# A Happy New Year! Why don't you solve these questions related to 2018?

## These questions are just for fun. I know the answer of the questions and I will post them after Jan.3. [Level 1] If the sum of four successive integers is 2018, what are the four integers? [Level 2] If the sum of more than four successive integers is 2018, what are the all possible sequences?

Dec 31, 2017

$503 , 504 , 505 , 506$ (level 1)

#### Explanation:

Let x be the first of the four integers, then the sum of x, x+1, x+2, x+3 is

$x + x + 1 + x + 2 + x + 3 = 2018$

that's

$4 x + 6 = 2018$

$4 x = 2012$

$x = \frac{2012}{4} = 503$

Then the numbers are

$503 , 504 , 505 , 506$

Dec 31, 2017

See below.

#### Explanation:

For the level $2$ question.

The sum of $m$ consecutive integers n, n+1, n+2, cdots, n+m  giving $2018$ can be equated as

$m \times n + \frac{\left(m + 1\right) \times m}{2} = 2018$

Solving for $n$ we will get

$n = \frac{4036 - m - {m}^{2}}{2 m}$

but $4036 = 1009 \times 4$ and if $n$ is integer then

$4036$ must be divisible by $m$ so $m = 4$ is the only positive integer solution, with $n = 502$

For negative values we have also $m = 1009$ with $n = - 503$ and $m = 4036$ with $n = - 2018$. Also for $m = - 4036$ we have $n = 2017$ etc.

Resuming

$\left(\begin{matrix}m & n \\ 4 & 502 \\ - 4 & - 503 \\ 1009 & - 503 \\ - 1009 & 502 \\ 4036 & - 2018 \\ - 4036 & 2017\end{matrix}\right)$

Jan 3, 2018

Thank you!

#### Explanation:

[Level 1]
Let $x$ the minimum integer of the four.
Four successive integers are $x , x + 1 , x + 2 , x + 3$.

$x + \left(x + 1\right) + \left(x + 2\right) + \left(x + 3\right) = 2018$
$4 x + 6 = 2018$
$4 x = 2012$
$x = 503$

The four integers are $\textcolor{red}{503 , 504 , 505 , 506}$.

[Level 2]
Let $a$ the first term of the successive integers, and $n$ the number of terms.
Then,
a+(a+1)+(a+2)+…(a+n-1)=2018

Use the formula of arithmetic progression sum.
$\frac{1}{2} n \left\{a + \left(a + n - 1\right)\right\} = 2018$
$n \left(2 a + n - 1\right) = 4036$

The product of two integers, $n$ and $2 a + n - 1$ is $4036 = {2}^{2} \cdot 1009$. ($1009$ is a prime.)

Note that $n$ and $2 a + n - 1$ have the opposite parity.
If $n > 4$, the possible answes are
$\left(n , 2 a + n - 1\right) = \left(1009 , 4\right) , \left(4036 , 1\right)$
$\left(n , a\right) = \left(1009 , - 502\right) , \left(4036 , - 2017\right)$

Therefore, the sequences are
(-502)+(-501)+…506=2018 ($1009$ terms)
and (-2017)+(-2016)+…2017+2018=2018 (4036 terms).