What Positive Number Doubles When The Last Digit Moves To The First Digit? Riddle For "Geniuses"

[@MindYourDecisions]

You guys are really sharp: a commenter immediately pointed out 0 would work. I missed it and even some of the sources I read missed this corner case. So a clarification: the number is positive.

[@MrDiarukia]

Be careful of the 0. For a programmer there are +0 and -0.

[@100najaja]

I think it would bo cool if you'd give the problems in a superprecise way becouse I often find some weird trick and kill my fun

[@Fexghadi]

Because when you remove the rightmost digit, all the other digits shift to the right.

[@Phroggster]

MindYourDecisions I suppose that now you're also going to say that it's a whole number, seeing as 1.05263157894736842 is quite a bit smaller than 105,263,157,894,736,842 but still satisfies all of the other rules set forth.

[@templarknight7]

Phroggster, your number won't work since you can't move the decimal point.

[@Barrabbazz]

Since the numeral system wasnt specified, i used Binary, and 10 is the double of 01

[@esmaystokhof]

But 1x2=2

[@skilz8098]

#include
#include
#include
int main() {
std::bitset value = 0x01;
std::bitset ans = 0x00;
std::cout

[@epictetusphilosophy]

@@skilz8098 are you a magician and is this an incantation

[@skilz8098]

@@epictetusphilosophy No; it's just basic C++ code! And it is a solution to the original question!

[@stabgan]

Genius

[@andrewcuber8968]

who else thought that this was a two digit number at first?

[@supercoolguygamer9239]

Andrew Cuber hahah me

[@andrewcuber8968]

IKR@@supercoolguygamer9239

[@thesos320]

Mi

[@e1woqf]

me too 🙄

[@jamespotter3777]

Just so you know the smallest positive number is 12

[@isaakvandaalen3899]

Damn, cant believe I counted all the way to 105,263,157,894,736,841 and gave up.

[@pieTone]

Hard times man

[@fifiwoof1969]

😂😂😂😂😂

[@brannythefrenchie8847]

Man, math is a lot more fun when you're not forced to learn it

[@Juuhimuuhi]

French Man I couldn't agree more

[@cae9838]

oui oui

[@msksha6379]

Really ....because I am a doctor, but I could enjoy it more 😍

Math should be fun and this way should be presented. Teachers who force math without fun are very bad and shouldn't teach math.

[@tomasbeltran04050]

Most things are fun when not forced

[@jackxer]

I love how Presh says at the end of every video "So, did you figure it out?" Regardless of how hard the question was.

[@dasguptaarup8684]

after so many videos ..... 🧐 maybe it's scripted

[@MrPLC999]

I'm so glad there are people who can tackle this kind of thing. I couldn't do it to save my life, and I had math in college all the way up to tensor calculus.

[@Brahma909]

Is not 24 and 48?

[@Matthew_MBG]

@@Brahma909 24 would become 42. 24×2 is 48, so no.
48 would become 84. 48×2 is 96, so also no.

[@StevenLoitz]

@@Matthew_MBG I just think "Nope!" and move on to the next video.

[@nicolask.3825]

9:57 "did you figure this out?" Take a wild guess...

[@adriandinu100]

:))

[@moinfahad]

IS there anyone who did?

[@MatthewDempsky]

@@moinfahad I solved it before watching. I used method 2.

[@taowang9735]

- did you figure this out?
- oh hell no lol

[@СергейВасильев-ы7с]

@@moinfahad lol yes. With a modified 2nd method. I used a big number calculator to check if (x mod 19) = 0, which isn't really forbidden but it's just not as cool as what they did.

[@nukegundie]

I came up with a really weird, complex, and time-consuming method at first (as I always tend to do), but I'm pretty sure it would've gotten a right answer:
It starts off as trial and error, but I'm looking for patterns.
I first started marking down 2-digit numbers that were close:
12 -> 21 is 3 too low.
13 -> 31 is 5 too high.
24 -> 42 (6 low)
25 -> 52 (2 high)
37 -> 73 (1)
38 -> 83 (7)
49 -> 94 (4)
and that's it... but for fun:
4'10' (50) -> 104 (which is 4 too high)
The reason I include the last example is because it fits the pattern: for each pair of numbers which is just too high and just too low, their combined distance from their double adds to 8.
As you can see in the last example, it fits. Why 8 though? Well, I kept going...
3 digits:
101->110 (92)
102->210 (6)
These add up to 98... 2 less than 100. 8 is 2 less than 10. I might be onto something.
112->211 (13)
113->311 (85)
At this point the pattern continues. However, there is another *very interesting* pattern I discovered at this point. In the 2-digit examples, all the "too high" numbers (3->6->1->4) actually is a continuous addition of 3, modulo 8, the sum! And for the 3-digits, 92->13 is an addition of 19, modulo 98! In fact, if you add 19 to 3, you get 6 (when you modulo by 8).
I tried many 3- and 4-digit numbers and kept getting 19 as the number which took me to the next "too high" number (and you could go reverse for the "too low" numbers).
And 998 was the sum for 4-digit distances.
I decided at this point that I would be looking for a number with distance 0 (obviously), but getting there would be the issue. Essentially, I could begin at the very first number pair of the x-digits and find out how many additions of 19 I would need to get 0 (when you modulo by [10^(digits-1) - 2]). I tried for 6 digit numbers and became discouraged after that. However, after watching the video, if I go immediately to 18 digits:
1...01 -> 11...0 (900...0002)
1...02 -> 21...0 (999...9996)
Where the sum is 99,999,999,999,999,998 (16 9's and an 8, exactly 10^17-2)
Now, the "too high" number (900...0002) must become 0 through addition of 19's. At this point I actually don't want to work this out xD
999...998 - 900...0002 = 999...9996. 999...9996 / 19 = 526,315,789,473,684 exactly. That is how many 19's to add, and tells us which number we want to use.
100,000,000,000,000,00X's "too high" is 900...0002 (as shown above), and
100,000,000,000,000,01X's "too high" is 900...0021 (by adding 19), and if we continue,
105,263,157,894,736,84X's "too high" will be 0 (once modulo'd). This means that our final digit (X) is the only question remaining. Finally, we get 2 (which is the only digit that works).
Thus, finally, my slow and steady (and way too complicated) method spurns out the correct answer!
But only after giving up and trying a different way xD
Something interesting that I haven't figured out yet: Why did 19 appear? It appears as the denominator in Method 2 in the video, which really intrigues me.
Also, why 8, 98, 998, etc? Some very interesting problems.

[@yugantbeniwal]

Would you like to be my friend?

[@warygg]

why 8, 98, 998? Well, If you have a number xyz, you can express it with the formula 100*x + 10*y+z. When the last digit is moved to the beginning, the number changes to zxy, which can be expressed as 100*z+10*x+y. According to the conditions of the problem, the new number is twice as large as the previous one, so we can determine the equality: (200-10)*x+(20-1)*y+(2-100)*z=0 or 190*x + 19*y - 98z = 0 Looking at this equality, it becomes clear that there are no three-digit numbers that we need, because the first two terms are multiples of 19, but the third is not (since z>0 and z

[@salmabiyad]

I can explain it to u ! Leave a reply if u still wanna know why

[@alinglingwannabeinthewild471]

@@salmabiyad I wanna know

[@FadkinsDiet]

All of those numbers are of the form 10^n - 2 which appears in the expression of the second method or actually half of it does, 5 x 10^n - 1

[@jollywood10]

Me thinking I can iterate through a range with a simple python script. Then finding out the answer is 18 digits. Rip CPU

[@samvictor6084]

Exactly what I did LOL!

[@numbdigger9552]

Laughs in gpu accelerated with rtx 2080

[@TheHbk1985]

@@numbdigger9552 so rip your gpu then ? :D

[@federicorios1140]

You can solve it with a python script by iterating over the exponent rather than the number itself. For example, you could check for which value of k you have:
(10^k-2)%19=0
You find the answer you're looking for after only 18 tries. I solved it in 10 minutes using this method

[@TheHbk1985]

@@federicorios1140 if you do the algebra that far you might as well just finish it :D. To point of an numeric approach is that you dont have to think about that at all... and its fun to test if its possible

[@decidiousforest4616]

Me thinking the answer was going to be two-digits long: “pfft, easy”
Me thinking the answer was going to be three-digits long: “Maybe it’s still doable...”
The answer is eighteen-digits long: “I... what?”

[@EHMM]

0×0=0

[@swipingeverymoment]

@@EHMM 0 isnt positive

[@ThinkHungry_1]

Theres only a select few options in the 2 and 3 digit number selections that would even make sense. As for 2 digits, only 12,24,36, and 48 would even be considered and they dont work. Because 50+ would require 3 digits to be doubled. The same can be said for 3 and 4 digit numbers. Theres only a handful that would even be considered until you need more digits to double. Which means whatever the number is, the first number will always be less than 5

[@Crystxllize]

@@ThinkHungry_1 37 would also be considered

[@ThinkHungry_1]

@@Crystxllize as would 25 and 49, i realized it immediately afterwards but couldnt find my comment😂

[@Luar77]

What's the Wi-Fi password?
It's in the back of the router
Back of the router:

[@arjungautam4369]

@@krelstm8241 but 0 is neither positive not negative

[@Neko_Necromancer]

@@arjungautam4369 Most people just agree it's positive. Just like how most people count 1, 2, and 3 as prime numbers.

[@ffggddss]

@@Neko_Necromancer In the Description, Presh remarks that many commenters quickly caught that 0 was a solution; so he says to restrict the question to positive solutions.
Secondly, 1 is not a prime number. Anymore. It was considered prime many years ago until a choice was made.
Because if you use the definition, 'no [integer] divisors other than itself and 1,' then 1 qualifies as prime.
The problem then becomes the complication this requires in stating the Fundamental Theorem of Arithmetic (FTA), that every positive integer is a product of primes that is unique apart from ordering the factors. Because you can put any number of "1"s into the product, breaking that uniqueness; so you have to amend the theorem to say that 1's are also excluded from the factorization, except when expressing 1 itself, where you're allowed only a single 1.
[The FTA is really what makes primeness even an important concept, so it's crucial to this.]
The other choice was taken some decades or more ago - amend the definition of a prime to be a positive integer that has exactly two divisors.
This excludes 1, which has only a single divisor.
Fred

[@susanjoy1686]

@@Neko_Necromancer 1 is not a prime no.

[@juniusluriuscatalus6606]

I immediately thought of zero, but began to think that would be "lazy" and began to think positive numbers. Then I thought negative numbers... And no, I didn't figure it out. I just broke my brain.

[@photoo848]

Zero is a positive number. It's also a negative number.
If the problem wishes to exclude zero as a solution it should limit itself to strictly positive numbers (i.e. |No)

[@0106johnny]

@@photoo848 By usual definitions zero is not considered positive, but just non-negative

[@photoo848]

@@0106johnny Not over here it isn't. From other comments I've since learned it can differ by region but I was taught that zero is both negative & positive. It's both in |N and in - |N

[@arandombard1197]

The problem never specifies positive numbers or even integers.

[@bucketofnubbins8532]

guess and maybe cheating but binary 01 to 10.

[@peabrainiac6370]

01 doesn't count... 01 is the same as 001 and 0001, but 100 and 1000 aren't twice as large

[@peabrainiac6370]

First digit can't be zero

[@isolatedsushi5996]

Would have to be a 2-bit number then no?

[@catman64k]

in binary there is no solution, if you exclude leading 0, as if you multiply any number by 10 (base2) you add another digit.

[@bucketofnubbins8532]

this only works if you allow the first digit to be 0 which it does not say in the initial setup that that is not allowed.

[@philippenachtergal6077]

9:14 Took me some time to understand why the equation could produce a "b" that doesn't work when a = 1.
The reason is that the equation is contingent on the number of digits of b. a=1 works if you accept to write b as a 17 digits numbers, that is to say with a leading 0.

[@JacobJNilsson]

Thank you! I was so confused by a = 1 since it got so close but with an extra 0. Cool that when calculating that b must be 17 digits long included, it includes leading zeros in the solutions.

[@samuelpeter9109]

The number obtained by writing b with a leading zero is smaller than the answer obtained in the video making Presh's answer incorrect.
052,631,578,947,368,421 x 2
105,263,157,894,736,842
Here, the first number has 17 digits whereas the second number has 18. The answer in the video is the same for the 2 methods because in the first method, there is a caveat that the 2 numbers have to have the same number of digits. Presh doesn't consider the possibility of a different number of digits for the 2 numbers, doing which would have yielded the smallest solution.

[@philippenachtergal6077]

@@samuelpeter9109 No it does not. The problem is posed in plain English and it is understand that a number is not written with unnecessary leading zeroes. In the same manner that it is understand to be base 10. Just like positive means >0 as this is assumed to be of US origin (0 is positive and negative rather than neither according to some definitions taught in some countries).
Solving a problem includes understanding what the plain language used means depending on the context (in a "trick" context, "1 and 1'' could be 11, in a boolean context it is 1, in pretty much any other, it is 2)

[@samuelpeter9109]

@@philippenachtergal6077 The problem here is that if you use the first method in the video and start with a 1, you'll get this answer directly. It seems to have been completely ignored. The seemingly out of the box thinking is only required for the second method.
But I get what you're saying about there being some implicit rules. Because the question only says that the answer is a positive number, one could play with decimals to get an even smaller number.
0.52631578947368421 x 2
1.05263157894736842

[@philippenachtergal6077]

@@samuelpeter9109 Yeah, he glosses over 1 not leading to a solution a bit too quickly. Which is why I wanted to check for myself what was happening :D

[@2Garins]

watches first 30 seconds
"25 × 2 = 50"
skips 8 minutes
"Set up an algebra equation"

[@Braden1005]

I thought it was 12....

[@katsuma-csgo8264]

Patriot10 why doesn’t 12 work

[@blazingrune8535]

@@katsuma-csgo8264 12 times 2 to make 24. If you moved the last digit to the first digit, you would get 21 which proves that 12 doesn't work

[@justarandomlol]

@@blazingrune8535 it clearly said a number where it's last digit becomes first digit when x2, the new number is also twice as large

[@countertopconfessions9975]

@@blazingrune8535 how is it 21? The question never said anything avout the first digit becoming the last aswell. Or in other words, switching digits. All it said was that the last becomes the first. The last digit doesn't need to be anything inparticular as far as the question goes.

[@carlstenger5893]

I haven't dealt with more complicated math in 40+ years. I managed to begin the problem set up properly, but failed to even consider how large the solution might be. (amateur error). Very interesting video. Thanks.

[@rueblimaster]

Same for me

[@zacharynguyen7286]

Hope everyone is doing good. Sending support and hearts! ❤️❤️❤️ Stay safe

[@JacobGoodman]

When I read your original problem, I mistook it for switching the first digit with the last digit (leaving all the rest preserved). I recommed giving it a try! I was able to prove that there are no solutions in that case.

[@photoo848]

Understood it that way first as well. Though there is a solution.
0 fits the bill. It's a positive number. You can move 0 to the front (because just like there's a bunch of zeroes behind the decimal point we don't usually write there's an equal amount of leading zeroes). When doubled the result equals the number if 0 is moved to the front.
Now if we were to limit the problem to strictly positive numbers (i.e. \mathbb {N} _{0} ) then you are correct that there are no solutions.

[@MyNameIsSalo]

@@photoo848 in what math dictionary is zero a positive number, its neutral, the only neutral number that exists, sits right at the middle

[@photoo848]

@@MyNameIsSalo Argument 2: Algebra (theory + exercises)
But since I doubt you're familiar with every math textbook out there I did 2 seconds of sleuthing and found the standard ISO 80000-2 which defines the natural numbers as beginning from 0 (Mac Lane & Birkhoff (1999, p. 15) include zero in the natural numbers)
Putting 0 in |N, a group meant for positive integers means it can't be neutral
In math we made a distinction between |N and |No (index 0) where the former is all natural numbers including 0 while the latter excludes zero.
It's fascinating that there's such a difference in definition possible for such a fundament of mathematics.

[@vitriolicAmaranth]

I wondered if that was the case and watched the first 30 seconds for clarification, glad I did.

[@pudy2487]

@@photoo848 I haven't seen the notation $N_0$. Generally if the ambiguity is important to the answer of the question, it's considered best practice to refer to either the positive integers {1, 2, 3, ...} or the nonnegative integers {0, 1, 2, ...}.

[@ezrahulsman392]

MYD: * Gives the incredibly difficult solution of the problem *
Also MYD: "Did you figure it out?"
Me: "Hold up, 108 quadrill-what now? I was still trying 46 as a solution :'/"

[@djedg10]

Ezra Hulsman I figured it out! With pen and paper. Although I hadn't proven it was the smallest solution. I just continued until the modulus went periodic. It's like Pisano periods, simple really... And fun to make music from!

[@phiefer3]

"When you are ready, keep watching the video for the solution"
There's a pause button for a reason.

[@ezrahulsman392]

DJEdg10 Haha well done, I tried it off the top of my head but I would've never gotten there XD Neither would I have come up with that with pen and paper, so yeah...

[@Ibakecookiess]

This puzzle definitely requires some mathematical education to solve. But anybody that has studied modular arithmetic and/or trained a bit for math olympiads would've gotten it.

[@hellofromc-1374]

Yeah, you've got '64' likes too.😎😎

[@kostiantnyn-faiuk]

316-->631
316*2 = 632
Close...

[@FotisValasiadis]

397*2= 794 dhvfjnbg

[@crazierchimp]

oof

[@strebicux6174]

@@crazierchimp oof indeed

[@manasgupta1093]

1052=2105
1052*2=2104

[@ferrari458960]

109 * 2 = 218
109 -> 910
*close*

[@mrnobodyjunior]

The answer is to work in binary. My smallest number is 01 (1 in decimal), which becomes 10 (2 in decimal), exactly double. And it is indeed the smallest positive number you can think of.

[@marioluigi9599]

Did YOU figure this out??

[@sethstuff921]

Twelve becomes 21 when you swap its first and last digit, which isn't equ to 24

[@therion8469]

@Thomas In Training no

[@pentamath]

No leading 0s allowed. That's the whole point.

[@mrnobodyjunior]

@@pentamath No it's not. Where does it say no leading 0s allowed? Besides, in no other system leading zeros have any impact whatsoever. If you tried that in decimal, you would get ten times which is not the answer.

[@DilipKumar-ns2kl]

The question asked is intelligent and thanks for it.
For a fun ,
2/19=0.105263157894736842
Simply remove decimals, to get answer .
There are other answers too but not smallest.
Just change the order of digits to get other
answers .

[@atarothe7456]

wait that's actually weird, if you were to try the same thing but for a number where moving the last digit to the first gave 3 times more, you will instead get your answer in the form of 3/29
i wonder if there could be a generalized rule about this
something in the line of : if moving a digit to first multiply your result by n times, you can find it by doing n/(10n-1)

[@bitmap4766]

Brilliant!

[@EaglePicking]

@@atarothe7456 It works indeed and the rule does continue.
4/39 = 0,10256410256410256
You take the period which is 102564 and for that number it works.
It often works with periods like this.
You can also do it reverse by using 4/19.

[@MasterQuestMaster]

That's insane!

[@maxgeopiano]

@@EaglePicking It actually makes a lot of sense. I first noticed something like that with n/7 which is
0.142857 or 0.285714 or 0.571428 etc. Depending on n.
So for n/19 doubling n results in the last digit of the decimal number moving to the front. It's pretty fascinating.
For n/49 it's probably the clearest as to why this happens:
1/49 = 0.02040816... so the digits are powers of 2 repeating.
So obviously doubling it will be the same but starting with the next power:
2/49 = 0.04081632... so it moved by 2 digits.
Multiplying 1/49 by 5 is the same as multiplying by 10 and then dividing by 2.
= 0.02040816... I /2
= 0.01020408... I *10
= 0.10204081...
So in total Multiplying 1/49 by 5 moves the last digit to the front.
I assume this rule works similarly for the other cases.

[@MrRyanroberson1]

I didn't understand the specific mechanism by which the last digit became the first until you showed that 25/52, but after that I came up with the same solution as in the video, and did it to all digits>2, which all get 18 digit numbers of the same pattern. So cool

[@danyst-gelais9505]

I made a entire excel sheet gathering formula and developping my intellect just to realise that excel can only go to 15 digits numbers -,-

[@bificommander7472]

Yep... which is when I headbutted my way through anyway by writing a program that calculated the formula using long-integers. This does work. (And fun fact, there are 8 17 digit numbers for which this is true, although the mentioned one is the smallest)
But it is lucky/clever that this problem's answer is just out of Excel's reach.

[@ShanksLeRoux_1]

SAME!!!!!!
I did the same formula as his...
Except for the modular thing.
I made a table for digits.... And got excited when when my excel spreadsheet had an integer value at 15 digits...
(although I was suspicious why rest of the numbers are integers as well after it but.... Was blinded by success.)
But alas... My bubble burst when I tried to use its figures but couldn't even able to add one more digit in the end.
SCREW YOU, EXCEL!!!
I WAS THIS CLOSE AT SUCCESS!!!!!!
But I am happy.... My approach was spot on.
Atleast... That's something.

[@OganySupreme]

Trick: use Google sheets. The numbers can go as large as you want if you format the cells as a number and extend the cells to easily fit the numbers.

[@sino-atrial_node]

multiplied by 2 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 105263157894736842
Last digit 3 157894736842105263
Last digit 4 210526315789473684
Last digit 5 263157894736842105
Last digit 6 315789473684210526
Last digit 7 368421052631578947
Last digit 8 421052631578947368
Last digit 9 473684210526315789
multiplied by 3 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 Calculation exceeded 100 Digit
Last digit 3 1034482758620689655172413793
Last digit 4 1379310344827586206896551724
Last digit 5 1724137931034482758620689655
Last digit 6 2068965517241379310344827586
Last digit 7 2413793103448275862068965517
Last digit 8 2758620689655172413793103448
Last digit 9 3103448275862068965517241379
multiplied by 4 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 Calculation exceeded 100 Digit
Last digit 3 Calculation exceeded 100 Digit
Last digit 4 102564
Last digit 5 128205
Last digit 6 153846
Last digit 7 179487
Last digit 8 205128
Last digit 9 230769
multiplied by 5 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 Calculation exceeded 100 Digit
Last digit 3 Calculation exceeded 100 Digit
Last digit 4 Calculation exceeded 100 Digit
Last digit 5 102040816326530612244897959183673469387755
Last digit 6 122448979591836734693877551020408163265306
Last digit 7 142857
Last digit 8 163265306122448979591836734693877551020408
Last digit 9 183673469387755102040816326530612244897959
multiplied by 6 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 Calculation exceeded 100 Digit
Last digit 3 Calculation exceeded 100 Digit
Last digit 4 Calculation exceeded 100 Digit
Last digit 5 Calculation exceeded 100 Digit
Last digit 6 1016949152542372881355932203389830508474576271186440677966
Last digit 7 1186440677966101694915254237288135593220338983050847457627
Last digit 8 1355932203389830508474576271186440677966101694915254237288
Last digit 9 1525423728813559322033898305084745762711864406779661016949
multiplied by 7 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 Calculation exceeded 100 Digit
Last digit 3 Calculation exceeded 100 Digit
Last digit 4 Calculation exceeded 100 Digit
Last digit 5 Calculation exceeded 100 Digit
Last digit 6 Calculation exceeded 100 Digit
Last digit 7 1014492753623188405797
Last digit 8 1159420289855072463768
Last digit 9 1304347826086956521739
multiplied by 8 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 Calculation exceeded 100 Digit
Last digit 3 Calculation exceeded 100 Digit
Last digit 4 Calculation exceeded 100 Digit
Last digit 5 Calculation exceeded 100 Digit
Last digit 6 Calculation exceeded 100 Digit
Last digit 7 Calculation exceeded 100 Digit
Last digit 8 1012658227848
Last digit 9 1139240506329
multiplied by 9 will yield same number with last digit moved to first
*******************************
Last digit 1 Calculation exceeded 100 Digit
Last digit 2 Calculation exceeded 100 Digit
Last digit 3 Calculation exceeded 100 Digit
Last digit 4 Calculation exceeded 100 Digit
Last digit 5 Calculation exceeded 100 Digit
Last digit 6 Calculation exceeded 100 Digit
Last digit 7 Calculation exceeded 100 Digit
Last digit 8 Calculation exceeded 100 Digit
Last digit 9 10112359550561797752808988764044943820224719

[@johannesvanderhorst9778]

@@sino-atrial_node For all those cases of "Calculation exceeded 100 Digit", there is a simple proof that there is no solution at all:
If the last digit of the solution is lower than the number we multiply with, the product needs to has one digit more. We can't achieve this since we only are rearranging digits and don't add another digit.

[@figloalds]

This was so cool and so far beyond my expectation, I was here innocently expecting the answer to be a simple 2-digit number or something

[@marclink0]

Nice math problem. However, I find the question quite confusing; trying by myself I interpreted it in a few ways. Having this question on a test would have been the end of me!
Take for example 102
Saying the last digit becomes the first digit, it could mean the '2' becomes a '1' -> 101.
It could also mean the last digit simply replaces the first, hence 102 becomes 202.

[@sststr]

Or could it mean a swapping of the digits? So for your example, 102 => 201 ? Or just move the right-most digit to be the left-most digit and shift everything down, so 210?
All this channel's riddles are equally ill-defined and vague. The only real riddle in every case is to understand the rules of the riddle. Once you've figured out what the rules are, the solution is usually trivial. You may find here in the comments someone who recognized for this problem that no base was specified, so he chose base 2 and then the solution was extremely trivial 01 => 10 is a doubling in base 2!

[@shawncowden3909]

​@@sststr I just tried to solve the problem with the assumption that the first and last digits were supposed to be swapped, and ended up proving that no such number can exist.
If 2[Bxxxxx...A] is set to equal [Axxxxxx...B], then B can't = 0 because the digit count would be different. B has to be even because 2*n is always even. And B can't be 5 or greater because then the double of the original number would have an extra digit (1) tacked on in front. So B must be 2 or 4.
A must > B , or else [Bxxxxx...A] would be larger than [Axxxxx...B]. Actually, A >= 2B, because any number plus itself will see its first digit at least double (if the digit count is conserved). But if A is merely double B, then every digit position of [Axxxxxx...B] must be twice the digit in the corresponding position in [Bxxxxxx...A], which can't be the case because all the x digits must be conserved respective to that position (and also because B would have to be twice A, but we know A>B). So A must = 2B+1.
But since B is at least 2, and A is 2B+1, A is at least 5. And you can't multiply a number ending in a digit of n >=5 by 2 without changing the value of the tenths place digit (which would have to be 0) in the resulting product. Therefore, no number can double itself merely via swapping its first and last digits.
Edit: actually that last paragraph isn't exactly true, since an endless stream of 9s would satisfy the relation because 9x2 +1 = 19, but then you'd run into a contradiction once the string of 9s stops.

[@Overseer10192]

I agree. I thought it was the latter interpretation and reasoned that the number must be a multiple of 10, as any other number doubled would change the last digit. Without messing around with different bases, there is no solution to this interpretation (I think).

[@Mehwhatevr]

That's why with these videos, I always watch a bit of the video first to make sure I understand the problem. Even if that means watching the first few seconds of the solution where he sets it up. No spoilers, just clarification on what we're looking for.

[@NeutroniummAlchemist]

@@shawncowden3909 I tried the same thing, and came to the same conclusion.

[@charlesmcmanus4229]

There is a difference between a “riddle” and a damn math problem.

[@dustonangeloyong8750]

This ones a math problem

[@mananshah7889]

@@BeanieRene1 I believe that is a solution too.

[@4pm462]

@@mananshah7889 21 does not equal 24

[@Channel-xy2wj]

Bright Kirby101 you don’t multiply though. It says just to move the last number to the front. So ur answer would be 21 not 24. And 21 isn’t 2 times of 12.

[@4pm462]

@@BeanieRene1 Actually, they are entirely correct. You only move the last digit to the front. There is no multiplying.

[@manojjoshi24x7]

In binary system (0,1).. Used in digital electronics.
0 is 00
1 is 01
2 is 10
3 is 11
In 2 digit format.
So if 1 is represented as 01
And right digit becomes left one 10 will be yielded. That is 2 in decimal.
So, Boom

[@grisly-bear585]

I don't get it at all but you get a like because you sounded genius

[@vibaj16]

But that includes leading zeros, so 01=001, and 001 becomes 100, which is not 10. That’s the problem with including leading zeros.

[@haakoflo]

in python: print(bin(2*0b01)) = 0b10

[@deharudragonsoul6238]

@@vibaj16 he's talking binary dude not leading zeros

[@lubosdostal8523]

.1 ?

[@jess.c.francis]

Late to the party, but I seemed to have had a similar approach as Method 2 and managed not to use guess and check on the modular arithmetic by using Fermat's Little Theorem.
Here's my solution; the beginning is very similar to Method 2, but I believe it is also cleaner:
Let "t" be the lowest possible positive integer that doubles when its last digit is moved to be the first digit
Let "a" be the last digit of "t"
Let "S" be the remaining number when "a" is removed as a digit from "t"
Let "n" be the number of digits in "t"
t = 10S + a
2t = 10^(n-1)a + S
20t = 10^(n)a + 10S
20t + a = 10^(n)a + 10S + a
20t + a = 10^(n)a + t
19t = 10^(n)a - a
19t = a(10^n - 1)
t = a(10^n - 1)/19
Because "t" is a positive integer, the last line implies that either "a" or 10^n - 1 is an integer multiple of 19. As "a" is a digit with a value ranging from 1 to 9, it cannot be an integer multiple of 19, meaning 10^n - 1 must be the multiple.
Modular arithmetic will be used to find the lowest value of "n" satisfying the above condition. As 10^n - 1 is an integer multiple of 19,
10^n - 1 ≡ 0 (mod 19)
10^n ≡ 1 (mod 19)
This is where Fermat's Little Theorem will be used; it states that if "p" is a prime, "b" is any integer, and "p" does not divide "b", then
b^(p - 1) ≡ 1 (mod p)
This is the same form as the last line, where b = 10, p = 19, and n = p - 1. Based on this matching, it is clear that
n = 19 - 1
= 18
Substituting in the value of "n",
t = a(10^18 - 1)/19
Now all that must be done is input the lowest possible value of "a" from 1 to 9 into the above equation. a = 1 is not quite a solution, but a = 2 is, so
t = 2(10^18 - 1)/19
= 105,263,157,894,736,842
Edit: It has come to my attention that this solution still requires checking as it is impossible to analytically prove that 18 is the smallest natural number that satisfies 10^n - 1 ≡ 0 (mod 19). However, it can still be proven by showing that n = 6 and n = 9 do not satisfy. If you want to know why this works, check out responses to the congruency question on math stack exchange (my question was flagged as a duplicate so the entry might disappear at some point). math.stackexchange.com/questions/4717125/how-can-i-prove-18-is-the-smallest-n-in-mathbb-n-such-that-10n-equiv-1

[@alexandre7946]

Hello!
Interesting! But I think I've missed something in your explanation. Because, from my understanding, when you used the Fermat's little theorem, you only proved that n=18 is a solution, not that it is the smallest one.
If indeed you didn't, do you have an idea to prove it without going back to an iterative solution? I'm stuck there as well.

[@jess.c.francis]

@@alexandre7946 Hey Alexandre, thanks for pointing that out. I'm not very familiar with number theory and have no idea how to prove that 18 is the lowest value satisfying the congruency. I just put the question (about the congruency) on the math stack exchange, so maybe we'll get an answer there.

[@antonlupanov7803]

@@jess.c.francis In general you can't. Remainders modulo 19 (except 0) form one big cycle and thus Fermat's little theorem generated smallest solution, but if we look at the number 11, for example, we will see that the cycle is only length two for remainders "1" and "10" (So Fermat's theorem doesn't always get us the smallest solution).
One thing you can do that's shorter than a full iterative method:
If n=18 is a solution, then the minimal solution have to be a divisor of 18 (since if actual cycle is shorter than 18 it has to repeat an integer number of times during our 18 multiplications). Thus it is enough to check divisors of 18 (2, 3, 6, 9) and if neither of them is a solution then 18 is a smallest solution.

[@zacharynguyen7286]

Hope everyone is doing good. Sending support and hearts! ❤️❤️❤️ Stay safe

[@ABW941]

Oh, at first i thought it ment that you have a number and you switch the first with the last digit.

[@HritikRoshan-uk5py]

Same here

[@anonygent]

Same. I was switching the numbers end to end.

[@ethanandrews3076]

I tried this too. Spoiler alert: there isn’t a number that has that property, which is why I was super confused 😂

[@anonygent]

@@ethanandrews3076 That's pretty much the conclusion I reached as well, but I couldn't figure out how to prove it conclusively.

[@ethanandrews3076]

@@anonygent so you know that the first number and the second number must have the same number of digits. This means the first digit of the number has to be less than 5 (because any number above 4 when doubled will add another digit i.e. 5*2=10, 6*2=12 etc.). Now you know that the first digit is either 4,3,2, or 1.
Now we will try to figure out what the last digit is. We know that any number multiplied by 2 is even. This means the last digit of the second number must be even. Because we get the second number by switching the first and last digits, the first digit of the first number must be even so that the seconds numbers last digit can be even.
This means the first digit of the first number is either 4 or 2 because it can’t be odd. Because we know that the second numbers last digit is either 4 or 2, we will try to see which numbers when multiplied by 2 will give a 4 or a 2 in the last digits place. The numbers are 7, 6, 2, and 1 (7 and 2 have four as their last digit, and 6 and 1 have two). This means that if the first digit is 4, the last digit is 7 or 2, and if the first digit is 2, the last digit is either 6 or 1. With this knowledge, you can get rid of some of the options. If the first digit is 4 and the last digit is 2, when you switch the numbers, you get a smaller number. You can’t multiply something by 2 and get a smaller number. This means it can’t be 4 and 2. The same argument can be made with 2 and 1. This means the two options are 4 and 7, or 2 and 6. If the first digit was 4, when you multiply the number by 2, you would get an 8 as the first digit. There is no way to make the 8 equal 7, so the first digit of the second number will have to be greater than 7.
This means the only possible solution would be if the first digit was 2 and the last digit was 6.
However, there is no number that starts with 2 and ends with 6, and that starts with 6 when you multiply by 2. We know this because to make the number as large as possible, you make the inner digits 9 (the largest digit). If we do this, we find a pattern where the second number is 5 followed by 9’s. 296*2=592 ; 2996*2=5992 ; 29996*2=59992 ; 299996*2=599992. Because the highest possible first digit is less than 6, that means the first digit cannot be 2 and the last digit cannot be 6.
We have gotten rid of all possibilities, meaning it cannot be done.

[@lightningstrikes9175]

Why doesn’t math learn to solve its own problems. I learned to solve mine.

[@dyson_sphere]

😁😁LoL

[@nathanderhake839]

Math solves mine for me, so I am returning the favour

[@scepticusverisimillimenonm8450]

Best reply

[@afloatgamer8582]

Lol

[@nickwilson3499]

Programming!

[@tabamal]

very interesting.. Thank you. I tried the same technique to find a positive integer which TRIPLE its value when its last digit moves to the first digit. It is this 28-digit number:
1,034,482,758,620,689,655,172,413,793.
what is remarkable is that the above sequence of digits is rather well equidistributed..

[@thedarkspeedninjashadittsux]

Ous Alam But can you find a non-zero number that becomes squared when the last digits moves to the first digit?

[@wombat4191]

Another remarkable thing is that if you continue to move last to first again and again, it will sometimes triple again. Actually it does that already with 3,103,448,275,862,068,965,517,241,379 and next time when 4 is the last digit. It does this for every number from 3 to 9 as the "jumping" digit, but only once for each. I figured this out when I solved Presh's number (I accidentally used the exact same method as he did :P), since that also has the same property, except that 2 can also do it.
And about the equidistribution.
The *2 sequence has two of all the numbers from 1 to 8 and one 0 and 9.
Your *3 sequence has three of numbers from 1 to 8 and two 0's and 9's.
I wonder if there is a *4 sequence that has 38 digits, four of each, except three 0's and 9's... And will it have the multiplication-by-four property for every number from 4 to 9 being the "jumping" digit?

[@wombat4191]

Apparently not. When I started from 4, I got 102 564 and that gets multiplied by 4 if you move the last digit to front!
Others that do it are:
128 205
153 846
179 487
205 128
230 769
For some reason 4 is special. There is no cycle, except for the ones ending to 5 and 8 and form a pair for whatever reason, nor is there very good equidistribution if we observe the whole pack. On to *5 sequence, if there is one.
Edit: 5 has its sequence(though 7 doesn't get to be the jumper), someone else can analyze it cause I'm tired. I think it would make sense that only primes have a sequence. 7 might be the next and last one as its the highest prime below 10.
102 040 816 326 530 612 244 897 959 183 673 469 387 755

[@chrisg3030]

Ous Alam I think I found a rule which also explains the results in this thread:
N = 1/(ab-1) where N is the Presh number, a is the proportionality (2 in Presh's example, 3 in yours, 4 in Wombat's), b is the base (ten in these examples), a

[@AA-100]

@@thedarkspeedninjashadittsux 1 works

[@Treviscoe]

This is really impressive, especially the proof that the 18-digit number obtained in the first part of the video is the smallest possible number that would satisfy the conditions.

[@obviouslyanonymous]

Wouldn't zero work?
Edit: Just noticed the top comment.

[@maniratnagubba7756]

The question asks for a positive number :P

[@fabijanskrobar1750]

@@maniratnagubba7756 1 maybe, yes

[@geometry1249]

0 is positive. The smallest one.

[@thesos320]

@@geometry1249 If you consider 0 positive then it can be negative too which leads to a contradiction that positive = negative

[@commanderkuplar3790]

Theseas Socratous no it cannot be “negative too” the negative is determined by its sign before the actual number and -0 doesn’t exist. Zero is always positive

[@jannegrey]

MYD: Did you figure this out?
ME: Are You taking the piss??

[@androlsaibot]

My first answer:
49.999~ * 2 = 99.999~
He said the last digit will become the new first digit, but not that the other digits are moved forward, so the old first digit is overwritten

[@TechAllByHarshit]

He also said even number

[@lunatik9696]

@@TechAllByHarshit He said whole number

[@sean_haz]

Even if you were allowed non integer numbers this wouldn't work... He gave the example of 102 becoming 210 so your number would go from 49.9999... to 94.9999...

[@poryg5350]

12x2=24 lol

[@Bob-jy4ju]

He never said it was a whole number. Nor did he say it was even. Plus, you cant talk about parity with fractions.

[@ankylosaurus2342]

The question doesn't specify that the rest of the digits stay the same, so for example 12 is an answer, since 12*2=24

[@redfluxbluedawn414]

Actually it did: if the first number becomes the last (12 becomes 21, the rest don't change, we're not at the doubling part of the question yet), the resulting number (21) is exactly twice as large (21 = 2 * 12).
In computer terms: IF this happens THEN this is the result
You're acting: IF this happens with this result THEN this is the result

[@stevendeans4211]

@@redfluxbluedawn414 As endianness is not specified, Ivo is correct. Also you stated a different problem then the one in the video. It is when the last digit becomes the first digit.
Poorly specified requirements gives unsatisfactory answers.

[@spannycat2]

That's exactly what I got

[@josephhodges9819]

If we use 12 as the number then if you take the 2 and make it the first number then that would be 21 which is not double the number of 12.

[@stevendeans4211]

@@josephhodges9819 But if you make the least significant digit 4, then that is twice 12. The 1 becomes 2, then choose the second digit to be 4. 12 x 2 is 24.

[@stevesmith9447]

How is the solver supposed to know that "becomes" implies move, and not copy or swap?

[@spyseefan975]

because 'becomes the first' does not in any way imply copy or swap(1 year later)

[@ffggddss]

The question does in fact, underspecify the operation in this problem.
You could make "the last digit become the first" by:
1. cyclically permuting the entire string of digits one position to the right
2. swapping the first and last digits
3. reversing the entire string of digits
or any other permutation of the digits that results in the last digit winding up in front, but these three ways are the most likely interpretations.
And while I agree that "becomes the first" doesn't mean that final digit is copied, leaving the original in place; I contend that it could very well mean swapping it with the last.
Although if that were meant, it should be stated with, "and the last becomes first." Which in turn, still admits both interpretations 2 & 3.
Yet another illustration why precision of statement is essential in mathematics.
Not sure whether interpretations 2 & 3 have solutions. Other than 0, that is, which works for all 3.
Curiously, all 3 interpretations are the same for 2-digit numbers.
For 3-digit numbers, interpretations 2 & 3 are the same, but different from #1.
For numbers of 4 or more digits, all 3 interpretations are different.
Fred

[@Tensquaremetreworkshop]

@@ffggddss Another interpretation: the last digit becomes (the same as) the first digit. So 2nnnn4 becomes 2nnnn2
Or nnnn42. or nnnn2. 'becomes' is not a mathematical function. It suggests a change over time, which is not what mathematical functions inherently do.(Yes, you can integrate over a time function, but that is not the same thing at all).
'Move the last digit to the front' is so much clearer, in the same number of words...

[@Tensquaremetreworkshop]

@@spyseefan975 It does not imply move either. 'Becomes' is a time based translation. A caterpillar becomes a butterfly. No movement required.

[@spyseefan975]

@@Tensquaremetreworkshop it does imply move, because if it becomes the first, it can no longer be the last, because then it would be two things at once.(when the number is bigger than 1 digit)
'also becomes the first' would be necessary for that.

[@TheWeizOne]

I have to agree with some other commenters. The phrase "the last digit becomes the first digit" doesn't really imply a swapping of the positions. Going by this wording, the process could be....
Take 25, if the last digit becomes the first digit it becomes 55, not 52. 12345 becomes 52345, not 51234.

[@treufuss-yt]

That's why he provides two examples making absolutely clear what he means.

[@TheWeizOne]

If he made up the question, that's fine. He has, as it were, artistic license. But if this is just his interpretation of the answer, I just wanted to let him know that the question is not as clear as he (and I'm guessing you as well) seem to think it is. If what he means is that the last digit of the number is moved to the front of the number, then he can say that, not that it "becomes the first digit".

[@treufuss-yt]

Check out his sources in the video description. The problem itself is well defined. I agree that his wording is ambiguous but the examples are not.

[@TheWeizOne]

Questions are important, and by that I mean what words we use. How you ask a question will change the answer you get. If you think the problem is well defined but the question isn't, then I'm going to have to thank you for proving my point. The question itself must be well defined if it's going to lead to the answer given. That you can see there is some ambiguity in the question, I think we can agree it would be better to get rid of that ambiguity rather than argue between you and I about what answer might come from the question as it's actually posed.

[@treufuss-yt]

Why do you only focus on the written wording? The examples are part of the video as much as the written text and what he says. Considering everything, there is NO ambiguity like you try to point out in your initial comment. That's all I wanted to say.
If you don't like that the written question alone does not accurately present the problem, that's fine. But saying it can be misunderstood, without taking the rest of the video into account, seems unfair to me.

[@20-sideddice13]

Even bigger problem : if you don't say that the number is positive, there is no solution because you can always find a smaller negative number that will have this property.

[@Mythologiga]

If you include negative then you could say the one with the lowest absolute value.

[@20-sideddice13]

Gautier Blandin of course you can easily resolve this problem, but that was the key point in the lack of the "positive" precision.

[@anitamcsd1]

I

[@yurenchu]

@20-sided dice, ehm... wouldn't the last digit be moved *in front of* the minus sign? Which means by shifting you'd always end up with a *less negative* number (while doubling a negative number would result in a more negative number)?
EDIT: Never mind, the problem description says that the last digit becomes the first digit... so I guess your assessment holds.

[@tychowozniaki9269]

or just take 0

[@WriteWordsMakeMagic]

People mention that 01 -> 10 works for binary which I'm not sure I agree with since 1 in binary is just written as 1, not 01, but there is a cool one on ternary which is 1012 (32) which becomes 2101 (64) when you move first digit to become last digit

[@WriteWordsMakeMagic]

For quaternary, 102 (18) becomes 210 (36)

[@WriteWordsMakeMagic]

For pentary, 102342 becomes 210234

[@WriteWordsMakeMagic]

And just because, in hexadecimal, 10842 becomes 21084

[@realitant]

The problem with this is that there are so many interpretations of the question. Does the first digit change into the last digit? Vice versa? Does the last digit physically move to the first digit and move every other digit over one place? Because of this ambiguity AND the fact that in all interpretations, 0 is the correct answer, I really feel that 0 should be the only acceptable answer.

[@Freestylefisch]

My 1st interpretation: first and last number change places.
My 2nd interpretation: last number stays, first number changes and becomes last number also
My 3rd interpretation was finally right. Then I started the video and he explained it in the beginning.
Don't know if non-positive-numbers were accepted in previous versions. But now the title explicitly says "positive", so zero ist out.

[@Tozzlt]

Which is why I used to hate pure maths tests, because they can be interpreted a number of ways and some dont specify much.

[@Navajonkee]

@@Freestylefisch I went through the *exact* same process.

[@defaultusername1145]

0 is not an answer because 0 is not a positive number

[@zoraackerman1684]

But you can't move over zero unless it's 00 and that's: cheating☺️

[@kaiwut]

0.
0x2=0
0 is first digit & last digit.
Problem solved

[@acaroMan]

Positive

[@purrplaysLE]

+0

[@dyson_sphere]

0 is neither positive nor negative
Thus your answer is wrong
Lol

[@nowonmetube]

@@dyson_sphere nahh

[@dyson_sphere]

@@nowonmetube What do you mean ?

[@pocarski]

I saw the thumbnail and decided to try this myself before watching. I ended up finding a generalized solution for if the number should become N times larger. You take the recurring decimal of 1/(10N-1) and shift its last digit into the first position. This works for every N unless it is a power of 10, in which case the number doesn't exist.

[@cobaltfolly]

Wow, ya.

[@Freestylefisch]

I like that.

[@zacharynguyen7286]

Hope everyone is doing good. Sending support and hearts! ❤️❤️❤️ Stay safe

[@combatwombat8581]

If anyone was wondering, the smallest possible number which becomes tripled when the last digit is moved to the front is 1034482758620689655172413793

[@tassay]

usually i dont post when i solve riddles on youtube but this time i m a little bit proud on solving this at 3 am in the morning when i allready should have been in bed for hours. i found the exact same number by nearly the same way i had 19a=(10^(n-1)-2)b sory youtube is not the best place to write down formulas. then used the modula funktion of my calculator to find it out it out. Thank you for the nice problem and sorry for my bad english its not my first language.

[@unrelatedK]

tassay still some decent grammar.

[@NomadUrpagi]

Kazakh?

[@tomasbeltran04050]

Omg ur calculator does module

[@BigDBrian]

good video, clever algebra gone through in an understandable and easy to follow way.
This is the type of content I enjoy on your channel -- but I think you should remove the clickbait part of the title (the 'Riddle for "Geniunses" ')
That way it can't be confused with your 'viral problem' videos and the like based on just the title.

[@oni_thefirst]

mrBorkD Viral problems are usually very easy or not clear, or have 2 solutions. For a problem to go viral it needs to have an answer that splits people into 2 or more groups that try to prove each other wrong.

[@BigDBrian]

Helgii A
I'm not sure you understood what I meant with my comment.
My complaint is that this video, which is NOT a viral/easy/uninteresting problem, is using the same type of clickbait title for it, whereas I'd think it to be a good idea to distinguish the two types of videos better.

[@coconutk6744]

You can use every digit from 1-9 as the final number to get an 18 digit solution. This is because as you double the number (using the reverse carrying method in the video), you go in a loop that contains all the digits. You just start at different points of the loop depending on the number with which you start.

[@sidkemp4672]

How about in binary? 01 becomes 10, In decimal, that is 1 and 2, and 2 is twice 1. It would be best to specify what base you are counting in!

[@WriteWordsMakeMagic]

This doesn't really work because 1 in binary is not 01, it's just 1.

[@sidkemp4672]

@@WriteWordsMakeMagic I disagree. on computers, bits are stored in bytes and words of various lengths. So 01 is a legitimate expression for the value one.

[@adamduvick]

1^2
But that’s probably cheating.

[@primmebox4734]

2^4

[@jaygreen7494]

1^2 = 1. Move last digit to first gives 1. Double 1 is 2. 2=/=1, so solution fails

[@themuffinman147]

@Michael Darrow In order for 2 to equal 1, it must be in the form of an exponent, where the exponent is equal to 0. Your base equation is invalid.

[@nowonmetube]

@Michael Darrow you're wrong

[@nowonmetube]

2^1
Becomes
1^2
Which is still the same and not twice as big
At least I tried...

[@pekuja]

I think I got it. I did some algebra and figured out what the ratio of the last digit and the rest of the number would look like, which is something like (10^n-2)/19, and basically since the last digit can't be divisible by 19, I kept trying larger powers of ten until I found an integer. That integer was 5263157894736842. 52631578947368421 wouldn't quite work as a solution though, so I tried multiples of 5263157894736842, and just by doubling it I found an answer: 105263157894736842.
Posting this before I finished watching, but I hope that's the right answer.

[@akshajtiwari5199]

Well done

[@eternaldreamerofsleep]

Cool you did it

[@sonalidasgupta3562]

..i got upto 19 | 10^n-2 ..😕(divisible by 19)

[@alexanderworth4352]

good job

[@DomaScholes]

I figured it was 12...
The last digit is 2 , and when it’s doubled it becomes 24, so 2 becomes the first digit...

[@prich0382]

12 would become 21

[@InquisitorThorn]

This works as given. The problem doesn't say anything about what happens to the other digit.

[@ivlev_channel]

It's an obvious result I figured out too. Since we are not restricted with the second number 12 should work

[@blechtic]

The wording is horrible and does not need to be interpreted the same way as in the video, but the real reason why 12 doesn't work is that it's backwards: The last digit must become the first (11 or possibly n1) and that has to be double the original number. Of course, the more not-explicitly-forbidden changes you start making, the more ridiculous the whole thing becomes.
Edit: Of course, "becomes" can be interpreted either way, but the causation has to be such that the change of digit doubles the number, not the other way around.

[@christiandufy]

Why can’t it be 12 ?
Last digit of 12 is 2
12 x 2 = 24
First digit of 24 is 2
D'un i miss something in the problem ?

[@jonaskoelker]

Here's a slightly different way of finding the smallest such integer:
When doing the addition algorithm, the inputs in each column is a digit (0 to 9) and a carry (0 or 1). There are 20 such combinations. Each combination has a uniquely defined next column. Grab a piece of paper, draw 20 dots (one per column input) and 20 arrows (from each column input to the next column input). For example, from "digit 7, carry 1" you draw an arrow to "digit 5, carry 1" (since 7 + 7 + 1 = 15, i.e. 5-and-a-carry), and from there to "digit 1, carry 1", and so on.
This will connect 18 dots into a big cycle, and leave two dots connected only to themselves: one is digit 0, carry 0 and the other is digit 9, carry 1. (Observe that 0 + 0 + 0 = 0 and that 9 + 9 + 1 = 19 = 9-and-a-carry). Since the number has to be positive we can rule out 0. Since the first column starts with a carry of 0 we can rule out digit 9, carry 1 as the start. So we must start somewhere in the big cycle.
Since the big cycle has length 18 all solutions must have lengths that are a multiple of 18. Since more digits means a larger value, we want to look for a number of length exactly 18 (and not 36, 54, etc.). The smallest such number is the one with the smallest leading digit (this relies on all numbers having the same number of digits). Since 0 can't be a leading digit, we pick 1 to be the leading digit. We can see in the cycle that 1 is the result of both digit 5, carry 1 and of digit 0, carry 1. Just like we picked the smallest leading digit, we pick the smallest second digit, i.e. the number that starts with "10..." rather than "15...".
To generate the sequence of digits, we can do this: the digit which "digit 1, carry 0" points to goes in the 1's place. The digit which _it_ points to goes in the 10's place. The digit which the prior digit points to goes in the 100's place, and so on. That is, 1 points to 2 points to 4 points to 8 etc., so the number must end in 842.
Sometimes instead of doing smart math you can just map out all the possibilities (in our case 20), look at them and be a little bit careful about what you do with the results.

[@charlesgaskell5899]

Brilliant! So 1012 (base 3) - 32 in base 10 - is a solution, as 2101 (base 3) is 64 in base 10
(There is no solution for base 2, because of the requirement for the leading digit not to be zero, otherwise the smallest solution is 01 base 2 which doubled becomes 10 base 2)

[@charlesgaskell5899]

An even smaller answer occurs in base 4:
2 x 102 (base 4) - 2 times 18 (base 10) - equals 210 (base 4), that is, 36 (base 10)

[@esyrim]

idk why I thought it was gonna be simple

[@Phoenixfeather100]

I kinda had the steps in my mind figured out but I realized it was long and got lazy and just skipped over to see if my method was right; not if I had the right answer because I never answered it anyway...

[@raismin739]

i think in math the real problem is how you do, not the correct answer. if you know the metod you only need time, if time is a problem you can use calculator, if you don't know the metod to find the answer you an't find the answer never, if you use brutal force to resolve this type of problem you life time is not enought to resolve the problem, i do the same exat thing you do, when i find it will be a 18 digit number i just stop

[@tomasbeltran04050]

I do that all the time

[@ShootMyMonkey]

The way the question is worded doesn't say anything about what happens to the other digits... only that the last digit becomes the first digit. By that reasoning, even 12 works. 12 * 2 = 24, so the 2 has moved from the last digit to the first.

[@TheFrenchPug]

Yes it does. It should equal the new number when the last digit is put in front if the original number was doubled. So 12=21 when swapping last number to 1st. Now does 21 equal 12x2? No it does not. That's how you read it.

[@Exachad]

I came up with 37 which is the closest reasonable answer.

[@Exachad]

And 0

[@mxn4719]

Dawn Ripper 😂

[@mxn4719]

Dawn Ripper 👏🏼

[@chrisg3030]

In hexadecimal (base 16) for example, the equivalent, though not quantitative equal, to 37 is 5B, which is just 1 more than half B5. This goes with 5B*3=111(16) just as 37*3=111(10).

[@spicyoof9083]

Also 12

[@jcn91]

Fairly certain this isn't a riddle.
It's a maths problem....

[@audiblemagician6751]

Fairly Certain it doesn't say riddle in the title

[@jcn91]

Audible Magician Fairly certain you are a moron... The title is, 'What Positive Number Doubles When The Last Digit Moves To The First Digit? Riddle For "Geniuses"'.

[@justinc2633]

the answer is 1, use common sense not the algebra he showed

[@orinscrivellostube4492]

@@justinc2633 1 * 2 = 2....

[@jimi02468]

It is a maths riddle.

[@ayejaybe]

This is a fascinating number as it contains all digits twice except 9 and 0 (which cannot appear twice or the chain would break). It is a recurring number (if you continue doing the multiplications) which means it shows all the nine possible solutions (ending with the digits 1 to 9)

[@Tracy_AC]

For the number to be as small as possible, we want the left-most digit to be 1. Therefore the right-most digit should be a 2 or a 3, since those are the only possibilities for the left-most digit after doubling a number with a left-most digit 1. (100*2=200, 199*2=398) Choose 2 to be the right-most digit. When 2 is doubled it becomes 4, so the next digit is 4 (42). When 4 is doubled it becomes 8, so the next digit is 8 (842). When 8 is doubled it becomes 16, so the next digit is 6, and we carry the 1 (6,842). When 6 is doubled it becomes 12, so the next digit would be 2, but we have the 1 from before, so the next digit is 3, and we again carry the 1 (36,842). Continuing like this, we find the number 157,894,736,842 which has the desired left-most digit 1. But we can't stop yet because at this point we are still carrying a 1 (from 5*2=10), so we continue. We then find the number 105,263,157,894,736,842 which has the desired left-most digit 1 with no carry. But what if we chose to start with 3? Luckily we don't have to check because the answer is already contained in the number we found. Simply take the 6 left-most digits and shift them all the way to the right to make the number 157,894,736,842,105,263. Both of these numbers satisfy the desired property, but since we want the smallest such number, we take the first number as our solution.
One final check we can perform to make sure that this is indeed the smallest possible solution to the problem is to notice that our solution contains every digit twice, once without a carry and once with a carry, except for 0 and 9 which each appear once. This means that we could start with any number as the right-most digit and double it both without a carry and with an imagined carry of 1, and we would find the same sequence of numbers as our solution, only shifted. So our number encompasses every possible solution to the problem, and our chosen solution is the smallest. As for the exceptions of 0 and 9: If the right-most digit were 0 and we doubled it without a carry, then we just get 0, which is the trivial solution. And if the right-most digit were 9 and we doubled it with an imagined carry of 1, then we get an infinite sequence of 9s, which is also kind of a solution to the problem, but it's infinitely big and it's not clear what it means to move the right-most digit to be the left-most digit when there is no left-most digit. ¯\_(ツ)_/¯
The full list of solutions from smallest to largest:
0
052,631,578,947,368,421 (if leading zeroes are allowed)
105,263,157,894,736,842
157,894,736,842,105,263
210,526,315,789,473,684
263,157,894,736,842,105
315,789,473,684,210,526
368,421,052,631,578,947
421,052,631,578,947,368
473,684,210,526,315,789
. . . 999 (kinda, sorta, not really)

[@Tracy_AC]

Yep
But I should also add that due to the cyclic nature of the solutions, we can also generate new solutions by taking a copy of a solution and appending it to the left. So 105,263,157,894,736,842,105,263,157,894,736,842 is also a solution. And we can keep appending more copies to the left to generate new solutions.

[@toriknorth3324]

The way I did the modular arithmetic I branched off at the point 19b = (10^m - 2)a so that I had:
10^m - 2 ≡ 0 (mod 19)
10^m ≡ 2 (mod 19)
2^m*10^m ≡ 2^m*2 (mod 19) (see note)
20^m ≡ 2^(m+1) (mod 19)
1^m ≡ 2^(m+1) (mod 19)
2^(m+1) ≡ 1 (mod 19)
so by Fermat's Little Theorem (a^(p-1) ≡ 1 (mod p)) we have the exponent m+1= 18*k for any integer k>0; in particular, m=17.
Note:
I found that 2^17 ≡ 2^-1 ≡ 10 (mod 19) by taking powers of 2 modulo 19 which is how I knew that 2*10 ≡ 1 (mod 19). Since 2 is smaller than 10 I could take the powers in my head rather than having to use a calculator.
2^4 ≡ 16 ≡ -3 (mod 19)
2^8 ≡ (-3)^2 ≡ 9 ≡ -10 (mod 19)
2^9 ≡ 2^8*2 ≡ 9*2 ≡ 18 ≡ -1 (mod 19)
2^17 ≡ 2^8*2^9 ≡ (-10)*(-1) ≡ 10 (mod 19)

[@hamed2800]

i know i'm 1 year late, but every time i see those nice and clever modular arithmetics like the one u did here, i have to admire it
they never occur to my mind, never, so every time i have to do big modular calculations, in this case 10^m mod 19.
the only comment that actually improved MYD's solution

[@jonathanhall7201]

This is the real answer

[@sonalidasgupta3562]

You did modular magic; sir 🍻 👏...this shift from 10^n to 2^n never occured to me

[@AdrianColley]

That was fascinating, and I think a little beyond my reach (I stopped when I realised I'd need to use a computer). More like this, please!

[@diaaeldintaha7432]

No need for a computer. You can do it by hand.

[@danielbreiner9830]

I saw this question in a science magazine like 20-30 years ago. They actually presented a mathematical proof of the solution rather than the 'easy' way. I can't even begin to remember how they did it though :)

[@tonyandraza3557]

I was just reviewing the question again and realized the confusion comes from the order of the elements of the question. Instead of, "What Positive Number Doubles When The Last Digit Moves To The First Digit?", perhaps it should be "What positive number when doubled, has the last digit become the first digit?"
The original question states that the 'action' is to swap the digits and the result is double. But the challenge is double a number such that the select digit is moved.

[@Zalacans]

We might also "SUSPECT" that it has to be an even number?
I'm out.

[@adamhuang8112]

2:14 how did we get to the conclusion that the numbers would shift by 1?

[@doctorkek]

hoooww???

[@stefankoppel3358]

When you change the Riddle from doubling (x2) to x5, then you get a smaller solution:
142857x5=714285

[@resoltion]

Check this out:
1/7=0.142857 (repeated)
5/7=0.714285 (repeated)
I think that's where you got that from.

[@tontonbeber4555]

@@resoltion Exactly, because of x5 and in base 10 you take 5*10 -1 = 49 = 7^2 . Then you find the development of 1/7 ... with x2 and in base 10 you take 2*10 -1 = 19. And the solution proposed is the development of 1/19 If you take x3, the solution will probably be in the development of 1/29 which is still longer ...

[@Nako3]

That was super easy. I dont get how people struggle with this.

[@hansjo666]

the smallest solution is 18 a.k.a. 102 (Quaternary numeral system)!
so, some people said "hey, he didn't specify the numeral system" and they're right.
But then they claimed that's why they use binary and the solution is "01 -> 10".
But that's wrong, as even in binary, you don't write down leading zeros. this actually means in binary you can't get any possible solution to this problem.
But you can do the maths for all numeral system and find out that 102 -> 210 in the quarternary numeral system is the smallest solution to the problem (and with it's values translated to the decimal system for comparison "18 -> 36" waaaaaaaay smaller than his solution)

[@barttemolder3405]

The smallest solution is 8 a.k.a 13 (quinary numerical system)!!
13(5) = 1*5 + 3 = 8, 31(5) = 3*5 + 1 = 16.
Honors go to Charles Gaskell, who found this a year ago (as I just saw).

[@realedna]

...or 26315 for engineers. It's close enough! ;)

[@xcashew7109]

I️ don’t get it

[@chrisg3030]

While Presh's exact non-zero solution requires at least 18 digits, I'm looking for other examples besides realedna's that are out by +or-1. 26315*2 is 52630, whereas putting the last digit first (LF) gives us 52631, 1 too many. The only 2 digit example seems to be 37, with 37*2 = 74, whereas LF is 73, 1 short. The 1 digit case is 1 since 1*2 is 2, but LF must be 1, so 1 short. The nearest 3 digit case I found thus far is 265, double is 530 and LF is 526, 4 short. Can anyone do better?

[@magikey7395]

@@chrisg3030 37.
Edit: Didn't read the whole comment, my bad

[@chrisg3030]

@Magikey At least you can tell me how to put the addressee's name in blue at the beginning, I just can't figure out how to do it. Your UA-cam name is auspicious.

[@magikey7395]

@@chrisg3030 If you're mobile, just tap the three dots next to the comment and tap "reply", if you're on computer, then I don't know

[@Walkerman379]

I got to the point where I realized I need 10^x = 2 mod 19 and “screw it, I’m not going through the rest of this.”

[@craigandsnowwadam4511]

Seriously!! I’m just happy to get one answer right ,…and I’m dyslexic!! So ,..1 answer,… ..left? ,..I pick the red one ,..on the right.

[@joshuawenninger6439]

4:58 And here I was trying to count through the numbers until I reached the answer

[@TehAarex]

My guess is 105263157894736842

[@kaboomwinn]

Aarex Tiaokhiao lol

[@dannyboy12357]

Me too, rather obvious.

[@thesos320]

It is just simple.

[@Juuhimuuhi]

Yea I thought of that immediately.

[@dominicjohnmutholil]

that's the trivial case luckily

[@vortexoku5570]

This one really had me thinking... after 25 minutes I gave up and I’m kinda happy I did😅 would be here for a loooooong time counting otherwise..

[@WriteWordsMakeMagic]

I got as far as searching for the prime factors of 499...9 to find one containing 19 but no site was able to keep going high enough to find it. I didn't think about modulo arithmetic!

[@cyclingcycles7953]

Me: Oh this is fairly simple, It's just gonna be a 2 digit answer
**Tries every 2 digit number**
Me: Maybe I'll look at the answer and try to understand the math
**Sees the answer**
Me: Nope, abort mission, ABORT ABORT!!

[@matthewlyons6544]

This never said the number had to be an integer...

[@angelgodplace]

There is no non integer solution

[@dirkb8410]

There is actually a non-integer solution. A number such as .10526315789436842 would become .21052631578943684 if either it were multiplied by 2 or if the last digit were moved to the place of the first digit. That's obviously a smaller number even though it's a bit of a cheesy solution.

[@gil1374]

wouldn't .10526315789436842 technically be 2.1052631578943684, since all decimals are represented as 0.2345678 and so on?

[@takinggaming6640]

Or you just completely take away the decimal and get the actual answer

[@godavid8130]

whole numbers are just positive integers plus 0

[@alzblb1417]

Can you generalize this and find smallest such numbers but 3 times larger, 4 times, 5 times, etc. ? Do they share any interesting properties?

[@__-nt2wh]

The general solution for N = a*(10^k - 1)/(10α - 1), where N is the whole number, ie. ba, α is the proportionality factor (2 times, 3 times, 4 times larger), k is the number of digits of N, and a is the last digit.
Note how 10α - 1 | 10^k - 1 for α > 1.
Therefore, generally, k = φ(10α - 1), where φ is Euler's totient function.
EDIT:
As yhtomit28 and others pointed out, the totient function only provides the upper bound for k, so k 1
therefore, a - α < 0 a < α is the condition.
Point me out if I missed anything.

[@yhtomit28]

I was really confused for ages, until I realised your N is effectively "ba" (using the notation) from the video. Which means your m is one greater than the one in the video (maybe it would have been better to switch to p or k or something?).
I'm very rusty on Euler's totient function (is it Fermat's Little Theorem you're using or something to establish that connection?), but I think using the totient function doesn't always give a *minimal* m. Some quick fiddling in Excel suggests the following m-values for given α. I've indicated where this is different from the totient of 10α-1.
2: 18
3: 28
4: 6 [ φ(39)=24 ]
5: 42
6: 58
7: 22 [ φ(69)=44 ]
8: 13 [ φ(79)=78 ]
9: 44 [ φ(89)=88 ]
So an example with α=4, m=6, a=4:
4*102564
=410256
Your little proof at the end is very nice though, and nice work on the lining up of digits in your example. So it's just a question of the totient stuff and how you find the minimal m.

[@ryuusel]

flashdrive Shouldn't it be 10^m-alpha?

[@alzblb1417]

theres a sequence for the numbers and for the number of digits in OEIS

[@ryuusel]

Lord Hamster What is it?

[@pentamath]

If you use ternary, the smallest possible answer would be 1012 x 2 = 2101.

[@shazebkhan2745]

It is interesting to note if a=1
Then number is
052631578947368421
(mind the first zero)
Which on multiplying gives
105263157894736842
Hence if that zero is taken into consideration it can become smallest number to satisfy property.

[@username17234]

You forgot to mention that works because 19 is prime, which implies it has to be a factor of one of the terms. Or else it could just be a combination of smaller factors from several terms in the numerator.

[@gmutubeacct]

10'52'631'5'7'8'94'73'6'842
This sequence (wrapped around) will give you all possible such numbers. The ' after a digit indicate that there is a carry associated with it. Just for this discussion, let us call the digits without an ' a pure digit. So 3 would be a pure 3, but a 3' would not.
Now, find a number ending in 3 that has this property, you would start with the pure 3 in this sequence and keep going left (with wrapping) and stop right before you hit that pure 3 again. That will be your answer - 1'5'7'8'94'73'6'84210'52'63
You would do the same thing for finding such numbers ending in each of the digits ( other than 0).
It is also interesting to note that to form numbers greater than 18 digits, you would have to wrap around the same pattern multiple times. Hence, such numbers would have to be 36/54/72 digits with the same pattern repeating. You cannot have any other numbers that satisfy this property!

[@Freestylefisch]

I wanted to ask why there can't be bigger numbers with this property. But I just hadn't read your whole comment. So, the sequence, repeated twice, also has the property: 105263157894736842105263157894736842 * 2 = 210526315789473684210526315789473684, lol.

[@luker.6967]

I came up with a way more complicated formula in the spirit of method 2, which works for all rational numbers with greater than one (non-zero) digits. I got stuck after coming up with the formula because it was too hard to work with. Where R is the original number, and R,k is the the last digit, we have:
2R = 1/10 * (R - R,k) + R,k * 10^floor(log(R))

[@andreytsankov6205]

Now I want the guy to read the answer in German and I am giving this man all my money

[@brandongalvan6603]

Except when you word the question incorrectly by neglecting to mention that the first digit becomes the last, so the answer actually is 12.

[@mxn4719]

The smallest number is 0 because 0 x 2 = 0. As well as that, 00 (0) if the last digit is swapped around u still get 00 (0).

[@atomkom773]

But we are trying to find smallest *positive* number and 0 is neither positive or negative so your statement is false

[@boltstrike2787]

You didn't specify in the problem description that the other digits in the number have to shift backwards one place, just that the last digit becomes the first digit. Remove the requirement of the other digits shifting backwards and 12 becomes the smallest positive number (Double 12 and you get 24, where the 2 goes from the last digit to the first digit).

[@kirthiramaniyer4866]

MindYourDecisions - thank you for posting interesting questions, that will really help many math lovers.

[@marctelfer6159]

1... in binary

[@MrDiarukia]

But 1 * 10 = 10. And the shift is 1 -> 1.

[@100najaja]

But it's with one leading zero :)

[@suprguy]

Oekanos where is it stated that it cant have a leading zero? hence leading zeros are standard in binary

[@100najaja]

I was replying to MrDiarukia. Answering your question - well, I think that if we're honest we can clearly see that this is not what the person giving the problem meant :)

[@MrDiarukia]

supr guy the leading zeros in binary are only present in computer field. The binary numbers as such have no leading zeros

[@miradarya]

I think this variation of the problem is even more interesting: What is the smallest number that quintouples if the last digit becomes the first digit? Other multiples give interesting results as well.

[@ipadair7345]

0

[@daneshannonsparks]

1 year later: wait, it's 142857. Oops.

[@idahawk5193]

"Can yo figure it out? Give it a try!"

[@toatrika2443]

This earns a special place on the top 10 anime betrayals

[@moinfahad]

@@toatrika2443 true

[@Jordan-yi8te]

What about 12? When it doubles the last one becomes the first

[@sakshamsinghal5418]

That's not how it works when the last digit becomes the first all the other digits stay the same in order

[@cowflick1180]

Lmao I only watched the thumbnail and I thought it meant that the last would become the first, but the first would remain the same, so like 25 would become 22, and I thought it was impossible

[@pianoforte5964]

Could the answer be 0??
As we can write 0 = 00
and 2x0 = 0
So.. is it possible?

[@pierluiginegro772]

Funny story: my mind was set on resolving this by bruteforcing it. Getting the answer was a race between the program I started and the video that resumed. Just one small problem: my program would have only taken a lot more than 761 thousands of years, so I guess it wasn't a race at all 😅

[@MisterFourSeventeen]

I can't believe we have to "resort to" 2/19 and 4/19.

[@danielglazar6811]

"The last digit... becomes the first digit."
I thought it was like 25 -> 55 or 102 -> 202. That would be impossible to solve, hence my puzzled look before watching.

[@Noswiatel]

Yeah

[@Pengochan]

a has to be at least two. After doubling a number the first digit can only be 1 when the result is longer. In fact the 17 digit number would work if a leading zero were allowed, and then its just a circular shift. Other solutions can be generated from the 18 digit solution by circular shifts such that the first digit is

[@wannabefunnyman]

12 satisfies that requirement, no? When doubled it’s 24. I believe this also qualifies as the smallest number that satisfies the requirement.

[@numbdigger9552]

Just no.

[@Mr_R_Oil]

Yeah, just no.

[@JBKiser]

No.

[@wonderful_whiscash9534]

Yeah, the riddle doesn’t specify it having to go on forever, so 12 works fine

[@CrabmanJenkins]

If only I understood the phrasing better lmao. I initially guessed 12