\\ \dfrac{2016}{2} = x - 1\\ x = 1009$

So, $2017$ lies in the ${1009}^{\text{th}}$ segment.

Now observe the last term of the first segment(1) is the first term of the original sequence.
The last term of the second segment(3) is the third term of the original sequence.
The last term of the third sequence(5) is the sixth term of the original sequence.
The last term of the fourth segment(7) is the tenth term of the original sequence.

So, we see that last term of the $y^{\text{th}}$ segment is the ${\left(\displaystyle \sum_{i = 1}^y i\right)}^{\text{th}}$ term of the original sequence.

So, the last term of the ${1009}^{\text{th}}$ segment(2017) = ${\left(\displaystyle \sum_{i = 1}^{1009} i\right)}^{\text{th}}$ term of the original sequence.

So, $2017$ is ${\left(\dfrac{1009 × 1010}{2}\right)}^{\text{th}}$ term of the original sequence.

$\implies n = \color{#69047E}{\boxed{509545}}$ .

Absolutely correct. ;) and nice and easy solution.

Abhay Tiwari - 5 years ago

Thanks! ʕ•ٹ•ʔ

Ashish Menon - 5 years ago

2017th edition

The answer is 509545.

1 solution

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