Can you solve the faulty odometer puzzle?

[@asterpw]

Just treating it like a base 9 number is by far the easiest. Did it in my head in like 15s

[@DoomRutabaga]

Me too, clicked because I wondered why it needed 12 minutes of explanation lmao

[@bretnufer7044]

Hardest part of the mental math was calculating 9^3 I work in automation and controls, there is a particular controller that I encounter often that uses octal addresses, immediately recognized the problem as a number base conversion 😊

[@malvoliosf]

Yeah, I did it on my calculator in like 10s. Which of us is smarter?

[@leojei]

Finally a problem which I can proudly finish in under a minute!

[@artex98]

I just looked at the thumbnail and calculated the answer only using my spinal cord (

[@R3_dacted0]

Normally when I see these videos pop up in my feed, I stare at the thumbnail and think about how to solve it until finally giving up and clicking on it to see the answer. But this time I immediately thought to use a base 9 to 10 conversion. It was a complete hunch that I wasn't sure was correct, but I'm happy to see that it was indeed one of the proper ways to do it.

[@JohnDlugosz]

Same about the thumbnail, but I was confidant base 9 was the answer, shifting down all the digits to make 9 the missing one, so adjust it by turning down every digit that's greater than 3 by 1.

[@MalachiTheBowlingGod]

Base 9 FTW!

[@TheBierp]

Same on all points here.

[@algorithms-memo104]

@@TheBierp My method was closest to method 3, realizing we need to calculate in base 9. Four-digit base 9 numbers have a 729's digit, an 81's digit, a 9's digit, and a 1's digit. So to get from 0000 to 2005, we just need to raise the first digit by 2 and the second digit by 4 (5 being the symbol for the fourth number). So it's just 729 x 2 + 4. I actually did one of these and got the right answer! I guess my feed will be filled with puzzles now.

[@RussellDuffer]

The base 9 method was my initial solution method.

[@jimdecamp7204]

2004 base 9 = 2 x 9^3 + 4 x 1 = 1462 base 10, by inspection. The odometer displays mileage base 9, not 10, with digits having value [ 4 5 6 7 8 ] represented by [ 5 6 7 8 9]. There is no digit representing the number nine, base 9. If you want to see the video before it goes public, become a patron. Just a note: I have not looked at the video, and see no reason to. I got this problem in the mail because I am a patron. You should be too. I have only seen the problem, I have not bothered to look at the solution, but it appears to be trivial. I could be wrong.

[@Awhan..]

bro the video is uploaded 13 sec ago , how do you have 3 days ago comment

[@Emka877]

Why is your comment older than the video itself

[@tigerinthejungle_14]

@@Awhan.. I think members have early access to vids

[@micke_mango]

Exactly

[@57thorns]

I forgout abuout the remapping and got 1463 in my head, good point.

[@verkuilb]

Another way, which does the math essentially like the second method but works the logic more like the first method, is to start at 2000, then subtract/eliminate the numbers with 4 in the ones position, by eliminating 10%, which is the same as multiply by 0.9. We’re now at 1800. Next do the same to eliminate the 10% of the remaining numbers which have 4 in the tens position. So we multiply by 0.9 again, and are at 1620. Do it once more for the remaining numbers with 4 in the hundreds place, and we have 1458. Now we just add 4 for 2001/2002/2003/2005, and we get the answer of 1462.

[@rogergeyer9851]

That seemed too obvious to me when I quickly thought about it. My age induces lack of practice and lack of confidence vs. when in college 45ish years ago and frequently working mind-stretching problems.

[@flmbray]

In addition to immediately seeing the base-9 solution, I also came up with a secondary solution that is essentially the same as what you are doing, however the 10% figure wasn't at the forefront of my mind. I started from 2005, and looked at each digit starting from the left. How many times did 2 bypass 4? None, so we are ok - that leaves 2005. Now the second digit - how many times did that end up skipping 4? Well, it's the number to the left of it - 2. So we'd subtract 2 from that position, which is equivalent to 2005 - 200 = 1805. Now the third digit... How many times did that skip 4? Well now that we've corrected the first two digits, we can see that it was 18 times. So we subtract 18 from that position, which is the equivalent of subtracting 180, so now we have 1805 - 180 = 1625. Now the last digit.. since we've now corrected the first 3 digits we see that it must have happened 162 times, plus one more since the 4 was skipped going from 3 to 5. So 1625 - (162 + 1) = 1625 - 163 = 1462. As I said it's essentially the same as plucking out 10% each time but from a slightly different perspective.

[@maxhagenauer24]

@@flmbray Everybody already did it the base 9 way and I think that way is harder and worse than this guys way.

[@maxhagenauer24]

Why on earth does 2005 - 2005(0.1) - 2005(0.1)(0.1) also not work then because that's the exact same thing?

[@flmbray]

@@maxhagenauer24 Cool your jets bro... I didn't say his way doesn't work or isn't as good. All I said was that I arrived at the base-9 solution (which to me is the easiest. but everyone can have their own opinion), and then I came up with a second solution to test/confirm my base-9 answer, and my second solution was "very close" to his 10% method. All is good!

[@joaocervi115]

I'm coursing Systems Development & Analysis, so I did the base 9 method right away... Very nice problem!!!

[@rogergeyer9851]

Damn. Very nice. I'm just an old man (70's COMPSCI major) keeping my mind active in various ways. I haven't watched past the statement of the problem yet. I hadn't even THOUGHT of just using base 9 math, but in hindsight, that seems VERY intuitively obvious, re an approach. I was thinking about how unpleasant (and SO not "in the book" re Paul Erdos) doing it by brute force would be... Brittleness in thinking is an unfortunate reality as we age...

[@angrytedtalks]

I studied computers from 1979 and went straight to the base method. Since _all_ the odometer wheels were faulty I just assumed it was a base 9 question.

[@anonymoushere7786]

​@@angrytedtalksWhat does this base nine mean? Please explain

[@angrytedtalks]

@@anonymoushere7786 We generally use base 10. Count on your fingers & thumbs of both hands and then count the number of hand pairs you have used "tens". Base 9 would be like losing a finger and counting to 9 before setting a 1 in your "nines" column and resetting your units. 9 in base 9 is "10". 2 in base 2 (binary) is "10". 16 in hexadecimal is "10". We call number characters "digits" because it means "fingers".

[@anonymoushere7786]

@@angrytedtalks I got it but how does base 9 come here? I have never heard of it before

[@user399878993]

Another way to do it is to visualize it mechanically; counting how many miles were lost due to “jumps” where the odometer missed miles. We start with the hundreds, and the odometer “jumps” twice, at 400 and 1400, and so 100 miles are lost each time. That leaves 1805 miles remaining. For that 1805 miles, it jumps 18 times in the 10s spot, so 180 miles are lost. That leaves 1625 miles, but of those, 163 miles (counting the partial revolution as one) were lost due to jumps. Subtracting 163 from 1625 gives 1462. It’s the same calculation as multiplying by .9 each time, but just wanted to provide this parallel solution since it’s yet another way of thinking about the problem.

[@garyrolen8764]

I did it this way.

[@larsvb8128]

I did it this way in like 2 minutes.

[@AbCat4]

I stopped studying maths at 16, and was able to solve this in 5 minutes 30 years later. Pretty pleased with that. :)

[@ren7a8ero]

I took more time understanding the first method than solving by the second one, but it is nice to see things in a different manner.

[@josephfredbill]

I love this. Just one point needs explaining in the base 9 to base 10 calculation imho - the diiference of one in the digits means all the digits > 4 map to one lower, which is why the 5 becomes 4.

[@crtwrght]

Part of what I like about this problem is it illustrates rather nicely how much is removed when you take out numbers containing a certain digit, or even multi-digit numbers. The harmonic series is famously divergent when summed to infinity but fewer people are aware that if you sum the reciprocal of all positive integers that don't contain a "9" the sum is convergent to about 22.92. Similarly the sum of reciprocals of all numbers not containing 123, 4444, or any finite string of numbers are convergent. Look up the Kempner Series if you want to see more.

[@flmbray]

Without looking it up, I suppose that implies that the sums of such numbers (i.e. sums of reciprocals of numbers that contain 9) must also be divergent to infinity because otherwise we would have an infinitely divergent minus a finite number, which would still be divergent. The fact that such sums would be infinitely divergent makes sense, at least for single digits, since the fraction of numbers that contain a particular digit approaches 100% as the range gets larger. I can kind of see how that would extend to multi-digits but can't get there in my head.

[@crtwrght]

@@flmbray The basic intuition for it is that when you consider, for instance, 100 digit numbers, almost all of them must contain every digit. That is to say the probability that a 100 digit number doesn't contain a nine is roughly (9/10)^100=0.00266%. The argument for multiple digits follows a similar line of thought, but obviously the limit of the sum increases drastically. Wikipedia gives the limit for sums of 1/n where n doesn't contain "42" as 228.4 and for not containing "314159" as 2,302,582.33.

[@Keldor314]

​@flmbray If you squint a bit, you can think of removing numbers with a given 2 digit sequence as being very similar to removing a single digit if you converted everything to base 100. Note that in the base 100 case, you're actually only removing half as many numbers as base 10, since the removed digits would have to be aligned with powers of 100. This is, suppose you were removing the sequence "11", and we're checking the number 9118. We can see that in base 10, there's a 11 in the middle, but in base 100, the number becomes 91 18, so it is not removed because neither digit (91 or 18) is 11. Thus, removing numbers with a given single digit in base 100 removes exactly half as many numbers as removing numbers with a given pair in base 10. Likewise, we can compare removing a 3 digit string to removing a single digit in base 1000, although in this case, we're removing 3 times as many numbers since there are two strings crossing the base 1000 word boundary for every string that's aligned. 4 digit strings remove 4 times as many numbers as single digits in base 10000, and so forth.

[@flmbray]

@@Keldor314 ok yeah converting to a higher power-of-ten base and then only removing "1 digit" is a perfect intuition. Thanks!

[@GameRoMan]

@MindYourDecisions Please change from 0:00 problem 1:14 method 1 7:23 method 2 9:09 method 1 11:12 simulation to 0:00 Problem 1:14 Method 1 7:23 Method 2 9:09 Method 3 11:12 Simulation

[@MichaelSmith-fj7di]

Since the number displayed is less than 4,000, I did it this way: Per thousand miles, there is a deduction of the following: -9*9=81 Occurances Where the Ones Place is Supposed to be a Four and No Other Four Is in the Number -10*9=90 Occurances Where the Tens Place is Supposed to be a Four and No Other Four Is in the Number -100 Occurances Where the Hundreds Place is Supposed to be a Four I then doubled each of these and subtracted it from 2,005 and then also subtracted another one. That got to 1,462.

[@toomanyhobbies2011]

Thank you for your description. All the previous commenters didn't bother explaining anything, they just made claims of solving in a few seconds, then criticized the MYD solution. I'm too lazy to actually find the solution, the methods are more important (typical physicist).

[@TWANDTW]

I found the solution right away with the third method. As a programmer, I saw it as a problem of converting between number bases almost instantly.

[@Peter-g8g]

Me and my friend just bought one brand new car each but I hacked mine to skip the digit 4. We drove together until my odometer showed 2005 miles and told my friend to check his odometer and it showed 1462 indeed. I was sceptic about the simulation method but I can confirm that one!

[@thomasharding1838]

First off, you should say "My friend and I". After all, if it was just you by your self, would you say "Me just bought..." Secondly, Sure You Did!!!

[@UmarFarooq-bb9qs]

I used combinatorics. Exclude 4 from each of units, ten, and hundreds, so that you get 9*9*9. Then you multiply that by 2 since the thousands digit is 2, and then you add 4, because the calculation assumes each of the digit is in effect 0. Not very rigorous but it got me the answer.

[@wyattstevens8574]

Exactly what I did- I actually followed this train of thought because of Numberphile's "almost all numbers contain 3" video!

[@Ou8y2k2]

I used combinatorics to get the wrong answer in < 30 secs.

[@MrNostril]

I loved the first way. I immediately jumped to converting to base 9 before I watched the video and when you started doing it that way I was like, "well, that's obviously the hardest way to do it." but the way the Venn Diagram worked with the union of set was very interesting so I'm glad it was included. I also thought it was interesting that the second method and 3rd method are similar in terms of the arithmetic you have to do. They both are different way to show that the answer is 2*9*9*9+4.

[@timecubed]

Much like a lot of people here, I did a base 9 to 10 conversion. I simply shifted the digits 5 6 7 8 9 down by 1 to become 4 5 6 7 8, and converted the resulting number to base 10. I also wrote a tiny little program to double check (it implements the logic of skipping the digit 4 regardless of digit position), and putting in 1462 gave me back the same number, 2005, back.

[@hughiekw]

I kind of cheated. Opened Excel, made 2 columns from 1 to 2005. Selected one column and replaced all the 4's with an "a" then sorted that column from smallest to largest. That sent all the cells with an "a" to the bottom. Selected them and deleted. Then looked to the other column to see how many were left. I know I'm not math smart like some of you guys, but I got the answer.

[@filipfabri4998]

ok but what is the number on the odometer is 208151419406351419362517206361916381910017161. How much would it take you to figure it out?

[@bencollier9423]

@@filipfabri4998 in that case: either that oddometer is faultless or has a diffent problem from the one posed, possibly involving transposing digits, or running in a base greater than 10. or you've been travelling through the universe at the speed of light for 2.5exp21 universe ages, so you probably have plenty of time on your hands!

[@somefoolishfool]

@@filipfabri4998The solution is to buy a new car

[@michaelmiller2210]

Resourcefulness is intelligence too. You got the answer, that's all that matters.

[@itsphoenixingtime]

As someone who worked with excel sheets for a while the sort and replace functions are a godsend

[@AkashG23]

Funny thing: I was working on a presentation about bases with my IRL friends for a math club. We used this exact problem at some point in the presentation.

[@wyattstevens8574]

Down to skipping 4 and ending at 2005?

[@AkashG23]

@@wyattstevens8574 Yes (we cited the competition)

[@jasejj]

As I'm a programmer (formerly assembly and embedded systems, latterly devops for an ISP) and network engineer, both occupations that require moving between different bases as a routine part of the job, the base 9 conversion was my immediate response to this. Despite having a degree in maths the first solution didn't even occur to me!

[@RadmanTheWise]

I finished scratching my head about hexadecimal notation about 15 years ago so I got this nearly instantly from the thumbnail.

[@GameRoMan]

Please change from 0:00 problem 1:14 method 1 7:23 method 2 9:09 method 1 11:12 simulation to 0:00 Problem 1:14 Method 1 7:23 Method 2 9:09 Method 3 11:12 Simulation

[@rextanglr4056]

So basically the odometer counts like a Chinese elevator?

[@thesun___]

I solved it by counting all the way up to 2005 while consciously skipping each number with a 4 in it.

[@jacktaylor9290]

I felt so proud for solving this until I looked at the comments...

[@notcoachfou7841]

I instantly recognized this as a base 9 problem and came up with the answer fairly quickly. However, I program PLCs and am constantly converting between base 2 and base 10, and also very familiar with base 8 (octal) and base 16 (hexadecimal). So removing 4 instantly jumped out at me as a base 9 problem.

[@petrkdn8224]

im a python noob, but i tried computing it using python, and got the correct answer on the first go. no errors at all. I bruteforced it as in your simulation. i kept track of every digit and skipped number 4. Got to 1462. I then asked microsoft copilot, explained the problem and asked it to come up with a solution. It also worked on the first try.

[@CTJ2619]

well done !! I liked the 3 related approaches to the problem

[@katy14ioana]

I'm glad that there are still people who can enjoy math videos 😁

[@KataisTrash]

Yeah I went with the base 9 idea right away as well, nice to see that worked out.

[@loiscampos4553]

I just see the 2004 number (by the offset error) and went with the base 9 number system and make the table: 9^4 9^3 9^2 9^1 9^0 6561 729 81 9 1 * * * * 2 0 0 4 So (4*1) + (0*9) + (0*81) + (2*729) = 1462 And to verify if the number is correct i just keep dividing the number by 9 taking and combining the residue starting form the last residue. 1462 / 9 56 162 22 4 162 / 9 72 18 0 18 / 9 0 2 2 / 9 2 0 Residue numbers are 2, 0, 0, 4 so 2004 is the number in base 10 system, which is correct.

[@JoeBorrello]

My approach was to simply treat the odometer as a base nine odometer where the displayed digits five through nine actually represented the digits four through eight, since numerals are arbitrary representations. That meant that the display 2005 merely represents what we would call 2004 base nine using our conventional numerals. Basically the same as what you did, but s slightly simpler logic.

[@Blahaj_IKEA]

I know how to solve it, but I'm half asleep, so imma let you do it for me

[@Henry-de4ox]

The way I did it was similar to the second method, but instead I counted how many "ticks" of the units digit it took to get to 2000, and then I added 4. It takes 9 ticks to go from 0 to 10, since you skip 4. It takes 9x9 ticks to go from 0 to 100. You count 9 ticks to increment the tens digit once, and you increment that one 9 times. It takes 9x9x9 ticks to go from 0 to 1000 because you increment the hundreds digit 9 times. 9x9x9=729 and displays as 1000, so for 2000 you'd need 729x2=1458 ticks. Then I manually added 2001, 2002, 2003 and 2005 to get to the total 1462.

[@verkuilb]

There are multiple possible answers. It’s a 6-digit odometer which may have rolled over-so “if the faulty odometer displays 002005” that could mean it is “displaying” 2005; or 1,002,005; or 2,002,005; etc. 1,002,005 (base 9) is 532,904 (base 10); subtracting 1 (for 1,002,004) is 532,903. For 2,002,005, the conversion and subtraction gives us 1,064,344. You can figure out more on your own if you like.

[@洋曾-v7t]

Wrong. The task says "...displays the number of miles...", not "...displays the last six digits of the number of miles...". You might think you're extra smart, but actually you're not.

[@DrMcFly28]

​@@洋曾-v7tif your odometer has six digits it will always display "last six digits".

[@flmbray]

One little nitpick - the proper procedure for using the base-9 method would be to remap the digits first (i.e. the mileage would be 1,002,004) and then perform the base-9 conversion. In this particular problem it makes no difference, since it's only the 1s digit that is > 4. However try this a mileage value where the other digits are > 4 and it becomes much harder if you don't do the remapping first.

[@verkuilb]

@@洋曾-v7tWrong. The sentence at the top is explaining what an odometer is, generically, in case some parts of the world use a different term. It’s not intended as a limiting definition of the problem. After all, if it was a limiting definition, it would be false, as the next sentence immediately says that yours is faulty-meaning it DOESN’T display the number of miles driven. The true limiting definition is the actual question at the bottom: “If the odometer displays 002005…”. Which would happen at 2005, and at any x+2005 where x is a positive integer.

[@elpme16]

I thought of base 9 but didn't actually work it out that way. What I did was go through each digit place from 1000's to 1's. The thousands column doesn't skip 4 at all, so we don't need to worry. The 100's column skips 4 twice, which accounts for 200 miles (400-499 and 1400-1499). The 10's place will skip 4 18 times, which accounts for 180 miles (4 in the 10's place occurs once in each of the two skipped cycles of 100, giving us 200-2*10 = 180 to avoid double counting). The 1's column skips 4 162 times before 2000, which accounts for 162 miles (4 in the 1's place occurs 10 times in each of the two skipped cycles of 100 and once in each of the 18 skipped cycles of 10, giving us 200-2*10-1*18 = 162 to avoid double counting). Finally, we skip 2004 before getting to 2005 accounting for the last mile skipped. So we take 2005-200-180-162-1 = 1462.

[@flmbray]

I used this exact method also... to verify my base 9 solution.

[@oldjoec3710]

Yes. That is a version of Method 1 that's actually quite competitive with both of the other methods, since intersections are eliminated before they occur. Only three easy subtraction steps.

[@coltranius]

This is the first one I’ve ever gotten right right off the bat, by converting it from base 9 and subtracting one, since we just passed 3 in the 9^0 place.

[@adamyaser7136]

I made a bit diffrent, that was fitting my understanding haha. 2000 has 2x1000, and every 1000 you have to subtract 100 so you are left with 1800. 1800 has 18x100, and every 100 you have subtract 10 so you are left with 1620. 1620 has 162x10 and every 10 you have to subtract 1, so you are left with 1458, and the 5 from 2005 becomes a 4 so you have 1458+4=1462

[@ruferd]

As soon as it said 2005, I thought : "this has to be from a math comp in 2005, like AMC" Was not disappointed.

[@KaeptnTerror]

I figured there were 9 numbers that aren't 4 in the ones, tens and hundreds. → 9x9x9 There are also two thousands. →9x9x9x2 And last but not least 4 numbers over 2000 that aren't 4 →9x9x9x2+4= 1462

[@lupus.andron.exhaustus]

Using base 9 was also my first inspiration. And I would call it the most logical solution.

[@jacoboribilik3253]

I used the second method.

[@xiaoshen194]

Hi Presh, thanks for the video. Hope you are doing well ❤

[@wesleydeng71]

Using the base 9 method to solve the question "how may numbers does not have 4 no more than 2005" would be smart.

[@arakien1]

There's an easier way. In every 100 numbers you get 9 leaps in the ones digit plus 10 between 40-49. So in every hundred digits there are going to be 19 leaps. In one thousand digits there are going to be 9*19+100 (400-499). This adds up to 271. You double this to get to 2000, so it gets 542. Then you add 1 for the 2004 leap. Then you subtract 2005-543

[@ck3908]

same here, can even do in your head this way.

[@heinrichmartin]

In the first method, skip the set arithmetic by not counting the numbers multiple times: #"4 at unit position" = 2*10*10, #"4 at tens position and not yet counted" = 2*10*9; ... 2*10*10 + 2*10*9 + 2*9*9 = 200+180+162 = 542

[@Smallpriest]

My favorite was method 2, so clean, so simple

[@EaglePicking]

Method 3 was cleaner and simpler, but am I right in assuming that you like method 2 because you're more of an engineer than a mathematician?

[@Smallpriest]

@@EaglePicking nah im not an engineer or mathematician, i just think method 2 is easier for most people to understand because it doesn't use base 9 (most people aren't familiar with non base 10)

[@MarieAnne.]

So the odometer simply shows a base-9 number that uses digits 0-3,5-9 instead of 0-8. We simply calculate as usual to translate from one base to another, except that any digit on odometer > 4 must first be reduced by 1. Example: 836 -> 735 (base 9) = 7*9^2 + 3*9^1 + 5*9^0 = 599 km 2005 -> 2004 (base 9) = 2*9^3 + 0*9^2 + 0*9^1 + 4*9^0 = *1462 km*

[@Llortnerof]

My only question would be where you got the unit from, since the original number is dimensionless.

[@TheBioRules]

The method I did that was similar to method 1, but in my opinion easier, was as follows: We start with the numbers from 0 to 99, aka the 2 digit numbers. Between 0 and 39, a total of 4 numbers are skipped (4, 14, 24, 34, aka the numbers ending in 4). Between 40 and 49, an additional 10 numbers are skipped. And finally between 50 and 99, a further 5 numbers are skipped (54, 64, 74, 84, 94). That means for every 100 numbers, 19 will be skipped. We then expand from 0 to 1000. For each hundred place from 0 to 3, 19 numbers will be skipped, so 19x4 = 76. For 400 to 499, all 100 will be skipped. And from hundreds place 5 to 9, another 5 lots of 19 will be skipped (95). So 76+100+95 numbers skipped in the first 1000, or 271. Double that for the next thousand up to 2000 for 542. Add 1 for 2004 being skipped, giving 543. Subtract that from 2005, and you get the answer of 1462. I find this "iterative" approach a lot more intuitive and scalable for this kind of problem. It also hearkens back to a Numberphile video on "how many numbers contain a 3".

[@luketurner314]

Like many others, I too immediately realized the base 9 method and used Python to do the math (as I usually do instead of a classical calculator): (9**3)*2+4 which yields 1462

[@clivegreen7139]

I managed to get there using probability: For the 1000 numbers {000 to 999}, the odds of randomly picking a number containing no 4 digits at all is given by (9/10)^3 or 729. Put another way, (1000 - 729) = 271 numbers in this range do feature at least one 4 digit. Apply identical logic for the range {1000 to 1999}. That's a total of (271 + 271) = 542 so far. Finally, in the range {2000 to 2005}, only one number has a 4 in it. That gives a grand total of (542 + 1) = 543. Then 2005 - 543 = 1,462.

[@stevenpurtee5062]

I'm proud to say that I immediately said "That's just base 9. Well, 5 is really 4, so I just need 2004 in base 9". Got the answer in less than a minute.

[@adamv242]

Nice puzzle, one of the few that I actually solved by the 'correct' method.

[@thomasharding1838]

The simplest way to determine the actual mileage on a car with < 2005 miles is to take it to the dealer and get them to replace the faulty odometer and determine the actual mileage so you won't get shorted on the warranty.

[@jaimecarter3988]

Subtract 1 right off the bat (2004). Then subtract 200 for all the 400s and 1400s, leaving 1804. For every 100 miles, 19 miles are skipped (all the 40s and every number with a 4 in the one's place). 19x18=342, 1804-342=1462.

[@sebastiannielsen]

Actually trying to think how an odometer could fail in such a way that it double count at number 4. The last digit could actually never get a such failure, since its the first wheel and directly actuated by the measuring sensor (gearbox wire or wheel axle wormgear), so the last digit will always be accurate, no matter what, unless that wormgear has missing parts, in which it would not double count at 4, but rather skip a mile there, so the meter would show less, not more. The 1 to 10 carry-over works by having 2 extra cog pins, sticking out to the LEFT of the 1's wheel at the position where the 9 would be visible, which will actuate a cogwheel that touches both the wheel for 1's and 10's . Im now starting to think what would happen if these wheels were accidentially mismanufactured, so it had 2 extra cog pins at position 3 aswell. This would mean it would carry-over both at the 9->0 position, and the 3-->4 position. That would make the number kind of 8 times larger actually. Only failure I could think of, is if the wormgear, that connects the gearbox to the odometer, had an even gear ratio (such as, one 360* rotation of the wormgear translates to one full rotation 0->9 on the odometer wheel) and also had missing parts in such a way, that it would skip all 4's, but that would actually mean the number 4 is visible double the distance, ergo, the faulty odometer displays LESS than the accurate measurement. And that would be valid only for the last wheel (the 1's). So still, interesting problem, but can't happen in real life lol.

[@tripnick555]

Wow. I said that there were 10 numbers missed in each hundred miles for the last digit, then another 9 missed for the 40s, so each hundred is actually 81. Missing out the 400 and 1400 means there are 18 x 81 to get to 2000, then add the additional 4. I got the right answer in my head in about 30 seconds, but it would have taken much longer for a larger number. The base 9 solution is so obvious I'm a little annoyed I didn't think of it, especially as I was basically working my answer out by multiplying 2x9 and 9^2. I even did that in my head as 2x9x9x9, so 2 x 729, and still didn't think of base 9¡¡ Great puzzle.

[@garyrolen8764]

So, I used the fact that each value for the tens, hundreds, and thousands column had a run of 10-1 in the lesser column. So, the thousands column being a 2 means the hundreds column skipped a 4, 2 times. The hundreds column being a 0 meant the tens column skipped 4, 18 times, and the tens column being 0, meant the one's column skipped 4, 162 times plus the 1 for now reading 5. So, 1+162+180+200 subtracted from the current reading of 2005.

[@Friend_of_the_One-Eyed_Ladies]

Pretty simple if you're used to switching between arithmetic bases. This is basically just translating base 9 to base 10. Less than 0x10 seconds to calculate. The only tricky part is you need to first go through the digits of the original number and subtract 1 from each digit which is greater than 4. (002005 becomes 002004). Then just treat the result as base 9, and convert it to base 10. 2(9³) + 0(9²) + 0x(9¹) + 4(9⁰) = 1462 miles Easy peasy.

[@Seegtease]

Base 9 was my go-to thought.

[@Tjs736]

If you understand what 2005 fundamentally represents namely 2 times ten cubed plus 5, it’s simple enough to work out that if it skips a digit every time the answer will be two times nine cubed plus four

[@PhilipMurphy8]

Interesting idea for a Maths question for sure

[@gavindeane3670]

This is just a number in base 9, so we have a units position, a 9s position, an 81s position, and a 729s position. The "5" in the units position really indicates 4 miles because "4" is not used. So we have 729×2 + 4 = 1462 miles. Can't really imagine how there can be 12 minutes of video to fill on this.

[@someblokeontheinternet]

I'm used to working with binary, so the base9 solution was my immediate thought - however in my smugness that I actually knew the answer to one of your videos for a change I forgot to correct the number sequence for the 5 in the final digit 🤦so was one away

[@metilaful]

Wow, that first method was the VERY long way. I just converted it from base 9. I loved to hear that that was the “genius” way 😂

[@loopcycle123]

The number is 2005, it has four digits, which means we could start counting from 1000 A thousand has 9 hundreds, excluding 400 A hundred has 9 tens, excluding 40 A ten has 9 units, excluding 4 2000 is 2 × 1000, so 2 × 9 hundreds = 18 hundreds 18 × 9 tens = 162 tens 162 × 9 units = 1458 units From 1 to 5, we skip the 4 (1,2,3,5), so we add four more units: 1458 + 4 = 1462 And that's how I thought for the solution for this one.

[@violetfactorial6806]

It did occur to me that it should be related to the value in base 9, but I couldn't convince myself that was definitely true. So I solved it like this: For every thousands digit, we must have cycled through 9 hundreds digits (0, 1, 2, 3, 5, 6, 7, 8, and 9 in the hundreds place before we increment the thousands place). Then, for each of those hundreds cycles, we must have cycled through 9 tens digits, and for each of those tens cycles we must have cycled through 9 ones digit. So it's 2*9*9*9 to get to 2000 on the odometer. Then just add 4 more to get from 0 to 5 in the ones place.

[@bluerizlagirl]

The units digit is going to count 0, 1, 2, 3, 5, 6, 7, 8, 9 and then the digit to its left will increase as it goes back to 0. So the "tens" digit is actually counting _nines_. It will behave the same way, so the "hundreds" digit will actually be counting 81s, and the "thousands" digit will be counting 729s. Since it reads 2, which is before the gap in the sequence, we can treat it as correct; but the 5 comes after the gap, so we have to deduct one to get the correct units. Then the answer is 2 * 729 + (5 - 1) = 1458 + 4 = 1462 miles.

[@CasualLifeExperiencer]

It's in base 9 where the digit 5 to 9 represent digits 4 to 8 so, for any number you subtract 1 from every digit greater or equal to 5 and convert in base 9

[@josephfredbill]

There must be all sorts of problems with clocks running fast or slow that could be solved the same way (the bases method) but usually are not.

[@gamefacierglitches]

In my head calculated all from 0-2000, the added the last 4 from 2004. 1st digits = 2×9×9×1 = 162 2nd digits = 2×9×10 = 180 3rd digits = 2×100 = 200 1 + 162 + 180 + 200 = 543 2005 - 543 = 1462

[@thomecq]

IMO there is much simple explanation. This is just base 9 representation of the number, in witch 0 corresponds to 0, 1 corresponds to 1 in decimal, 2 correspomds to 2, 3 to 3, 5 to 4 (because 4 is missing) . So we have 2 ^ 3+ 0^2 +0^1 + 4 (because 5 corresponds to 4).

[@jonathanhibberd9983]

Method 2 and 3 are exactly the same. 2 * 9 * 9 * 9 is 2 (9^3). If it were a number like 2716, you would have to calculate how many combinations are < 2000, then how many are < 700, then how many are < 10, then deal with the 6 in the ones spot. Which calculates to 2*9*9*9 + 7*9*9 + 1*9 + (6-1). Or 2*(9^3) + 7(9^2) + 1(9^1) + 5(9^0). Or 2715 in base 9.

[@vidyashankars]

Start with 2000. Subtract 200 for the 4s skipped in the 100s place (400 and 1400 would have gotten skipped). 1800, now subtract all the 40-49 in each of these 1800 = 1800 - 180 = 1620. Subtract all the units place 4 from 1620. 1620-162 = 1448. Now add the 4 miles from 2001 to 2005. 1448 + 4 = 1452.

[@marcelosalgado9729]

Numbers with 4 between 0 and 2000: - Numbers with 400s: 0400 and 1400. So there are 2x100= 200 - Numbers with 40s: 9x10x2= 180 - Number with 4s: 9x9x2= 162. The total is 542. - Number 2004: 1 Big total 543. So the actual odomemer: 2005- 543= 1462. Never came to my mind the base 9 thing.

[@हिंदुस्तानीगर्व]

Here you can use exploding dots , let's imagine all dots in last box So ans is 2*9*9*9+4=1462

[@Cryptocurrency69]

Damn, this is the first time I'm hearing about exploding dots. Can you explain this method please?

[@guyhoghton399]

Correctly functioning odometers use a fourier sequence.

[@ck3908]

0 to 100 skip 9+10(all 40s skipped) = 19 times, 100 to 1000 skip 19*9 + 100 (all 400s skipped) = 271, so 0 to 2000 skips 2*271 = 542 and add 1 for 2004 = total 543 skipped => so ans 1462, fairly direct approach

[@rinkio9044]

It counts in "base 9" with the digits 0 1 2 3 5 6 7 8 9 When the meter reaches 10, it means 9 100 means 81 1000 means 729 2000 means 1458 2003 means 1461 2005 means 1462

[@cheeseparis1]

I did it the fast way (base 9) and it took me 12mn40 : 5 seconds to get the result + 12mn35 to enjoy your video.

[@SiqueScarface]

To me, it was just 2005 written in the nonal (base 9) system, 2*9³+4, as we now have only 9 working digits per (non-decimal, but nonal) place.

[@K9Megahertz]

You do enough computer programming, you learn to switch bases quite often. Like many others I could tell this was a base 9 problem since you're just dropping a digit.

[@shadowpriest2574]

My solution was to think that 4 is always skipped , so 100 loses 19 numbers , so 81, then you every 1000 loses 100 numbers because of 400, so with this , its 2005 - 200= 1805 ( subtracted the 400 twice per thousand ) then every 100 is 81 , so 18 * 81 + 4 because of the 5 from 2005. That is 1458 + 4 = 1462

[@platypi_otbs]

Great video. I thought base 9 right away.

[@florianlipp5452]

It's really quite simple: 2x9x9x9+4. Did it in my head in 10 seconds. And then I saw the lenght of the video and thought there must be some catch I missed. So it took me another 2 minutes to double check my logic...

[@xeuszzz]

I solved this the following way. Below 2005 we have 2×100=200 numbers with 4 in the hundreds place. Let's subtract 200 from 2005 to discount all numbers with 4 in the hundreds place. 2005-200=1805 Among these remaining numbers we have 18×10=180 numbers with 4 in the tens place. Let's discount those. 1805-180=1625 Finally we have 163 numbers remaining with 4 in the units place. We subtract once more to reach 1625-163=1462, which is the correct answer.

[@wyattstevens8574]

Easy: 1462 Split the numbers from 0 to 2005 into 0-999, 1000-1999, and "loose case" 2000-2005. You know there's only one value in set 3 (2004) that has a 4. Now sets 1 and 2 are the same except for what's in the thousands position. Within each set, let's count how many numbers *don't* have a 4 in them. There are 9 potential values for each of the last 3 positions that aren't 4, making 9³=729 actual miles until a 4 *digit* rollover. 729*2 (as I said, sets 1 and 2 are almost identical: the thousands digit is the only difference) for everything sub-2001. For 2001-2005 there are 4 options. 729*2+4=1,462 actual miles at labeled mile 2,005.

[@cvkline]

All my days of assembly language programming in not only base 10 but also base 8 (octal) and base 16 (hexadecimal) had different base systems in the front of my mind, so I went immediately to base 9. The only trick there is what value the last digit has to have in this puzzle, since 4 is skipped. 🙂

[@joepiazza3756]

I did it the base 9 way first but I already knew the 2nd way was the super easy way. The 1st way is just over-complicating things.

[@TofuBug24]

I did the base 9 method but used it to calculate the missing numbers instead and then subtract. Basically calculate ALL the numbers ### starting with 4, then all the numbers ## starting with 4, and finally all the numbers # starting with 4. By knowing every larger calculation already covers 10% of the next smaller calculation the factor reduces by 10 while the count per increases by 9^(n + 1). Then we just multiple it by how many complete #### from 0001 to 1000 numbers we see and finally subtract 1 for the lone 2004 . 2005 - 2( 9^0 x 100 + 9^1 x 10 + 9^2) - 1

[@karl7736]

having learned to count binary on my fingers a long time ago I used the third method counting in base nine.

[@justjoshingaround]

Minor comment, the "base 9" method is titled "method 1", which I think you meant to call method 3.

[@manudev0004]

Amazing as always

[@henrydenner5448]

I immediately used base 9 principles, regarding numbers 0 to 3 as the same and 5 to 9 as (number - 1).

[@benvanrensburg4261]

Did it in my head, but I nevertheless consider it a marvellously good idea for a junior or first-round olympiad problem.

[@nightfox6738]

So it's essentially base 9 with each digit above 4 being one higher than it actually is... So the actual traveled distance is 2 * 9^3 + 4 * 9^0 = 2 * 729 + 4 * 1 = 1458 + 4 = 1462 miles

[@jacobgoldman5780]

Every shown group of 100 miles is the same. For the first hundred miles we see it skips mile 4, 14, 24, 34, 40-49, 54, 64, 74, 84, and 94 which is 19 total miles so when you reach mile 100 you are actually on mile 100-19=81. We observe this wont change for miles between 101-200, 201-300, but then we have to be careful because it is 301-399 (400-499 dont appear on this broken speedometer) so far we have 81x4-1 since there are only 99 miles when considering the 300s, and this will continue until 1300, then 1300-1399 we travel 80 miles again, and then for 2001-2005 it is only mile "2001", "2002", "2003" and "2005" so 4 more miles. Since we exclude the 400s and 1400s there are 20 hundreds minus 2 hundreds to consider or 18 hundred so 81x18-2+4=81x18+2= 1460 miles, why is this wrong?

[@naphackDT]

Didn't go through it with a fine tooth comb, but that -2 seems sus. 81×18 is the same as 2(9^3). The solution for base 9 conversion is 2(9^3)+4, which is exactly the same as your solution except it lacks that -2.

[@ashitnayak1912]

I calculated like below. 1. Every 10 count 1 missing. 2. For each 100 count 10 missing numbers. 3. However for 10th place entire 10 numbers missed. 4. So 10+10=20, but 44 is calculated twice. So 20-1= 19 missing number for each 100. 5. For number upto 2000, there are 20 times 100. 6. But 400 and 1400 series not to be calculated. So 200 numbers will be missing. 1 missing number from 2001 to 2005. 7. 19*18=342, 342+200=542, 542+1=543 Correct value 2005-543=1462

[@MikeyMacc]

This video is anprime example of how to make an easy problem hard. Also there is another solution, the answer being what it reads or minus 1 because skipping 4 could mean it skipped it, but still counted correctly in the internal comluter and only the display ja wrong.

[@marcusscience23]

Much like some of the others in the comments, my first thought was to consider it in nonary (base-9), with the digits 5 - 9 notating the number one below. I wouldn’t pronounce nonary 2004 as “two thousand and four” though, because that’s decimal terminology and we shouldn’t read non-decimal bases in terms of decimal. “thousand” specifically refers to 10^3 in decimal, so it wouldn’t be appropriate to use it for 9^3 - it’s a different number!