January 30, 2019

# Mathematical fanatical 2019 – the answers.

Hi folks!

In yesterday’s post, you’ll recall how I gave you, dear readers mathematicians, a mathematical brainteaser: how to get ‘2019’ using the four main arithmetic operations [+, -, ×, ÷], plus parentheses [(, )’], and the figures 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1; and not in any order but in the order given [10 down to 1] – and with no joining up of the numbers to make bigger numbers.

Budding mathematicians among the readers of my blog in Russian sent in their answers. Here are some of the more elegant runners-up among them:

( 10 * 9 * 8 – 7 * 6 – 5 ) * ( 4 – 3 + 2 * 1 ) = 2019 (by __Skarbovoy__);

( 10 + 9 ) * ( 8 + 7 + 6 ) * 5 + 4 * ( 3 + 2 + 1 ) = 2019 (by __eve_nts__);

(10 + 9 * 8 * 7 – 6 – 5 ) * 4 + 3 * 2 + 1 = 2019 (also by eve_nts).

A big thanks to everyone, btw, who sent in answers! I had great fun picking out the right ones, checking for mistakes – and coming up with some formulations of my own.

But back to the podium…

The bronze medal goes to __eve_nts__ for a meticulous investigation of all possibilities for the full 10 numbers. Your prize: a KL backpack!

Silver medal – to ‘Skarbovoy’ for his multiple correct answers. A KL backpack goes to you too!

And the gold goes to Yana Baruskova for her beautiful solution to the hardest part of the task – and for also pointing a mistake in my examples! For you – an action camera!

(Prizes also to some other runners-up: Marat Husainov for this: 9^{3}+8^{3}+7^{3}+6^{3}+5^{3}+4^{3}+3^{3}+2+1 = 2019; voffka_33 for his commendable effort; and sir_derryk for his digital poetry.)

All our efforts combined, a total of 45 answers were found to the brainteaser. Details of all 45 – coming up below!

A few of the answers given turned out to have violated the rules, so were disqualified; for example:

( 10 * ( 9 – 8 ) * (( 76 ) – 5 – 4 ) + 3 ) * ( 2 + 1 ) = 2019 *// ’76’*

Putting numbers together like that to make a double-digit number: not allowed.

More cheating, but elegant cheating:

(1098 – 76 – 5 – 4 – 3) * 2 – 1 = 2019 // my own handiwork!

Ok, but all that was using all ten numerals (10 – 1). But what about using 9 (9 – 1)?

**===== 9 =====**

More difficult? Sure is. Doable? Sure is! In more than 20 ways, we discovered – probably more. Here are some highlights:

9 * 8 * 7 * (6 + 5 – 4 – 3) + 2 + 1 = 2019 (by *Skarbovoy*, similar to his answer last year for 2018)

( 9 * 8 * 7 – 6 + 5 ) * 4 + 3 * 2 + 1 = 2019 (*eve_nts*)

And the elegant cheater:

9^3 + 8^3 + 7^3 + 6^3 + 5^3 + 4^3 + 3^3 + 2 + 1^3 = 2019 (by Marat Husainov).

**===== 8 =====**

Well, well. Yet more difficult… Still doable; for example:

8 * 7 * 6 * (5 + 4 – 3) + 2 + 1 = 2019 (by *Skarbovoy*)

**===== 7 =====**

Woah. Getting to ‘2019’ with 7-6-5-4-3-2-1 and basic arithmetic only is… impossible! Even if you multiply everything together you get 5040, and I can’t see any of us living that long…

To get 2019 from this set of seven numbers, you need to at least drop one of them; for example – 2. But even multiplying all the remaining figures you get 2520, and simply 2.1 – no matter how magical the arithmetic – won’t give us 2019. If we lose the 3… then it needs to be multiplied by the remaining figures (remembering that 2019= 673*3). Try getting 673 from 7-6-5-4-2-1… it’s not gonna happen.

In short, to solve the question for seven figures and below one needs a more advanced mathematical set of tools, but more on that later on. For now, check out some of the arithmetical wizardry of our winners, all the while perhaps trying to come up with your own solutions.

Btw, I can tell you that all answers were obtained ‘manually’; that is, without a calculator or computer. In fact, fifth-graders can manage it with their brains alone. If you’re older – or a lot older – there’s nothing quite like a spot of arithmetic to brighten your day and/or comfort you that you’re not losing it – yet ).

Ok; let’s do this…

But first – a few bits of advice, tips and rules…:

**Rule 1.** You can check your attempts on various online calculators (like this one) – which can handle pretty much anything mathematical you throw at them.

**Rule 2. **Examples are written out in alphabetical order, with numerals first, then brackets, then arithmetical symbols (order: * / + -); this is done so that it’s easier to check the novelty of your own results.

**Rule 3. **Tautological arithmetic results in an alphabetical analogue; for example, 3.1. Tautology like ‘n*1’, ‘n/1’, and ‘n^1’ is calculated as one variant and is written as ‘n*1’. It would be the same if ‘a-b=1’ ->

10 + 9 * 8 * 7 * ( 6 – 5 ) * 4 – 3 * 2 – 1 = 2019

10 + 9 * 8 * 7 / ( 6 – 5 ) * 4 – 3 * 2 – 1 = 2019

– only the first variant is taken into account.

Moreover, variants like ‘3*2*1’ or ‘3+2+1’ don’t count tautology-wise:

10 * 9 * ( 8 + 7 + 6 * 5 / 4 ) – 3 * 2 * 1 = 2019

10 * 9 * ( 8 + 7 + 6 * 5 / 4 ) – 3 – 2 – 1 = 2019

3.2. Tautology like ‘a/b/c’ or ‘a/(b*c)’ is also considered as one variant and is written as ‘a/b/c’ ->

10 * 9 * 8 * 7 / 6 / 5 * 4 * 3 + 2 + 1 = 2019

10 * 9 * 8 * 7 / ( 6 * 5 ) * 4 * 3 + 2 + 1 = 2019

Or:

10 * 9 * 8 * 7 / 6 / 5 * 4 * 3 + 2 + 1 = 2019

10 * 9 * 8 * 7 / ( 6 * 5 / 4 ) * 3 + 2 + 1 = 2019

– also only the first variant works.

3.3. Of course, you can’t have (-a)*(-b), since that’d be ‘a*b’.

**Rule 4. **If new methods of solving the task come about in the future, they’ll be added to this here list. Therefore, it’s worth checking back here occasionally to check out the newbies.

**Advice, methods. **The ‘2019’ conundrum (and all like it), is split up into several methods:

**Method 1: **The breaking up into simple selections of variants. ‘2019=3*673′, from ’10-…-1’ – you need to add a 3 and the rest.

**Method 2. **Simple sum. ‘2019 = 20??’ (or thereabouts) + N, where N is made up of ’10-9′ and/or ‘4-3-2-1′, and from the remainder you get ’20??’ or thereabouts. Am I explaining this clearly? Hope so…

**Method 3.** A complex sum. ‘2019 = A +/- B’. We look for A and B, the sum or difference of which gives ‘2019’.

**Method 4. **A multiple. Selection of a number, a multiple of 2019 from ’10-…-5-4-3′ (4038, 6057, …), then dividing it by 2, 3, 4 or other variants that you can make up from the remaining numbers.

**Method 5. **Luck. ‘It came to me in a dream!’

**Method 6. **There are probably more variants for searching for solutions. I just don’t know about them right now.

So, the variants for 10-9-8-7-6-5-4-3-2-1 =>

**===== 10 =====**

*– 3*673*

Many variants, as could be expected.

( 10 * 9 * 8 – 7 * 6 – 5 ) * ( 4 – 3 + 2 * 1 ) = 2019 (by *Skarbovoy*)

(( 10 * 9 * 8 ) – ( 7 * 6 + 5 * ( 4 – 3 ))) * ( 2 + 1 ) = 2019 (by *me*; then, independently, also by *voffka_33*)

( 10 + (( 9 * ( -8 + 7 + 6)) * 5 – 4 ) * 3) * (2 + 1 ) = 2019 (also *me*)

(( 10 + ( 9 * 8 * ( 7 + ( 6 * 5 )) / 4 ) ) – 3 ) * ( 2 + 1 ) = 2019 (*moi*)

( -10 * 9 + ( 8 * ( 7 + 6 ) + 5) * ( 4 + 3 )) * ( 2 + 1 ) = 2019 (*Skarbovoy*)

( -10 * ( 9 + 8 ) + 7 * 6 * 5 * 4 + 3 ) * ( 2 + 1 ) = 2019 (*Skarbovoy*)

((( -10 + ( 9 – ( 8 – 7 * 6 ) * 5 )) * 4 – 3 ) * ( 2 + 1 )) = 2019 (*Skarbovoy*)

*– 1995+24*

Surely there are other 19??+2?

( 10 + 9 ) * ( 8 + 7 + 6 ) * 5 + 4 * 3 * 2 * 1 = 2019 (by *eve_nts*)

( 10 + 9 ) * ( 8 + 7 + 6 ) * 5 + 4 * ( 3 + 2 + 1 ) = 2019 (by *Skarbovoy*)

*– 2000+19*

10 + 9 + 8 * ( 7 * 6 * ( 5 + 4 – 3 ) – 2 ) * 1 = 2019 (*Skarbovoy*)

10 + 9 – 8 * ( 7 – ( 6 + 5 ) * 4 * 3 ) * 2 * 1 = 2019 (*Skarbovoy*) (you can also switch the plus-minus after the 9:)

*– 2010+9*

Automatically you get a solution for 2020 and 2021; will have to remember that next year and the year after!…

10 + 9 * ( 8 * 7 * 6 – 5 + 4 ) / 3 * 2 – 1= 2019 (*eve_nts*)

10 + ( 9 * 8 * 7 * 6 – 5 – 4 ) / 3 * 2 – 1 = 2019 (*eve_nts*)

*– 2012+7*

Brackets (2+1) and a variant for 2021.

(10 + 9 * 8 * 7 – 6 – 5 ) * 4 + 3 * 2 + 1 = 2019 (*eve_nts*)

*– 2016+3*

2016 – a very ‘correct’ number, the factorization of which into prime numbers contains eight figures!

(2*2*2*2*2*3*3*7) – that’s why there’s such a quantity of variants. There are probably more.

10 * 9 * 8 * 7 / 6 / 5 * 4 * 3 + 2 + 1 = 2019 (*friends*; independently – *eve_nts*)

10 * 9 * ( 8 – 7 ) * 6 / 5 * ( 4 + 3 ) + 2 + 1 = 2019 (*mine*)

10 + 9 * 8 * 7 * ( 6 – 5 ) * 4 – ( 3 * 2 ) – 1 = 2019 (*mine*)

10 – 9 + 8 * 7 * 6 * ( 5 + 4 – 3 ) + 2 * 1 = 2019 (*eve_nts*)

( 10 – 9 ) * 8 * 7 * 6 * ( 5 + 4 – 3 ) + 2 + 1 = 2019 (*eve_nts*)

(( 10 * 9 + 8 ) * 7 + 6 – 5 * 4 ) * 3 + 2 + 1 = 2019 (*Skarbovoy*)

10 + 9 * 8 * 7 / ( 6 – 5 ) * 4 – 3 * 2 – 1 = 2019

*– 2020-1*

Hmmm… 2020-1 also has a ton of variants. Btw, practically all of them are the solution for 2020.

10 * ( 9 * 8 * 7 / 6 + 5 * 4 – 3 ) * 2 – 1 = 2019 (*eve_nts*)

10 * ( 9 * 8 * 7 / 6 + 5 + 4 * 3 ) * 2 – 1 = 2019 (*eve_nts*)

10 * ( 9 * 8 + 7 + 6 * 5 * 4 + 3 ) – 2 + 1 = 2019 (*eve_nts*)

10 * ( 9 * 8 + ( 7 + 6 ) * ( 5 + 4 + 3 – 2 )) – 1 = 2019 (friends)

10 * ( 9 – 8 + 7 * 6 * 5 – 4 – 3 – 2 ) – 1 = 2019 (*eve_nts*)

10 * ( 9 – ( 8 – 7 * 6 * 5 ) – ( 4 + 3 + 2 )) – 1 = 2019 (*Skarbovoy*)

10 * ( -9 + ( 8 + 7 * ( 6 + 5 + 4 ) – 3 )) * 2 – 1 = 2019 (*Skarbovoy*)

10 / 9 * ( 8 – 7 * 6 * 5 ) * ( -4 – 3 – 2 ) – 1 = 2019 (*Skarbovoy*)

( 10 + ( 9 + ( 8 + 7 ) * 6 ) * 5 ) * 4 – 3 + 2 * 1 = 2019 (*Skarbovoy*)

(( 10 + 9 ) * 8 * 7 + 6 – 5 * 4 * 3 ) * 2 – 1 = 2019 (*eve_nts*)

(( 10 + 9 ) * ( 8 + 7 + 6 ) + 5 ) * ( 4 + 3 – 2 ) – 1 = 2019 (*Skarbovoy*)

(( 10 + 9 ) * ( 8 – ( 7 + 6 ) * 5 + 4 ) – 3 ) * ( -2 ) – 1 = 2019 (*Skarbovoy*)

(( 10 + ( 9 + ( 8 + 7 ) * 6 ) * 5 ) * 4 – ( 3 – 2 )) * 1 = 2019 (*Skarbovoy*)

(( 10 – 9 + 8 + 7 ) * 6 + 5 ) * 4 * ( 3 + 2 ) – 1 = 2019 (*Skarbovoy*)

( -10 + 9 * ( 8 + 7 ) * 6 * 5 * ( 4 – 3 )) / 2 – 1 = 2019 (*Skarbovoy*)

( -10 + ( 9 * 8 * 7 + 6 + 5 )) * 4 * ( 3 – 2 ) – 1 = 2019 (*Skarbovoy*)

-10 * ( 9 – 8 ) * ( 7 * ( 6 – 5 * 4 ) – 3 ) * 2 – 1 = 2019 (*Skarbovoy*)

-10 + 9 – ( 8 – 7 * 6 * 5 ) * ( 4 + 3 * 2 ) * 1 = 2019 (*Skarbovoy*)

*– 2025-6*

10 * 9 * ( 8 + 7 + 6 * 5 / 4 ) – 3 * 2 * 1 = 2019 (*eve_nts*)

10 * 9 * ( 8 + 7 + 6 * 5 / 4 ) – 3 – 2 – 1 = 2019 (*eve_nts*)

*– other magic:*

( 10 * ( -9 + ( 8 + 7 ) * 6 ) * 5 / 4 – 3 ) * 2 * 1 = 2019 (*Skarbovoy*)

-10 – 9 – ( 8 – 7 * 6 ) * 5 * 4 * 3 – 2 * 1 = 2019 (*Skarbovoy*)

*– Multiple of 2:*

( 10 * 9 * ( 8 + 7 – 6 ) * 5 – 4 * 3 ) / 2 * 1 = 2019 (adapted from 2018, (*Skarbovoy*))

( -10 * 9 * ( 8 + 7 – 6 * 5 ) – 4 ) * 3 / 2 * 1 = 2019 (*Skarbovoy*)

Total: around 45 variants of getting 2019 from ’10-9-8-…-1′. Excellent result!

Now let’s try dropping the 10!…

**===== 9 =====**

To me, the most elegant solutions were:

( 9 * 8 * 7 – 6 + 5 ) * 4 + 3 * 2 + 1 = 2019

9 + ( 8 * 7 * 6 – 5 + 4 ) * 3 * 2 * 1 = 2019

( 9 * 8 * 7 + 6 – 5 ) * 4 – 3 + 2 * 1 = 2019

Here are all 22 of them:

*– 2010 + 9*

9 + ( 8 * 7 * 6 – 5 + 4 ) * 3 * 2 * 1 = 2019 (*eve_nts*, *Skarbovoy*)

9 + ( 8 * 7 * 6 – 5 + 4 ) * ( 3 + 2 + 1 ) = 2019 (*Skarbovoy*)

9 + ( 8 * 7 * 6 – 5 + 4 ) * 3 * 2 * 1 = 2019 (*Skarbovoy*)

9 + ( 8 + 7 ) * (( 6 + 5 ) * 4 * 3 + 2 ) * 1 = 2019 (*Skarbovoy*)

9 – ( 8 + 7 ) * ( 6 – 5 * 4 * ( 3 * 2 + 1 )) = 2019 (*Skarbovoy*)

*– 2012 + 7*

( 9 * 8 * 7 – 6 + 5 ) * 4 + 3 * 2 + 1 = 2019 (*eve_nts*)

*– 2016 + 3*

9 * 8 * 7 * (6 + 5 – 4 – 3) + 2 + 1 = 2019 (*Skarbovoy*)

9 * 8 * 7 * ( 6 – 5 ) * 4 + 3 * ( 2 – 1 ) = 2019 (*Skarbovoy*, and a friend, independently)

9 * 8 * 7 * ( 6 + 5 – 4 – 3 ) + 2 + 1 = 2019 (*Skarbovoy*)

9 * 8 * ( 7 – 6 + ( 5 + 4 ) * 3 ) + 2 + 1 = 2019 (*Skarbovoy*)

9 * 8 * ( 7 + ( 6 + 5 – 4 ) * 3 ) + 2 + 1= 2019 (*Skarbovoy*)

9 * ( 8 + ( 7 + 6 + 5 ) * 4 * 3 ) + 2 + 1 = 2019 (*Skarbovoy*)

-9 * 8 * ( 7 – 6 – 5 ) * ( 4 + 3 ) + 2 + 1 = 2019 (*Skarbovoy*)

*– 2018 + 1*

( 9 – 8 * (7 – ( 6 + 5 ) * 4 * 3 )) * 2 + 1 = 2019 (*Skarbovoy*)

*– 2020 -1*

( 9 * 8 * 7 + 6 – 5 ) * 4 – 3 + 2 * 1 = 2019 (*eve_nts*)

( 9 * 8 * 7 + 6 – 5 ) * 4 – 3 / ( 2 + 1 ) = 2019 (*Skarbovoy*)

( 9 * 8 * 7 + 6 – 5 ) * 4 * ( 3 – 2 ) – 1 = 2019 *Skarbovoy*)

-( 9 – ( 8 * ( 7 + 6 * 5 * 4 ) + 3 )) * 2 – 1 = 2019 (*Skarbovoy*)

-( 9 – 8 * ( 7 + 6 * 5 * 4 ) – 3 ) * 2 – 1 = 2019 (*Skarbovoy*)

*– 2025 – 6*

9 * ( 8 + 7 ) * ( 6 + 5 + 4 ) – 3 * 2 * 1 = 2019 (*Skarbovoy*)

9 * ( 8 + 7 ) * ( 6 + 5 + 4 ) – 3 – 2 – 1 = 2019 (*Skarbovoy*)

*– prime 2*

( 9 * ( 8 + 7 ) * 6 * 5 – 4 * 3 ) / 2 * 1 = 2019 (*Skarbovoy*)

*– cheating:*

None, yet; but I’d like to see some!

**===== 8 =====**

Of course the number of solutions is lower. Only three variants were put forward:

*– 2016 + 3*

8 * 7 * 6 * ( 5 + 4 – 3 ) + 2 + 1 = 2019 (*Skarbovoy,* and *eve_nts, *independently)

*– 2020 − 1*

// Standing ovation. Bravo! Oh well done that man. Managed it – against all the odds:

-( 8 – 7 * 6 * 5) * (4 * 3 – 2) – 1 = 2019 (*Skarbovoy*)

-( 8 – 7 * 6 * 5) * (4 + 3 * 2) – 1 = 2019 (*Skarbovoy*)

All righty, now we get to the real mathematical fireworks: getting 2019 from 7-6-5-4-3-2-1 using factorials, degrees, roots, elations, binary shifts (x!, x^y, √x, x<<y), and other logarithms.

With such a strengthened arsenal at our disposal, the first thing we can say is that things have gotten a little easier. Let’s see some answers:

**===== 7 6 5 4 3 2 1 =====**

**(7 + 6 * (( 5 << 4 ) + 3) << 2 ) — 1 = 2019** (by *Skarbovoy*)

**7! * 6 / 5! * 4! / 3 + 2 + 1 = 2019** (*friends*)

**7! / 6 / 5 * 4 * 3 + 2 + 1 = 2019** (*Skarbovoy*)

**(7 * 6! / 5 + 4 — 3 ) * 2 + 1 = 2019** (*Yana Barsukova*)

**-7 + ((( 6 + 5 + 4 ) * 3 ) ^ 2 + 1 ) = 2019** (*Skarbovoy*)

Which makes five different right answers. There are probably more.

Cheating (you can’t combine numbers):

**( 7 * 6 + 5 ) * 43 — 2 * 1 = 2019** (*Skarbovoy*)

Next!

**===== 6 5 4 3 2 1 =====**

Four operations are added for 6 on down: multifactorials, superfactorials, subfactorials, and primorials ( n!!..! , sf(n), !n, n# ).

It goes like this:

**6! / 5 * ( 4 + 3 )!!!!! + 2 + 1 = 144 * 7 * 2 + 3 = 2019** (adapted from last year’s ‘… +2*1 = 2018’)

**6! / 5 * ( 4!! + 3! ) + 2 + 1 = 2019 **(*Yana Barsukova*)

Wait… no! It’s possible without new newbies – factorials, shifts and degrees!

**6 / 5! * (4! / 3)! + 2 + 1 = 2019** (*Skarbovoy*)

**-6 * 5 + (4 << (3 ^ 2)) + 1 = 2019** (*Skarbovoy*)

Next!…

**===== 5 4 3 2 1 =====**

Here we’ll be needing super-sub-factorials and primorials, and also different well-known numeric sequences.

And we are allowed here to use even more mathematical devices:

!n = subfactorial

# n = primorial

sf(n) = superfactorial

And standard numerical sequences like:

F() = Fibonacci number

Fm() = Fermat number

C() = Catalan number

L() = Leonardo number

M() = Mersenne prime

And assorted other widely-known numbers on the internet.

So, what do we have for 5-1? Straight off the bat we have this one, adapted from 2018:

** ( 5! — 4! ) * F( F( 3! )) + 2 + 1 = 96 * 21 + 3 = 2019**

Also:

**5 * C (( M ( F( 4 ))) — M ( M( 3 )) + 2 — 1 = 2019** (*Yana Barsukova*)

Plus one of mine, without any ‘big name’ number sequences:

**5# — ( 4! * sf( 3 ) ) — 2 — 1 = 2310 — 24*12 — 3 = 2019**

Oh yes! Primorial-factorial-superfactorial – and it’s done! Which means there’s still more room for maneuver. Not all the possibilities of mathematical operations have been checked and used completely.

Cheating (numbers stuck together):

**-5! + F( 4 ) * ( 3! )! — 21 = 2019 **(*Skarbovoy*)

**===== 4 3 2 1 =====**

It’s getting still more interesting: the less there are numbers, the harder the task – just how we like it ).

**C( 4 ) * F( (3!)!!!! ) + 2 + 1 = 14 * 144 + 3 = 2019** (an adaptation from a solution from 2018)

From *Yana Barsukova*:

**C( 4 ) * F ( sf( 3 )) + 2 + 1 = 2019
sf( 4 ) * M( 3 ) + 2 + 1 = 2019**

And as everyone knows, with multiple factorials you can get to practically any figure. Therefore, their application needs to be limited. Multiple factorials can be used either once, or their meaning can exceed half a mulitiple-factorialized number. That is, either once, or 32!!! Kindly multiply by every three. That all clear?!

Well, if so, then it’s possible to approach this here exercise: **4 3 2 1 = 2019**. And we get:

4!*sf(3) = 4*3*2 * 3!*2! = 24*12 = 288

288!!!…281-fold factorial…!!! = 288*7 = 2016 // How convenient and pleasant when such good and correct numbers are so near :).

**( 4! * sf( 3 ) )!!!…281-fold factorial…!!! + 2 + 1 = 2019**

And if we take a closer look… multiplying by seven could have been done earlier, and that bunch of exclamation marks – that’s it ->

**( 4! ) !!!!!!!!!!!!!!!!! * sf( 3 ) + 2 + 1 = 2019**

**===== 3 2 1 =====**

Updated from one of last year’s answers:

**L(L( 3 )!!) + !L(M( 2 )) + F( Fm( !1 )) = L(15) + !L(3) + F(3) = 1973 + 44 + 2 = 2019**

Here’s another:

**3 + (( M( 2 ))! )!! * C(Fm(1)) = 2019** (*Yana Barsukova*)

**===== 2 1 =====**

With just a 2 and a 1 we need to make use of the Mian–Chowla sequence. Heard of it? I hadn’t either! I found it here, btw.

2019 = 2584 — 565

2584 = F(18) = F(6*3) = F(6!!!) = F( (3!)!!! ) = F( ( M(2)! )!!! )

565 = MC(21) = MC( MC( 3! )) = MC( MC( Fm(!1)! ))

Which gives us:

**F( ( M(2)! )!!! ) — MC( MC( Fm( !1 )! )) = 2019**

…If I’m not mistaken.

Here’s another approach to **2 1 = 2019**. As already mentioned, with a multi-fold factorial you can get to practically any figure there is. Well let’s try get to 2019…

2019 = 673*3 = 673!!!…!!! // 670-fold factorial.

673 = 672 + 1 // we’ve used the 1 and dropped it; the 2 remains.

I’ll take the simplest, tried and tested route. From the 2 we get to 0 (for example, with a subfactorial), and then a Fermat turns the 0 into a 3.

!2 = 1

!1 = 0

Fm( !(!2) ) = 3.

3! A mere 2 no longer! A 2! Now we add fertilizer – simple and multiple-fold factorials…

3! = 6

6!! = 6*4*2 = 48

48!!!…!!! (46-fold) = 96.

96!!!….!!! (89-fold) = 96*7 = (and, using a calculator) = the required 672, which = 2^{5} * 3 * 7

Hurray comrades!

**((((( Fm( !(!2) )! )!! )!!!…46-fold…!!!)!!!…89-fold…!!!) + 1 )!!!…670-fold…!!! = 2019**

Check it!

A little later I thought of something else:

The 2 seems… unreasonable! I’d forgotten about primorials! Using them, things are even simpler, with a Fermat and factorial 3! Removing…

And on that basis…

**((((( 2# )!! )!!!…46-fold…!!!)!!!…89- fold…!!!) + 1 )!!!…670- fold …!!! = 2019**

Which just goes to show: there’s no limit to our perfection and exercises for self-development.

**===== 1 =====**

And finally… the ultimate – and our favorite – exercise.

To get from 1 to 2019, it’s time to turn to another set of lesser-known gizmos ‘generalizations of Fibonacci numbers’ known as Tetranacci numbers, henceforth, *Tcci* (my abbreviation). Let’s go…

Tcci(12) = 673

12 = 6!!!!

6=3! — and the 3 we’ve been getting from a 1 for a while already with the finesse of a Fermat: !1=0, Fm(0)=3. Which gives:

**Tcci(( Fm( !1 )! )!!!! ) !!!…!!! (670-fold factorial) = 673*3 = 2019**

Voila!

Phew. Done. Thanks all for your attention. Hope you enjoyed the ride!

That’s it: the brainteaser solved. But I’d still love to see some other elegant solutions!