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).