Question

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?

Answer 1

`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

`4x+6=2018`

`4x=2012`

`x=2012/4=503`

Then the numbers are

`503, 504, 505, 506`

Answer 2

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 xx n +( (m+1) xx m)/2 = 2018`

Solving for `n` we will get

`n = (4036 - m - m^2)/(2 m)`

but `4036 = 1009 xx 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

`((m,n),(4,502),(-4,-503),(1009,-503),(-1009,502),(4036,-2018),(-4036,2017))`

Answer 3

Thank you!

Explanation:

Here is the answer.

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

`x+(x+1)+(x+2)+(x+3)=2018`
`4x+6=2018`
`4x=2012`
`x=503`

The four integers are `color(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.
`1/2 n{a+(a+n-1)}=2018`
`n(2a+n-1)=4036`

The product of two integers, `n` and `2a+n-1` is `4036=2^2*1009`. (`1009` is a prime.)

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

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