Sample Problem on Arithmetic Sequence

Question:

Given 4 consecutive integral numbers a,b,c,d >0. Under what circumstances will their sum be a multiple of 3 or 9?

Solution:

a+b+c+d = 4a+6 =4b+2 = 4c-2 = 4d-6

For some integer N,M, P, Q, R >0

a+b+c+d = 2b+2c = 4a+4d =3 N

if and only if a = 3M or b=3P,

but 3M + 1 is NOT divisible by 3.

so, we have b+c = 3Q and d=3R.

for a+b+c+d = 9S for some integer S>0

d=a+3 = b+2 = c+1

then for some integer W, U>0,

we have a+d=b+c= 3 + W = 9S/2

since S is an integer,

W must be a multiple of 3 such that

a+b+c+d = 6 + 2W =9S

such that W = 2a.

It follows that a, d must be a multiple of 3 such that 9S/2 is an integer,

i.e. min(S)=2

S is always even,

plug a = 3 or 12 then we yield S = 2 or 4

Erratum:

a+b+c+d = 2b+2c = 2a+2d =3 N