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Here are two matches dates that I hold with love in my heart:

The numbers 1970 and 1997 written with matches

The current sum is 1970 + 1997 = 3967.

You must requisition at most 10 matches so that the sum is "as big as possible". We will call this sum $S$. You can remove in both figures with your own distribution as long as the number of matches isn't greater than 10. You must also ensure that all final digits are the same size. e.g. you can not have figure 1 with only one match, it must contains 2 matches.

Multiplier

With the matches that you get, you must create a Multiplier $M$ that also have the same digits size. The final aim of the puzzle is to maximize M times S.

Exemple

Suppose that I remove the one from 1970 and a nine from 1997. I requisitioned 8 matches and get $S = 970+197=1167$ Suppose that I make the number 10 with my 8 matches. My final score will be $10S =11670$ which can easily be improved :D

Happy puzzling and optimizing!

lateral-thinking tag on!! However, cutting matches are not allowed.

My non lateral-thinking best is

$$42466242$$

My lateral thinking best is

$$\simeq 1.09046121\mathrm{e}{15}$$

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  • $\begingroup$ rot13(Pna jr znxr gur "R" qvtvg? r.t. vs jr unir 9 zngpurf, jr pna znxr 1R1, "R" pbafvfgvat bs svir zngpurf.) $\endgroup$
    – ophact
    Commented Feb 21, 2022 at 8:32
  • $\begingroup$ @ophact I did not think of that and this is great for a lateral-thinking answer :) $\endgroup$
    – JKHA
    Commented Feb 21, 2022 at 9:22
  • $\begingroup$ Does the formula have to be M*S, or is there a tad more liberty and we can just add "things" to S $\endgroup$
    – Auribouros
    Commented Feb 21, 2022 at 9:38

4 Answers 4

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Lateral Thinking

Take 10 matches out by removing both 7s and both 1s, that leaves us with an $S$ equal to 189, a very small sum. With the 10 matches, make a pentagon with side length 2 (matches) around $S$ (or the number 189 written with matches), that gives us a number so large, writing it down would need several generations of people.
@Evargalo also suggested making a decagon around $S$, which might be possible, and if it is, it'd make a term a lot bigger.

Additional Info

According to Steinhaus-Moser notation, a number $n$ inside a triangle is equal to $n^n$.
A number inside a square is equals to $n$ nested triangles, meaning it would then be equal to $n^n$, then adding the second triangle, which makes $(n^n)^{(n^n)}$. And so on until $n$ triangles are nested.
Pentagons have nested squares, Hexagons have nested pentagons...

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  • $\begingroup$ This is a very nice approach, but it is easily beaten by (rot13) qenjvat n qrpntba bs fvqr 1 vafgrnq bs n cragntba bs fvqr 2 , dwarfing your score ! $\endgroup$
    – Evargalo
    Commented Feb 22, 2022 at 12:27
  • $\begingroup$ Although I considered that approach, the usage of such a notation is a tad more esoteric than the one I use, hence me not using it. Feel free to make an edit though. $\endgroup$
    – Auribouros
    Commented Feb 22, 2022 at 12:30
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Mildly lateral:

(1911+19111)x7^7^11 approx 2 x 10 ^ 1671034959

enter image description here

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Step by step:

No lateral thinking

Without lateral thinking, you simply have to remove 10 matches and make M equal to 11111, and in doing so maximize the two numbers so as to maximize the sum. This is accomplished by making both numbers 1911, so MS = (1911 + 1911) * 11111, equal to 42466242.

Lateral thinking, part 1

Notice that by removing the top and bottom parts of a zero, we can make two 1s. Make the first nine of the first number a seven, and remove the top and bottom parts of the zero, amounting to the removal of five matches from the first number, and make the second number 1911 by removing five more, and then MS = 11111(17711+1911) = 218020042. This can further be improved by making the first number 19111 and making the second one 1411, for MS = 228019942.

Lateral thinking, part 2

As confirmed by the OP in the comments, it is also allowed to make a letter E with matches: this uses up five matches. The match configuration is equal to that of an eight, but with the two right matches removed. With 10 matches, the largest scientific notation number that can be made is 1E7. Using the "best possible regular pair of numbers" above, we see that MS = 1E7(19111 + 1411) = 205220000000, or 2.0522e11.

Lateral thinking, part 2bis

If allowed, we can also form scientific notation numbers by utilizing the spaces between numbers. By keeping the first number 19111 and making the second digit of the second number a "1", and turning the third digit into three parallel lines, then the second number and third number form an "E". Thus the second number becomes 1E7, and having removed ten matches, we can also set M = 1E7, therefore MS = 1E7(19111 + 1E7) = 100191110000000 = 1.0019111e14.

I sadly did not manage to beat the OP's best, but this answer will be updated as I get better results.

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No lateral-thinking

I can reach

$238 019 842$

By removing

* The upper and lower bars of the $0$ in $1970$, changing it into $11$ ;
* three matches to change the first $9$ of $1997$ into $7$ ;
* four matches to change the second $9$ of $1997$ into $1$ ;
* one match to change the $7$ of $1997$ into $1$,
for a total of ten matches in hand.

Then S=

$19711 + 1711 = 21422$

The missing matches are arranged to form

M= 11111

Finally

$ S * M = (19711 + 1711) * 11111 = 238 019 842$

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