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The 28th of december 2018 will exactly be the 28th month since I date my fabulous girlfriend, by this time she'll be 18 years old. Here's a puzzle I'll give her this day :

Classical part

Maximize $\ \ \ \ 2\ \ \ \ 8\ \ \ \ 1\ \ \ \ 2\ \ \ \ 2018 + a\times18$

with two constraints :

  • You have to insert exactly $12-a$ symbols within

$+,\times,/$ and $-$

  • The final result must contains the digits 2 and 8 at least one time. (so 2328 is a possible maximized value, 2323 isn't)

Maximize means you have to find the greatest number. "$($" and "$)$" (braces // embrace) are allowed in an infinite number of time.

Concatenate numbers is not allowed.

Magical part

If you guys are just too strong for the classical part, there is a Magical part where you have to insert exactly $12-a$ symbols within :

$+,\times,/,-,!, |n|$ (absolute value of the integer $n$, counted as 1 symbol) and $!!$(double factorial)

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    $\begingroup$ May we use "()" or "|x|", for the equation?" $\endgroup$
    – Goose
    Commented Sep 11, 2018 at 20:14
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    $\begingroup$ @QuantumTwinkie Oh yes, good remark, as "(" and ")" are braces (parallelization with embrace), you can use them in an infinite number of time :) $\endgroup$
    – JKHA
    Commented Sep 11, 2018 at 20:16
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    $\begingroup$ @QuantumTwinkie I added the absolute value in the magical symbols :P $\endgroup$
    – JKHA
    Commented Sep 11, 2018 at 20:19
  • $\begingroup$ What if your girlfriend is on-site and looks up the answer here before you ask her? $\endgroup$
    – Wais Kamal
    Commented Sep 13, 2018 at 15:15
  • $\begingroup$ @WaisKamal I know she won't because we are French and she rarely visits English websites ^^ $\endgroup$
    – JKHA
    Commented Sep 13, 2018 at 15:28

6 Answers 6

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I'm pretty sure the maximum is

1,312,848

By

$2*(8+1)*2*(2018+8)*18$

I only could make 4 numbers higher and none of them have both a 2 and an 8. Here they are.

$2*8*(1+2)*(2018+8)*18 = 1750464$
$2*8*((1+2)*2018+8)*18 = 1745856$
$2*(8*(1+2)*2018+8)*18 = 1743840$
$(2*8*(1+2)*2018+8)*18 = 1743696$

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Classical part

My result is:

(2 + 8 + 1 - 2) * (2018+8)*18 = 328212

Since the first constraint:

4 operands so a = (12 - 4) = 8

And the second constraint:

final result has two '2' and an '8'

Magical Part

My result is (approximated):

(2 * 8! * 1 * 2) !* (2018 + 6) * 18 = 1.3669210815446285 × 10^769843

Since the first constraint:

6 operands so a = (12 - 6) = 6

And the second constraint:

final result has several '2' and '8' (as shown in the most significant ciphers we have two '2' and two '8')

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  • $\begingroup$ Welcome to puzzling.se! That's a very well written and formatted answer! Hope to see you around in the future. In the meantime feel free to take the tour, browse the Questions, and make yourself comfortable here! $\endgroup$
    – Christoph
    Commented Sep 13, 2018 at 15:51
  • $\begingroup$ Welcome to Puzzling.stackexchange ! This is a nice answer and currently the greatest one for Classical part :D $\endgroup$
    – JKHA
    Commented Sep 13, 2018 at 20:50
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Maybe I’m getting this wrong, but I know

$28122018+12 \times 18=28122342$ might work, for $a=12$. It’s also possibly possible to do $28122018!!^{(12-a)} + (a \times 18)$ which ought to be huge numbers too..

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  • $\begingroup$ You were not supposed to concatenate numbers, it's true I should have specify it in my puzzle (it's done now) ^^ However, I +1 you because it's a nice answer if concatenate is allowed :) $\endgroup$
    – JKHA
    Commented Sep 13, 2018 at 20:48
  • $\begingroup$ For the magical part, it is not possible to add a number $(12-a)$ neither to exponent it $\endgroup$
    – JKHA
    Commented Sep 13, 2018 at 20:58
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A possible result for the classical part, might be against the rules, though:

$2\color{red}\times8\color{red}{{}\times{}(}1\color{red}+2\color{red}{){}\times{}}2018\color{red}{{}\times{}(}{+7}\color{red}{){}\times{}}18=1\color{blue}{2\,2}04\,\color{blue}864$
Adds $12-7=5$ operations and $2$ sets of braces, and uses a unary plus. However, I am unsure if adding operations (and not just braces) inside the $2018+7\times18$ is allowed.

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A maximized possible value for the classical part that I hope you will beat :

$8218$

Because :

$2+8\times(1/2)\times2018+8\times18 =8218$

First constraint :

$a = 8$, $12-a=4$ symbols that are $+$, 2 $\times$ and one $/$

Second constraint :

1 two and 2 eights

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  • $\begingroup$ Shouldn't switching the second operation into its opposite yield a larger number? That number also fits the constraints with 2 and 8. Alternatively, the third into its opposite. $\endgroup$
    – nishuba
    Commented Sep 13, 2018 at 13:59
  • $\begingroup$ @nishuba unless I didn't understand your propositions, I get two numbers that don't contain any 8 with your operations. You suggested to transform the division by a multiplication, right ? $\endgroup$
    – JKHA
    Commented Sep 13, 2018 at 14:26
  • $\begingroup$ Yeah, you are right, I thought I could multiply the whole result by 4 in that way... $\endgroup$
    – nishuba
    Commented Sep 14, 2018 at 9:05
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My answer is INCORRECT I HAD A RANDOM FACTOR OF 2 IN MY CALCULATOR:

3,487,248

Made by:

((2*8*(1+2)*2018)+8)*18

Which makes a:

Equal to 8, as I added 4 symbols; **+*

And the result has:

two 8s and one 2.

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  • $\begingroup$ Wow, shame you had this factor of two because it'd be a really large answer !! $\endgroup$
    – JKHA
    Commented Sep 13, 2018 at 14:24
  • $\begingroup$ Yeah I typed it all out and then i put it in my calculator again to add a factorial to the end (replacing the 8 with 7 for the additional symbol) to work out the magical one but then realised :( $\endgroup$
    – AHKieran
    Commented Sep 13, 2018 at 14:26
  • $\begingroup$ That's not cool, but I have a solution for your answer, even if it's not as good as you firstly found : you could replace your second * by a / ;) $\endgroup$
    – JKHA
    Commented Sep 13, 2018 at 14:30

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