 # Digital 2018 – pt. 2

Hi folks!

Quite a bit of motivation is needed to solve interesting brainteasers. Thankfully I’ve never had any trouble mustering motivation. But more about that in a bit…

First up, as per the requests of many, two brainteasers that don’t require a calculator or computer – it’s quicker using a trusty old pencil and pad. All righty…

Brainteaser No. 1

There exists a really beautiful 10-digit number. The first (left-most) digit in it is the overall quantity of 0s in this number. The second digit – the quantity of 1s. The third – 2s. And so on. The last digit is the quantity of 9s. What is that digit?

It’s not as hard as it may at first seem. To solve it you need merely (i) a head, (ii) a brain inside it, and (iii) the ability to use it. So good luck!

The second riddle is a little more difficult. Even if you have a head, a brain and ability, not everyone will get it. This one’s solving is probably reserved for arithmetic geniuses – the sort that are able to multiply large numbers in their heads. Let’s see…

Brainteaserdestroyer No. 2

Does there exist a natural (whole, nonzero, positive) number that gives upon multiplication by 2018 a result that consists of a number made up of 10 1s and/or 0s? (everyone’s a programmer here: it’s all about the 0s and 1s:). In other words, is it possible to multiply 2018 by something whole and positive so that the result of the multiplication only has 0s and 1s in it – and is 10 digits long? If yes – let’s see it! If there are many – which is the smallest, and by how much? If there are none, explain the reason why.

Ok all you smart alecks, and Alexandras, thinking caps on! For the best/funniest answers – prizes!

And now a bit on how last week’s riddle was solved:

Digital 2018 – pt. 1

How to get 2018 out of the sequence 10-9-8-7-6-5-4-3-2-1 and its truncations: 9-…-1, 8-…-1 and so on?

Here’s how:

10 – 9 + (8*7*6*(5+4+3))/2 + 1 = 2018

Of course it’s not the only way of doing it. I think there’ll be many ways. Here’s another answer:

(10*9*(8+7-6)*5 – 4*3)/2 – 1

Any more ways?

Next – we drop the 10:

9*8*7*(6+5-4-3) + 2*1 = 2018
8*7*6*(5+4-3) + 2*1 = 2018

Without the 10, 9 and 8, a factorial is needed:

((7 * 6! / 5) + (4 – 3) ) * 2 * 1 = 2018
6! / 5 * (4+3)!!!!! + 2*1 = 2018

– where the ‘!!!!’ is: a multifactorial

Hmmm. With a multifactorial you can practically any answer you want…

3 2 1 -> 3!=6 -> 6!! = 48 -> 48!!!…!!! (41-) = 48*7 -> and so on, until you get 32*9*7 + 2*1 = 2018

Ok. No let’s try without such perversions…

I was given another alternative:

(9*8 – 7 ) * (6*5 + 4 – 3 ) + 2 + 1
8*7*6 *(5 + 4 – 3) + 2*1
-7 + ((6 + 5 + 4)*3 ) ^ (2/1), or a little micro-cheating: 7*6*(5 + 43) + 2/1
-6*5 + 4^3! / 2*1 or -6*5 + (4 << (3*(2+1)))

Hmmm, (4 << (3*(2+1))): crafty :).

We continue:

5 4 3 2 1 = 2018
4 3 2 1 = 2018
3 2 1 = 2018
2 1 = 2018

And the favorite:

1 = 2018

Here are a few answers I was given:

(5! – 4!) * F(F(3!)) + 2*1 = 96 * 21 + 2 = 2018
– where F() = a Fibonacci number

From 4 I managed myself:

4 3 2 1 = 2018 ->

C(4) * F( (3!)!!!! ) + 2*1 = 14*144 + 2 = 2018
– where:
C() = a Catalan number; and
F() = a Fibonacci number

3 2 1 = 2018 ->

L(L(3)!!) + !L(M(2)) + 1 = L(15) + !L(3) + 1 = 1973 + 44 + 1 = 2018
– where:
L() = a Leonardo number;

M() = a Mersenne prime; and

!n = a subfactorial.

2 1 = 2018 ->

first we deal with ‘2’

M(2) = 3 // a Mersenne prime
M(3) = 7
!7 = 1854 // a subfactorial

then we develop ‘1’

!1 = 0
Fm(0) = 3 // a
Fermat number
M(3) = 7
L(7) = 41 // a Leonardo number
41!!!…!!! = 41*4 = 164 // 37th factorial

=> 1854 + 164 = 2018

Therefore:

!M(M(2)) + L(M(Fm(!1)))!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = 2018

And here’s the funniest bit. How to get 2018 from 1? Here’s how:

!1 = 0
Fm(0)=3
W(3)=23 // a
Woodall number
23!!!!!!!!!!!!!!!!!!!!! = 46 // 21st factorial

As(46) = 1009 // what antisigma is shortened to I don’t know, so I made up ‘As’ )
1009!!!…!!! = 1009*2 = 2018 // 1007th factorial. Like I’ve already said, you can get practically any figure using a multifactorial.

(As(W(Fm(!1))!!!!!!!!!!!!!!!!!!!!!))!!!!!…..!!!!! = 2018

And there you have it. Festive arithmetic brainteaser solved. Here’s looking forward to 2019’s!…

Bye for now!… #### Brian

I think I got the 1st one.
8100000000
Still working on the second one. #### ¯\_(ツ)_/¯

But now you have one 8, making your number 810-000-0010, which isn’t correct again ;) #### Jorge LUIS

Ok Eugene number is nice but !!!
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