Question 1045922
Luckily we are in the 2010's, and luckier yet, Zaina wants the digits 2, 0, 1, and 6 to appear only once. That makes it easier.
It is easier to think through than to explain, but I will try.
The first year number is, of course 2016. That is {{{1}}} year number.
Next come the 5-digit numbers that we can make by adding one of the remaining 6 digits (3, 4, 5, 7, 8, and 9) to the back of 2016. That is {{{6}}} more.
Then come the 5-digit numbers that we can make by adding one of the remaining 6 digits to the front of 2016.
That is another {{{6}}} numbers, for a total of {{{2*6=12}}} 5-digit year numbers.
After that, we have 6-digit year numbers formed by adding a 2-digit sequence to 2016.
There are {{{6^2=36}}} 2-digit sequences that can be formed with the digits 3, 4, 5, 7, 8, and 9,
and there are {{{3}}} ways to add them to 2016. We can attach the last 2, 1, or 0 of the digits in a sequence to the back of 2016, and put the remainder of the sequence at the front.
That gives us {{{3*36=108}}} 6-digit year numbers.
We can also make 7-digit year numbers that will be larger than all the 6-digit year numbers.
We make them by adding to 2016 3-digit sequences made from the digits 3, 4, 5, 7, 8, and 9.
There are {{{6^3=216}}} such sequences, from 333 to 999,
and we can add them by including he first 0, 1, 2, or 3 digits to the front of 2016, and the rest of the sequence to the back of 2016.
That is {{{4}}} ways, and gives us {{{4*216=864}}} 7-digit year numbers.
So far, we have {{{1+12+108+864=985}}} year numbers with 7 digits or less.
The largest of those is 9992016,which is The 985th year number.
There are a lot of 8-digit year numbers, and among those will be the 2016th year number.
To get to the 2016th year number, we just need to form the smallest {{{2016-985=1031}}} 8-digit year numbers.
We make 8-digit year numbers by adding a 4-digit sequence to 2016.
With the 6 digits 3, 4, 5, 7, 8, and 9, we can make {{{6^4=1296}}} such sequences,
from 3333 to 9999.
The {{{1296}}} 8-digit year numbers by adding a 4-digit sequence to 2016
formed by adding all digits in those 4-digit sequences to the back of 2016,
20163333 to 20169999 are the smallest.
The 8-digit year numbers formed by including one or more digits from those 4-digit sequences at the front 2016 start at 32016333, which is larger than 20169999.
The question now is which of the 4 digit sequences from 3333 to 9999 is the 1031st.
If we count the first 8-digit year number, 20163333, as our element {{{red(0)}}} ,
we are looking for element {{{red(1030)}}} .
In a base 6 system, using only the 6 characters 0, 1, 2, 3, 4, and 5.
the base 10 number {{{red(1030)}}} is written as {{{green(4434)}}} ,
because {{{green(4)*6^3+green(4)*6^2+green(3)*6+green(4)=4*216+4*36+18+4=864+144+22=1030}}} .
We cannot use the digits 2, 0, 1, or 6, so we would use.
{{{3}}} instead of {{{green(0)}}} ,
{{{4}}} instead of {{{green(1)}}} ,
{{{5}}} instead of {{{green(2)}}} ,
{{{7}}} instead of {{{green(3)}}} ,
{{{8}}} instead of {{{green(4)}}} , and
{{{9}}} instead of {{{green(5)}}} .
The first 8-digit year number, our element number {{{red(0000)}}} in our usual base 10 system,
is element number {{{green(0000)}}} in the base 6 system using the digits  {{{green(0)}}} to {{{green(5)}}} ,
and is element {{{3333}}} in base 6 when our characters are 3, 4, 5, 7, 8, and 9, and those are the last 4 digits of the first 8-digit year number.
The 1031st 8-digit year number, our element number {{{red(1030)}}} in a base 10 numbering system,
element number {{{green(4434)}}} in a base 6 numbering system using the digits {{{green(0)}}} to {{{green(5)}}} ,
is written as {{{8878}}} in base 6 when our characters are 3, 4, 5, 7, 8, and 9.
Those are the last 4 digits of the 1031st 8-digit year number.
So, the 2016th year number is formed by adding 8878 to the back of 2016.
The 2016th year number is {{{highlight(20168878)}}} .