1+1.......?

As a starting approach one can try to maximise the number of $1$ 's in our expression: $1111 \times a + 111 \times b + 11 \times c + 1 \times d = 8102$ .

It is most efficient to use the most $1111$ 's as possible. Hence we want to maximise $a$ s.t. $8102 - 1111 \times a \geq 2018 - 4a$ . This gives $a = 5$ .

$8102 - 5555 = 2547$ .

We can continue by maximising $b$ s.t. $2547 - 111 \times b \geq 2018 - 20 - 3b$ . This gives $b = 5$ .

$8102 - 5555 - 555 = 1992$ .

We now look for $c$ s.t. $1992 - 11 \times c = 2018 - 20 - 15 - 2c$ , and $c = 1$ satisfies this.

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Currently we have $1111 \times 5 + 111 \times 5 + 11 \times 1 + 1 \times 1981 = 8102$ . With $1991 ( = a + b + c + d - 1)$ pluses.

If $a = 4$ and $d > 1981$ , we have $< 2018 - 1981 - 16 = 21$ 1's left. So $b < 7$ and we can't reach $8102$ . Similarly if $a = 5, b = 4, d > 1981$ , we have $< 2018 - 1981 - 12 - 20 = 5$ , so $c < 2$ and we can't reach $8102$ .

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One should note that there is a similar solutions for years $2013$ to $2019$ . Assuming we are summing the $1$ 's to equal the year reversed. All have $1991$ pluses.

E.g. $1111 \times 6 + 111 \times 4 + 11 \times 1 + 1 \times 1981 = 9102$ as we swapped $111$ for $1111$ which gains $1000$ and uses one more $1$ . Likewise if we swap $1111$ 's for $111$ we can do earlier years:

$1111 \times 0 + 111 \times 10 + 11 \times 1 + 1 \times 1981 = 3102$ .

$2102$ is possible. We can't have any $1111$ or $111$ and $2102 = 11c + (2012 - 2c)$ , so $2012 + 9c = 2102$ .

$c = 10$ and $2102 = 11 \times 10 + 1 \times 1992$ with $2001$ pluses.

This is a matter of solving the linear programming problem to maximize $a+b+c+d-1$ for nonnegative integers subject to the constraints $1111a + 111b + 11c + d \; = \; 8102 \hspace{2cm} 4a + 3b + 2c +d \; = \; 2018$ the maximum value of $\boxed{1991}$ is achieved with $a=5$ , $b=5$ , $c=1$ , $d=1981$ .

Formulation: Find values of $(a,b,c,d)$ such that $a*1111 + b*111 + c*111 + d = 8102$ , $4*a + 3*b + 2*c + d = 2018$ , and $(3*a + 2*b + c)$ is minimized.

Let us approach the problem from this perspective: If we start with $d = 2018$ , we have $2017$ " $+$ " symbols but the sum is only $2108$ .
If we change a " $1 + 1$ " to an $11$ , we have one fewer " $+$ " symbol, but the sum increases by $9$ . If we change " $1+1+1$ " to $111$ , we have two fewer " $+$ " symbols, and the sum increases by $99$ If we change " $1+1+1+1$ " to $1111$ , we have three fewer " $+$ " symbols, and the sum increases by $999$ .

So how many " $+$ " symbols do we remove going from $d=2018$ to the correct solution? Well, for each " $1111$ " we remove $3$ " $+$ " symbols, and for each " $111$ " we remove $2$ " $+$ " symbols, while for each " $11$ " we remove $1$ " $+$ " symbol. Thus the target function is $3a + 2b + c$ .

Our heuristic is to maximize $a$ , then maximize $b$ , then maximize $c$ . Each " $1+1+1+1 \rightarrow 1111$ " transition increases the sum by an average of $333$ per " $+$ " removed, while each " $1+1+1 \rightarrow 111$ " transition increases the sum by $49.5$ per " $+$ " removed, and each " $1+1 \rightarrow 11$ " transition only increases the sum by $9$ per " $+$ " removed.

Now note that $a < 6$ , because $6666 + 1994 = 8660$ , which is too much.
If we let $a = 5,$ then $b < 6$ by a similar calculation.

Finally, observe that if $a = 5$ and $b = 5$ , then if we let $c = 1$ and $d = 1981$ , we have a solution. These values of $a$ , $b$ , and $c$ require removing only $26$ " $+$ " signs. And no other sum that removes only $26$ " $+$ " signs will achieve the same sum.