Pigeonhole principle to prove that given $4$ numbers the difference between two numbers is divisble by $3$ "Use the Pigeonhole Principle to show that among any four numbers one can find two numbers so that their difference is divisible by $3$." 


I am struggling with this supposedly basic question on one of our past paper as its only worth $3$ marks.
Hint : A number can be of the form $3k , 3k+1$ or $3k+2$. Since there are $4$ numbers , at least two of them leave the same remainder when divided by $3.$

What remainder would their difference leave when divided by $3 ?$
There are 3 possible remainders when we divide a num by 3.thus by pigeon hole principal we have 4 num.two of them must have same remainder when we divisible by 3.so we can write these n1 And n2 as
$N1=3q1+r,$
$N2=3q2+r,$
$N1-N2=3q1+r-3q2-r =3(q1-q2)$
Which is divisible by 3