The answer is 1991.

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how do I know that the maximum value

Dorra Hamza
- 1 year, 11 months ago

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Try using the Simplex Method for solving such problems. There are plenty of online implementations to be found.

Mark Hennings
- 1 year, 11 months ago

how to calculate a, b, and c ?

Daniel Sugihantoro
- 1 year, 11 months ago

$5*1111+4*111+3*11+1980 = 8012$ ,

I got, $a = 5, b = 5, c = 1, d = 1981$ .

Alex Burgess
- 1 year, 11 months ago

Why maximize a + b + c + d? That's not the right target function. The solution you have gives a + b + c + d - 1 = 1991. If I change b to 4 and c to 13 I get a + b + c + d - 1 = 2000. But I have many fewer '+' symbols.

What you want to do is
*
miinimize
*
3
*
a + 2
*
b + c.

Richard Desper
- 1 year, 11 months ago

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This is wrong. We are creating $a+b+c+d$ numbers, namely $a$ copies of $1111$ , $b$ copies of $111$ , $c$ copies of $11$ and $d$ copies of $1$ .

- They have to add to $8102$ and so $1111a + 111b + 11c + d = 8102$ .
- We have to use exactly $2108$ copies of $1$ , and hence $4a + 3b + 2c + d = 2108$ .
- These numbers have to be joined together with $a+b+c+d-1$ pluses, so that is the quantity we need to maximize.

Mark Hennings
- 1 year, 11 months ago

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I've realized you're not wrong and I'm not wrong. We're just counting the number of "+" symbols diferently. My objection is incorrect because I didn't account for the change to d. But with the constraints we have, maximizing a + b + c + d is equivalent to minimizing 3a + 2b + c.

Richard Desper
- 1 year, 11 months ago

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@Richard Desper – Yes, Number of pluses $= a + b + c + d - 1 = 2017 - 3a - 2b - c$

Alex Burgess
- 1 year, 10 months ago

2018* copies of 1. ;)

Alex Burgess
- 1 year, 10 months ago

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.

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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.

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 $8102$ .

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.