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?

3 Answers
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

4x+6=2018

4x=2012

x=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 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))

Jan 3, 2018

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).