If AA + BB + CC = ABC, What Are A, B and C?

[@Selicre]

You can also solve this algebraically, and it works out to be very neat: 11(a+b+c) = 100a + 10b + c 89a = 10c + b As all the variables are integers and in the range of [1, 9], a can only be 1, and so the right-hand expression turns into "cb" as a number, which gives you the solution of 198.

[@PerDanielRosin]

Smart! Better than the solution given I the clip.

[@mkt3857]

I + Like + The + silk = SHIRT HOW TO SOLVE THIS

[@adiraaz7736]

easier than mine.

[@MarieAnne.]

@@mkt3857 I + LIKE + THE + SILK = SHIRT 4 + 8472 + 502 + 1487 = 10465

[@twistytastylife2074]

I made some more equations and got b=10 c= -1and a =0 Ha ha

[@patrickwienhoft7987]

AA means 10*A + A = 11*A So the "actual" equation is 11A + 11B + 11C = 100A + 10B + C. 11A + 11B + 11C = 11(A+B+C) is divisible by 11 and the result is a 3 digit number. A 3 digit number that is divisible by 11 has the property that the first and third digit add up to the 2nd digit, so A + C = B. You can put that in the equatoin to get 22A + 22C = 110A + 11C 11C = 88A C = 8A As all numbers are smaller than 10, so 8A needs to be smaller than 10 as well. So A = 1. C = 8 then. As A + C = B, B = 9. 11 + 99 + 88 = 198. edit: after watching your solution, I feel mine is more elegant. I feel smart now, haha :D

[@turun_ambartanen]

thats a nice way of solving the problem! I know of the fact of multiping with eleven, didn't think about it when solving the peoblem tho.

[@pieterluyten3559]

you missed the trivial solution ;) 8.A=C and C

[@DragonHunter926]

One statement is incorrect. Counterexample: 616 = 56*11, but 6+6 != 1.

[@anshluthra941]

Thats what i exactly thought but didn't bother to type. :)

[@FarSeenNomic]

+Pieter Luyten the numbers need to be distinct.

[@philippepoulin7842]

Easy. Units: If A + B + C = xC, then A + B = 10. Tens: If A + B + C + 1 = xB, then C + 1 = B Then the hundreds leftover is 1, so A is 1. If A is 1, then B is 9 (1+9=10), which means C is 8 (8+1=9).

[@one2play4]

+Philippe Poulin That was literally the same process I used, too. Noice

[@nadavzeira389]

Any chances you can you explain the last two moves? Why's A + B + C + ONE = xB Where did the one come from?

[@philippepoulin7842]

Because the leftover from the units row is one (A+B is 10 and C is less than 10, which means that the tens row has + 1). Same logic for how I then found that A is one.

[@philippepoulin7842]

Well C is 8 and xB was 19, so what you say stands too, C+1 = xB-10. If A+B+C=xC, and A+B+C+1=xB, that means xC+1=xB, which means C+1=B.

[@odysseuszeus1]

Exactly the same process for me too, more use of simple logic then higher math equations.

[@BatPhil]

You don't need to consider 2 options - start with the units: A + B = 10 (no other possibility if it is to generate C in the units column) B = C + 1 (from the carry over to the tens column) So, A + B + C + 1 < 20 gives A = 1 (it can't be more than 19 - think about it!) This then leads to B = 9 and C = 8 Quite simple really and can be done in your head.

[@justaskurtinaitis3716]

I just wrote equation like this: 11A+11B+11C=100A+10B+C From there its quite easy to conclude the answer.

[@duichersie1]

I get the right side, but why the 11's on the left side of the equation ?? :)

[@jacksmith7430]

+duichersie1 11xA=aa(not multiplied) because it's 10a + 1a

[@duichersie1]

Thanks :) thats very clever thinking, Sir !

[@quarkyquasar893]

+Justas Kurtinaitis Also, we need to know that LHS is a multiple of eleven, so A+C =B

[@justaskurtinaitis3716]

+Red Sniper Sure, but when you reduce the equation to: 89A - B - 10C=0 You can conclude that A can only be 1. And 89-B-10C=0 Has an obvious answer for B and C.

[@cloverpepsi]

A=1, B=9, C= 8. Here we go- A+B+C= C with something leftover. By this logic, C in the second row MUST be smaller than B. You must have different values because each letter is a different number. You have A leftover in the hundreds, so that means you had A leftover in the tens. Therefore, C+A=B. Using these two equations, we can extrapolate that A+B+C = A//C (// means concatenation.) Subtract C from each side, A+B= 10A. Subtract a from each side, B= 9A Here's what's been established. C+A=B, B=9A. Therefore, C+A=9A, C=8A. The only multiples of 8 and 9 between 1 and 9 are, well, 8 and 9. Therefore, C=8 and B=9. And, obviously, 1=A. Plug this in, you get 11 99 +. 88 ----------- 198 Perfect!

[@M.Melkonyan]

AA+BB+CC=ABC 11A+11B+11C=100A+10B+C 11A+B+10C=100A B=89A-10C If A is a any different digit than 1 the equation makes no sense because B can't be a digit between 0 to 9. So A is 1 B=89-10C To make B possible between 0 to 9 and not negative C is 8, so B=89-80=9 A=1, B=9, C=8:

[@notoriouswhitemoth]

I figured it out from the other end - starting with the fact that each summand is a multiple of 11, and that, as such, the sum has to be a multiple of 11. Given that, I started listing three-digit multiples of eleven, with an emphasis on those with three unique digits.

[@acediamond5399]

+notoriouswhitemoth Very good thinking!

[@ntv820]

awesome bro

[@mageddie5974]

I did it similarly. First I figured out A had to be 1 through assuming it's 2. Then I concluded it's a multiple of 11 and mentally noted all multiples of 11 from 110 to 198 and figured out through testing it's 198.

[@maxfalcon2531]

Oooh, I like that.

[@xisumavoid]

Yay i got it! I feel smarter :-D

[@Gapiedaan]

+PeltedPancake lol me neither

[@marc8561]

What the f is going on o.O

[@adityadeb4099]

IT IS FF

[@ramuk1127]

+PeltedPancake Why not? You would have similar interests to him because of the content you watch and thus this video would be in both of your recommended feeds.

[@mb98765]

WHAT X I NEVER THOUGHT ILL SEE YOU HERE LOL

[@shreyanshk8186]

ALTERNATE ----------------------- A+B+C=10+C »A+B=10 OR A+B+C=20+C »A+B=20 But the second case is wrong , 9+8=17(highest values for A andB) because 17≠20 AND , A+B+C+1=10A+B »9A-1=C where A can't be 2 as , then C will become a two-digit number. Thus, A=1 »B=9 »C=8

[@taiweihao98]

A+B=10 A+C=9 A+B+C+1=10A+B 9A-C=1 Solve simultaneously. A=1 B=9 C=8 Done.

[@JohnLeePettimoreIII]

I got this one. I don't always figure out the solutions, but i still enjoy learning these new ways of arriving at answers.

[@kamoroso94]

My solution: 10a+a+10b+b+10c+c=100a+10b+c 11a+11b+11c=100a+10b+c b+10c=89a a=1 b=9 c=8

[@Medreth]

+Zorokmk You can just look at the equation. 89 = 10 * 8 + 9.

[@PianothShaveck]

+Zorokmk a = 1 => x + 10y = 89, it's a linear diophantine equation.

[@Minecraftster148790]

+Zorokmk well u can do a method using matrices that I don't know or u can do an extended version of a simultaneous equation. Make an equation for a using the first equation. Substitute this into the others. Use this to make an equation for b. Substitute this into the last equation. Solve the last equation for c and substitute the result into the equation for b. Use these results for the equation for a. Long and complicated but it works

[@PianothShaveck]

+Minecraftster148790 Absolutely not necessary actually. If you set a = 1, you get x + 10y = 89. It's a linear diophantine equation, so you can solve it like that. But you can also skip pretty much any calculation by noticing that x + 10y is units digits + tens digits. So x = 9 and y = 8. Easy.

[@thomasr.jackson2940]

+PianothShaveck Nice. Even simpler and faster. Some here were disappointed that this was so easy, but examining the multiple solutions of even simple problems can help skill and alacrity.

[@AREmrys]

I varied the procedure slightly in the third step: Knowing that A+B+C leads to C in the right column and at the same time to B in the middle column, we can conclude that B = C+1, as we only have to add the carryover from the right column. That gives us the two triples 11/99/88 and 22/88/77, and by checking them we see that only the first one is valid.

[@insertphrasehere15]

I figured this out the same way but in reverse. My first step was actually working out that no 2XX number would work, since to get into the 200s you'd need 22 , 99 and 88 (no other combination gets to 200 and also has 22), but that number doesn't work. Then I worked out B=C+1.

[@nsnick199]

I did it about the same way. After realizing A+B+C=10+C ==> A+B=10, I looked to the tens column and used the carry to find A+B+C+1=10+B ==>B=C+1.

[@hahalolxyz0685]

I did it in a different but more brain teaser ish way. My logic was that since A+B+C equals C then it would be in the teens so then B is one bigger than a since you do the same thing over but this time add the carried over 1 99 plus 88 plus 77 doesnt work so A has to be 1 so that leaves A as 1 then C as 8 and b as 9

[@DaKnightsofawesome]

well, if the last digit must be c, then that means that a + b = 10. Since we know that the second column is a + b + c + 1 >= ab, we know that 11 + c = ab. So 20 >= ab >= 11. there's only one number in that range where the digits add to ten: 19. so a = 1 and b = 9. now we know that 11 + c = 19 so c = 19 - 11 so c = 8. test it out: 11 + 99 + 88 = 198.

[@sagar110]

Did it almost the same way as you but figured out a and b a little differently. I knew a could only be 1 or 2, and so b 8 or 9. since 11+c could not equal 28 i knew a and b were 1 and 9. that left c = 8.

[@calcustom5026]

I found the max possible (99, 88, 77 = 264) then realized the only possible way to get 9's and 8's in the final number was to change the 77 to 11. I planned to take that logic down for each number, but luckily that worked immediately.

[@iWantToDetonate]

The "logical" approach to solve A is much easier. If we just assume B and C are equal to 8 and 9 (sums up to 17), for A being able to be 2 it has to be 3 at least, which is impossible. So we can rule out the case A=2 without even looking at the other numbers. Now we know that A+B+C=AB is actually 1+B+C+1=10+B. Subtracting B and 2 gives us C=8. Solving for B is now trivial.

[@Coldheart322]

I got A = 1 straight away. Going by the units column, A + B + C = xC, subtract C from both sides gives us A + B = x0, and the largest number ending in 0 possible is 10. Which immediately gives B = 9. It saved me working out an impossible solution.

[@NeatNit]

This was an especially easy/simple one (nothing wrong with that). If we start with the units digit, A + B + C = C. That means that A + B = 0, or 10, or 20 etc. But two single-digit non-zero numbers can only do 10. Therefore: A + B = 10 Now the tens. We carry over the 10 from the units, and we have: A + B + C + 1 = B. Replace A+B with 10 and we get: 10 + C + 1 = B, but the 10 is irrelevent. Therefore: B = C + 1 As for the hundreds, we already know that A+B+C+1 gave us a single 1 to carry over. Therefore: A = 1 We have A now so we can use it in the previous equations: 1 + B = 10 Therefore: B = 9 9 = C + 1 Therefore: C = 8 And we got the correct (and only) solution, 198, without any guesswork and IMO in a much more straightforward way than yours. I was honestly expecting a catch or a different solution (different digits) to turn up in the video, since that happens to me often with your videos.

[@GraveUypo]

+NeatNit yay someone used the same method as me. but i have to say, there was actually a tiny little bit of guesswork. you can't be 100% sure that the number carried over to the 3rd digit is 1. all we know at that point is a+b = 10 a+b+c < 20 and we're looking for a+b+c+1 c could be 9 that way, it would carry a 2. we couldn't know that it wasn't for sure at that point. all you can brag about is that your guesswork had a 9:1 (90% - C had to be a number between 1 and 9 ) chance of being right in the first try, while his had a 40:29 (58%. simplified ratio of numbers of number with non-repeating digits between 100~199 vs same story only between 200 and 264) chance

[@ste1l1]

+GraveUypo I did it somewhat similar, and in my solution you are 100% sure A is 1; i) A+B = 10 (like in the video) Since A+B+C = 10+C, which is maximum 19 and minimum 11, we get the second equation: A+B+C+1 = B + 10n A+C = 10n-1 Only solution is when n=1 (A+C is maximum 19), hence ii) A+C = 9 Since the tenths always add up to a number between 10 and 19, A is equal to 1. From this simply solve i) and ii) and obtain the result.

[@no_pane_no_gane]

+GraveUypo Actually you can be sure that the carry to the hundreds is one. This is because we know that A+B = 10. The only possibility to get a carry of 2 to the hundreds is if we assume C = 9, but this is not possible since it would result in B = 0 which is not allowed.

[@dominikstepien2000]

My solution 11A+11B+11C=100A+10B+C A+B=10 (so last number stay C) A+C=9 (B will stay if it is equal to ten, but there is 1 overflow) And after calculation you get answer A=1, B=9 and C=8.

[@snowfloofcathug]

I solved it the same way as you did

[@marcotavano4681]

Me too

[@flmbray]

There is no need to test two different answers. The units column gives you a relation A+B=10 (as you say). But that means that in the tens column, A+B+1+C = 10A+B, or C=9A-1 (the +1 on left-side of first part is result of carrying one from unit column, since we know 10 < A+B+C

[@Pining_for_the_fjords]

This is one of your few puzzles I actually got. 11a+11b+11c = 100a+10b+c Subtract 11a from both sides: 11b+11c = 89a+10b+c Subtract 11b from both sides: 11c = 89a-b+c Subtract 11c from both sides: 89a-b-10c = 0 If a was 2, 89a would be 178, and you can't subtract a single-digit number and a two-digit number from 178 to get to 0, so a must be 1. Therefore: 89-b-10c = 0 b+10c = 89 And since the only single-digit number and multiple of 10 can you add to get 89 are 9 and 80, b and c must be 9 and 8 respectively.

[@carloft]

The equation missing in the explanation is A+C = 10-1 You know that A+B+C=B, which means that A+C=10, but you have to subtract a 1 that would be carried forward from the sum in the first position on the right. Then it becomes a very straight forward solution of equations. I love your channel! I am an entrepreneur of an engineering company and every once in a while when I come to the office I test my employees/engineers with your problems :-)

[@ghffrsfygdhfjkjiysdz]

Essentially the same way, but a bit simpler solution is to write out the 2 equations and 1 inequality, rather than guessing for C. units gives equation: A+B=10 Tens gives equation A+B+C+1=B+10 (one is a carry from units; in theory RHS cold be B+20 as well, but that would yield A+C=19 and with A

[@adityadeb4099]

no, it's FF

[@saxbend]

I did this algebraically. From 11(A+B+C) = 100A + 10B + C I got C = (89A-B)/10. From there I saw that the only single digit number you can subtract from 89 to get a multiple of 10 is 9. So if A is 1 B must be 9 and C must be 8. Then I had a valid solution. If A is greater than 1, then 89A is a three digit number which rules out all other possibilities since either B or C would have to be greater than 10.

[@leksa8442]

I solved it faster finding three conditions for the three variables. 1) A + B = 10 from the rightmost column of the sum 2) B = C+1 comparing the rightmost column of the sum and the next one on the left.3) the result ABC is a number which is a multiple of 11. Such numbers have the properties that the sum of the odd digits equals the sum of the even digits. Therefor A + C = B. So ultimately we have a linear system of three equations and three variables: A + B = 10, B = C + 1, A + C = B. Much more straightforward and elegant solution

[@balvantpatel1327]

Turn on caption... Then look at 0:01

[@teav7432]

Hello my name is fresh towel walker Someone must have made a big mistake Poor deaf people

[@Cyberlisk]

You can also derive B the same way you get C. With A + B = 10, and C

[@erinzoretich]

I liked the video but thought your explanation was too confusing. Why are we even working through the possibility of carrying a 2? Many other commenters have great solutions that use better algebra then I would have thought of. But here was my logic: Because C passes right down to the sum, A + B must add together and end in 0, so it must be that A + B = 10. Two different numbers valued 9 or less cannot ever equal 20 or more. Because we now know that A + B is equal to 10, A + B + C must be equal to or greater than 10, so it's absolutely going to carry something over to the next column, the tens column. Now, three different numbers valued 9 or less can absolutely equal more than 20. We could be carrying a 1 or a 2 here. BUT! We already know A + B + C = C (all units) from the ones column. We can surmise that if C were large enough to make A + B + C equal to 20 or more, it wouldn't be the UNIT that would be left over in the ones place of that very same sum. The only two possibilities for a three number sum are 9 + 8 + 7 = 24, and 8 + 7 + 6 = 21. In neither of those does the ones place of the sum have the same unit as ANY of the possible addends. We're carrying a 1, not a 2. Next, the tens column of our problem now essentially reads "1 + A + B + C = B". You can simplify that to 1 + A + C = 0 or (we're in units, remember) 1 + A + C = 10, further to A + C = 9. We just figured out A + C = 9. Looking at our tens and hundreds column, there's no way even with the carried 1 from the ones into the tens column those numbers can equal 20 or more. A + C = 9, remember? 9 + (carried) 1 + B cannot equal 20 or more. B is a unit, and can only be 9 at most! Let's go ahead and assume A = 1 and see how it all plays out. A + B = 10 1 + B = 10 B = 9 A + C = 9 1 + C = 10 C = 8 Go back, does it work? Yup! 11 + 99 + 88 equals exactly 198. That wasn't so hard after all.

[@sssdddkkksss]

In the units: A + B = 10 was the first thing that jumped out at me. It directly means A + B + C C + 1 = 9A Because everything is single digits => A = 1 => C = 8 => B = 9 => 11 + 99 + 88 = 198

[@OkanZagorAkkoyun]

It is actually crazy simple. You pick C as a natural number less than nine and then do this: To find B: C+1 To find A: 9-C It always works, try it. You'll thank me later.

[@jakubfabisiak9810]

there is a simpler solution: consider the 3rd column. We can deduce, that A+B = 10, since the most you can get from adding two single digits is 17, so 10 is the only possible candidate that will leave C in the units place. From the second column we know, that A+C+1 = 10, since that is the only way to leave B in the tens place (two different single digits + 1 is at most 18), and that automatically tells us, that A=1, and it's a simple road from there (knowing B=C+1)

[@corbness]

By looking at the units column, you can see right away that A+B=10. Then if you look at the 10s column, you know that A+B=10, but we carried the 1 over from the units column, so it must be the case that B=C+1, then we carry 1 over to the 100s column, so A must be 1. A=1 implies B=9 implies C=8.

[@leoitshere]

I did it by parametrizing the equation and then logically deducing the only possible values for the parameters You have 11A + 11B + 11C = 100A + 10B + C Reorganize the terms so that A is on one side and B and C on the other. -89A = -B - 10C Now solve for A A = (B + 10C)/89 Now A is a function of B and C, but we also know that A is an integer. The first possible values of B and C so that A is an integer is when B + 10Z = 89, so that the division reaches just 1. However, we cannot choose any C and B because they all need to be one digit. So whatever you choose C to be, it needs to be big enough to allow B to be an integer. For example, if you choose B = 7 then C must be 19. This is no good, but if you choose B = 8 then C = 9 and that satisfies our needs. All we need is to find A but fortunately it is just a matter of plugging and chugging. A = (80 + 9)/89 = 89/89 = 1 And 11 + 99 + 99 = 100 + 90 + 8 = 198 EZ PZ

[@williamtraynor4350]

I don't often solve these Presh but this was easy... comparing units and tens columns A+B equalled 10...B was 1 greater than C...and the value of A in the answer line is 1 after adding 10 to a single number

[@Evultz]

Given that 11A+11B+11C=100A+10B+C As A, B and C are positive digits: 0

[@frabert]

I made a somewhat simpler reasoning: AA+ BB+ CC= ABC If we use column addition, for the first digit, A+B+C=C, so we must have a sum of C and a remainder. This takes to A+B = 10. Then we have A+B+C with a remainder of 1, so it is A+B+C+1=B, so A+C+1=10. Then A must be 1, because you can't get a higher remainder than 1 using 3 single digit numbers added together. If we solve the other equations, we get A=1 B=9 and C=8

[@Neehize]

11a + 11b + 11c = 100a + 10b + c b = 89a - 10c, adding the constraint b in [1, 9] we rule out a = 2, and the only solution for a = 1 is c = 8, b = 9

[@TZPlayer]

there is an easier method once you discover that a+b=10 (in decimals) you can see that a+b+c would be 19 at maximun so if a+b+c=c (in units) than a+b+c=b (in decimals) a+b+c+1=b so c-1=b if a=1 b=9 c=8 if a=2 b=8 c=7 so you just check and see wich one is right

[@tryhardofdoom7682]

Well my solution was less elegant in some points but there is one thing I can add to make your solution perfect, which is the proof that B must be odd. AA + BB + CC = ABC can be rewritten into 11A + 11B + 11C = 100A + 10B + C which can be rewritten into B = 89A - 10C. Since 89A ist always odd, no matter if A is even or odd, and since 10C is always even, no matter if C is even or odd, B = (2x + 1) - (2y) = 2(x + y) +1. Thereby B must be odd. Together with your conclusion that A must be either 1 or 2 and B complementarily must be 9 or 8, there is only one choice, so this exercise can be solved without experimenting. Nice exercise, I like searching for variables, especially if it is as easy as this :D And its nice learning from other attempts on the same exercise, I learned much from your video and the comment section!

[@potato-hj9nm]

I completely mksunderstood this I thought you had to times them together.

[@marttielvisto3519]

same here

[@beng4186]

+potato 123 Google alphametics. They're a common type of math problem used in schools.

[@JohnLeePettimoreIII]

You don't "times" numbers, you "multiply" them.

[@mattjw16]

Same

[@anon6514]

1.)from 3rd column: A + B + C = XC and A+B < 20 .'. A + B = 10. 2.)from 2nd column: (X=1) carries over .'. C = B - 1 3.)and from 1st column: (A+B+C) < 20 .'. A = 1. which implies (from 1) that B = 9 and (from 2) that C = 8

[@pedroff_1]

this one was particularly easy: AA+BB+CC starts with c, A and B can't be both 0, so, they must add up to 10. on the second digit, you have again it being C, but, instead, it is added by one (since the previous number went over 10), therefore, B=C+1. and, since the third digit couldn't be higher than 1, A was . With A=1, and A+B=10, B=9. with B=c+1, C=8. 11+99+88=99*2=100*2 - 2 = 200-2=198.

[@guest_informant]

Suggestion: Why not Level these questions according to how hard you found them? Level 1: Something like this puzzle Level 10: Oh I don't know, maybe the Riemann Hypothesis :-) Level 1 is not necessarily a bad thing: it's nice to be able to get something right once in a while. SPOILER . . . . . . . . . My method A+B=10 C+1=B A is probably 1 so it's 11+99+88=198 Yes it is.

[@denny141196]

Linear algebra! 11a + 11b + 11c = 100a + 10b + c. Therefore 89a - b - 10c = 0. We get the matrix: [89 -1 -10 | 0] Where the first column represents a, the second b and the third c. This matrix has rank 1, but there are three unknowns. Set b=s and c=t, giving: 89a - s - 10t =0 Therefore a = s/89 - 10t/89. This means that (a, b, c) = s(1/89, 1, 0) + t(10/89, 0, 1). To get an integer value for a, use s=9 and t=8. This gives the same result: a=1, b=9 and c=8.

[@davidsmith6859]

OMG Presh, you made a meal of this. The largest carry over from adding three positive digits is a 2 but if A is 2 then that's not enough to generate a carry over of 2 even if B and C are 9 and 8 so A must be 1. Therefore B must be 1 more than C. Also from the units column, A + B must = 10 so B = 9 and C = 8. Simples.

[@jishnuviswanath]

A is 1 from the straight looking at it, i.e. if you just take the last digits A B C added to XC which means A+B = X0 while X can be only 1 since two single digits can not be added to 20. So A+B = 10 and from second digits A+B+C+1 = AB and again A+C+1= A0.. Since you can not get 20 with 2 single digits +1. So A =1 and so B = 9 and C= 8

[@ngarcia103]

Also, since 28C is guaranteed to be greater than 264 (the highest possible sum you worked out in the beginning), you know A = 2 and B = 8 is not a solution.

[@HeartistMurali]

I went for straight algebra. Since aa,bb,cc are multiples of 11, then 11(a+b+c)= abc, which is, 11a+11b+11c=100a+10b+c, which is, c=b-89a+11c, which is, b=89a-10c, therefore, since 'a' cannot be greater than 1 and 'c' cannot be greater or less than 8 so as to make 'b' a single digit or a positive number, it will be a=1 and c=8 which in turn makes b=9.

[@jeffreyblack666]

Another option: Start at the units place. We have A+B+C=C. That means A+B must equal 10 (technically any multiple of 10, but as two 1 digit numbers can only add to 18, we are limited to 10). Now look at the tens and hundreds. We have A+B+C+1=10A+B This means C+1=9A, and we know that C+1 is 10 at most, this means 9A is 10 at most and thus A is at most 1. Thus A=1, and thus B=9 and C=8. No need to try a brute force method of listing possibilities and checking each one.

[@lokegustafsson247]

I'm lucky enough to be competing in the national(Sweden) finals for junior highschoolers(hope thats correct). Channels like these sure help in fueling the interest, so thanks a ton for what you do to your subs and wish me luck!

[@lokegustafsson247]

Whoops, i forgot to mention that this whole thing happends on friday.

[@OEpistimon]

+Loke Gustafsson Good luck!

[@aungthansein]

Find the numeric value of (A, B, C, D, E) from the following three formulas. However, A, B, C, D, E represent numbers between 0 and 9. Default: AC x 1E = 4CD (1) ADE x 1EC = AB (2) AB + E9 = 11A (3) Example: If F = 1, G = 2, then 3FG would be 312.

[@luisisaurio]

My solution: AA+BB+CC=11(A+B+C)=ABC. Therefore 11 divides ABC therefore A+C=B. Also lemma 1. A+B=10, AA+BB+CC==A+B+C==C mod 10, therefore A+B=10. Biggest case is when A=B=C=9, so A is at most 2. We have 2 cases to value. A=2 and A=1. From where only A=1, B=9 and C=8 are solutions.

[@JesusAlbertoPinto]

The sum of the first column: in terms of units A+B+C=C means that A+B=10 as he said Then the second column is A+B+C+1=B, that leads to A+C+1=10 in terms of units, again. The 1 comes from the 10 in the first sum that has to be carried on to the second column. Finally, the sum of the whole number is (A+10A)+(B+10B)+(C+10C)=100A+10B+C Using these equations, the results leads to A=1, B=9, C=8

[@alexanderweisberg5062]

I got the same answer, but with a completely different method: When starting at the right column, we see that either A+B+C = C, A+B+C = C+10, or A+B+C = C+20 which we can simplify into A+B = 0, A+B = 10, or A+B = 20. Since A+B can’t equal 20 or 0, we know that A+B = 10. Moving on to the next column, we know that 1+A+B+C = B+10a, which we can simplify to 11+C = B+10a. We can isolate A by making the equation 11+C-B = 10a. If we assume that B = 1 and C = 9 to find the highest possible value of 10a is 19, so A

[@JMScibra]

I did A+B=10, and A+C=9, since A+B=10 in the unit column carries over only 1 (C can be at most 9, making A+B+C at most 19). Similarly, A=1 since A+C=9, meaning A+B+C is at most 18, meaning a carry over of 2 is not possible from the tens to the hundreds. So we get 1+B=10, so B=9, and 1+C=9, so C=8, and we get 11+99+88=198.

[@Jimpozcan]

The digits A, B and C correspond to integers, _A_, _B_ and _C_ (respectively), greater than or equal to zero and less than ten (or whatever number you're using for a base, e.g. twelve in dozenal, but let's ignore that). When you add the units column, you get some multiple of ten plus _C_. Thus _A_ + _B_ is a multiple of ten. A and B are different digits, thus the maximum _A_ + _B_ could be is 9 + 8 and the minimum is 0 + 1. So, 0 < _A_ + _B_ < 18. There is only one multiple of ten in this range and that is ten. _A_ + _B_ = 10 ... (equation 1) You write down the C and carry the one. So, when adding the tens column you have the sum _A_ + _B_ + _C_ + 1. This gives some multiple of ten plus _B_. Thus _A_ + _C_ + 1 is a multiple of ten. As with _A_ + _B_, the maximum _A_ + _C_ could be is 9 + 8 and the minimum is 0 + 1. So, 1 < _A_ + _C_ + 1 < 19. Again, there is only one multiple of ten in this range, which is ten. _A_ + _C_ + 1 = 10 ... (equation 2) So, you write B in the tens column and carry the 1. You write a 1 in the hundreds column. Thus _A_ is 1. From equation 1, _B_ is 9. From equation 2, _C_ is 8.

[@kevinfoflygen1627]

Here's how I solved it. A+B=10, since the LSD of the sum is C. A+C+c0=10, where c0 is the carry from the sum of the LSDs. But c0=1, since A+B+C

[@JohnRandomness105]

The solution (hard to explain in a short comment): A=1, B=9, C=8. 11+99+88=198. A+B must be a multiple of 10 -- giving a carry. A+C+carry must also be a multiple of 10 -- giving A as the 2nd carry. A little work shows that both carries must be the same, and only A=1 works. B must be C+carry = C+A.

[@1Darco1]

The final trial and error part wouldn't have been necessary. We could say that A + B + C = DE (expressed formally: 1A + 1B + 1C = 10D + 1E). By looking at the units, we see that in the result C = E. Now, for the tens, there is the exact same thing going on, but moved one digit to the left (thus, multiplied by ten: 10A + 10B + 10C = 100D + 10E). Again, we see that in the result A = D. And the middle digit in the result B is just the sum of D from the first paragraph and E from the second paragraph. B = D + E We deduce, because we know A (which equals D) and B, that C = E = B - D = B - A.

[@markmetzger794]

for a in range(1,10): for b in range(1,10): for c in range(1,10): q = 10*a+a+10*b+b+10*c+c w = 100*a+10*b+c if q == w: print(a) print(b) print(c)

[@Tyrian3k]

I did it like this: For the third digit to be C, A+B must be 10. Since you carry one over, B must be C+1. AA+BB is 110 and CC can't be more than 88, which in sum cannot be over 200. Therefore A is 1, which means B must be 9 and C is 8.

[@fernandopizarrovillagarcia6992]

A+B+C=C (units). That means A+B=10. A+B+C+1=10A+B (tens). That means 10+C+1=10A+B. By Diophantine equation, A=1. That means B=9. Finally, C=8. Replacing the variables 11+99+88=198.

[@TigruArdavi]

I did it without equations by mere logic (and lastly trying two cases at the end as well): 1. A =1 or 2, as 3 two-digit numbers can be no more than 200something (as Presh did). But then furthermore: 2. Regarding the units column, as A is no larger than 2, with two other numbers, you get no more than 19 and thus no carry >1, and exactly 1, as no 0 involved as stated. So in the tens colum you get the same result as in the units + carry 1, but it cannot be 20, as no 0 involved. That means, in the hundreds column: A=1 ! 3. B+C must be >9, as A=1 and we need get a carry and no 0 at the end 4. As we get carry 1, B must be exactly one greater than C, to get the tens column right. Thus B and C must be neighbouring numbers adding to >9, so pairs (3,2) (4,3) (5,4) are out 5. As (6,5)+1 or (7,6)+1 would get us a 2 or 4 at the end (excluded in step 4) only pairs (8,7) (9,8) remain. Then it's quickly tested as Presh did, 11+99+88=198 is the only correct solution.

[@vaishaliraut5369]

got the answer by thumbnail 😃 Used the trivial method... 10+c=a+b+c (becoz carry over ) So a+b=10 A=1 and b=9 Satisfies this condition and also the aa+bb+cc=abc

[@ociemitchell]

I started with a+b=10 from the units column since that allows c in the units of the sum. In the same way, the 10s column gives a+c+1=10. Carrying the 1 gives a=1 and substituting in a=1 gives b=9,and c=8.

[@supersatangod4460]

Let's start from the end: Since A+B+C=10+C we get A+B=10. Moving on, A+B+C+1=10+B (the +1 comes from the A+B=10) gives us A+C=9. So we know that A+C+1=10 therefore, the first digit (A) must be 1. We have three equations now: 1) A+B=10 2) A+C=9 3) A=1 This doesn't require any fancy mathematics, and we arrive at the values for A, B and C as 1, 9 and 8. 11+99+88=198, so the initial requirement is met. This can be done by hand in less than 2 minutes (if you write down every step) and is way more elegant, at least in my opinion.

[@johnytest464]

A = 1, B = 9, C = 8 A+B+C = C, which means A+B = 10 A+B+C+1 (as the first column from left gives 10) = B, which means C+1 = B as A+B = 10 and as A+B = 10, hence A = 1; 1+B = 10 => B = 9 C+1 = B => C=8

[@alok2047]

Case1: A+B=10 ( so we get C in addition of very first step) Case2: A+B+C+1(carry)= 10A+B Or A+C+1=10A C+1=9A A,B&C are Integer and single digits So A=1 which gives C=8 and by Case1,B=9 Answer Verification 11+99+88=198

[@Phlebas]

I figured this out in a slightly different way by thinking of this in terms of grade-school addition. First I realised that A+B=10 because of the units column equaling C. Then I realised that A+C+1=10 for the tens column to equal B (+1 because if A+B=10 and C

[@1996Pinocchio]

This is an easy one^^

[@thingysandstuff9242]

Correct...

[@sumitjangir9946]

@@thingysandstuff9242 zs, A's# 😑 😆😇😕😑

[@ja72wol]

. a+b+c = 10*x + c (where x is integer and >= 0), so a+b = 10*x and 0 < (a+b) 0,b>0 . a+b+c+x = 10*a + b, so (from ) 11 + c = 10a +b and 11

[@cryptexify]

Here's a simple solution: A+B+C = C (mod 10) implies A+B=0 (mod 10) implies this generated a carry of 1. Take carry and add it to second column. A+B+C+1 = B (mod 10) and from the previous fact it implies C+1 = B (mod 10). We also got another carry of 1. Since third column has nothing but the carry of 1, A = 1. Implies B = 9 which implies C = 8. So we have 11+99+88=198.

[@Maxman013_]

*Before watching the solution* So in the ones (units) column, A+B+C = C That means that A+B = 0 or 10. For A+B = 0, A and B must both equal 0 because they can't be negative, which is impossible as both aren't distinct. So, A+B = 10. A,B possible pairs: 1,9 or 2,8 or 3,7 or 4,6 or 6,4 or 3,7 or 8,2 or 9,1. If you add the largest three two-digit numbers together, with each of the digits in the number being the same (eg 11, 22, 33) and each one being distinct, you have 99+88+77. 99+88+77 = 264. That means the largest number A can be is 2. This eliminates 3,7 and 4,6 and 6,4 and 7,3 and 8,2 and 9,1. A,B possible pairs: 1,9 or 2,8. That means that if A=2, the top number is 22. B must equal 8, so the middle number is 88. 22 88 +CC ------- 28C ------- For 2+8+C to equal 28, C must be 18. Even with the carry over of 1 from the ones (remembering that A+B+C = C + 10, and the 1 carries), C must be 17. Since C can only be a single digit, this is impossible, so A cannot be 2. This eliminates 2,8. Now the only possible A,B pair left is 1,9. That means A = 1 and B = 9. The top number is 11 and the bottom is 99. 11 99 +CC -------- 19C -------- Now let's look at the tens column. 1+9+C = 19 (the carry over to the hundreds). So 10+C = 19. Remember that there is a carry over from the ones of 1, so the tens would actually be 1+9+C*+1* = 19, or 11+C = 19. Simple algebra will reduce this down to C = 8. So A = 1, B = 9, C = 8 11 99 + 88 --------- 198 --------- *After* Oh yay I got it right! :)

[@funicksystem]

Got the same kind of resolution as Patrick Wienhöft, it differs in the end actually : Same start : AA = 11A BB = 11B CC = 11C ABC = 100A + 10B + C Which gives us an equation : 11A + 11B +11C = 100A + 10B + C Equivalent to : 89A - B - 10C = 0 or 89A = 10C + B Once Here, the solution appears to be pretty simple (at least to me).

[@chtoffy]

I start by observing that A+B=10 on the right since you keep C as a result. You carry the one. That means 1+A+C=10 since you keep B as a result in the second column. You carry the one. So A=1. And therefore C=8 and B=9. No need to consider multiple cases.

[@matj12]

You can rewrite it to equation 11A+11B+11C=100A+10B+C. You can express A as (10C+B)/89. Since A is a natural number, 10C+B must be divisible by 89. B and C must also be smaller than 10. The only Solution is C=8 and B=9, therefore A=1.

[@karoolc]

Hello! Not always am I able to solve your problems, but this time in 3mins in my head :) I remember we had similar in about 9th grade. Cheers!

[@lakshu8768]

A=1 B=9 C=8

[@ItsCugel]

unit colum: A + B + C = C -> A + B = 10 since A = B = 0 is not distinct -> gives a carry to tenth column tenth column: 1 + A + B + C = B -> 1 + A + C = 10 -> A + C = 9 -> gives carry to hundreth column hundreth column: 1 = A This gives 1 + C = 9 -> C = 8; 1 + B = 10 -> B = 9 Soultion: 198 (super simple)

[@ItsCugel]

+ItsCugel Just to be clear, the tenth column 1 + A + C cannot be equal to 0 since A and B are positive numbers.

[@itorres008]

Don't know why go through examining the possibility of A=1 or A=2. From ones col, A+B=10. From tens col, A+C+1=10 and from these two B=C+1 Also from tens col, which is A+C+1+B =X carry + B remainder, since we already know A+C+1=10 you have to be carrying 1, so A=1. You get the others A+B=10, so B=9. B=C+1, so C=8.

[@LongfellowLP]

I suppose a more direct solution would involve reparsing it into "mathy" terms and solving for A, B, and C like equations. But this is exactly the kind of problem that appeared in the Sideways Arithmetic at Wayside School books when I was a kid, and I'm glad to see you solved it just like I would have back in the day.

[@alfonsorodriguez2739]

Most of these videos make the problems look more difficult than they are. I solved this problem in 20 secs in my head. The following were my mental steps: 1. A+B+C = C so A+B =10 2. A+B+C=B so C+1=B because A+B=10 and we had a 1 carried over from previous step 3. Letter A must be 1 because it is the carry over from step 2 So Because A=1 then B=9 since A+B=10. Lastly, C must be 8 and thus we have 198. That is how I solved it mentally.

[@ShyamSundar-dj6ez]

My way of solving is AA = [10A]+A= 11A similarly ABC=[100*A]+[10*B]+C So equating 11A + 11B + 11C = 100A+10B+C 89A= B+10C Assuming A=1 B+10C=89 As each is a whole number the digit in the ones place is B and the digit in the Tenth place is C Which gives the solution B =9, C=8 So 11+99+88=198

[@srinjoyghosh8229]

Found out a more intuitive solution applying a few elementary concepts of number system. AA, BB, and CC are all multiples of 11. So let's consider AA=11p, BB=11q, CC=11r. And ABC= 100p + 10q + r (The intuition is if AA=33 (say), 11*3 = 33, so here p=3 and hence A=3) So, 11p + 11q + 11r = 100p +10q +r Simplifying, 89p - q - 10r = 0 From here it is easily observed that if p=1 then r=8 and q=9 fits. So AA=11 , BB= 99 and CC= 88 Bingo!

[@deltagenium3246]

A can't be 0 infact if a =0 B+C = 10 + C So B = 10 and it is impossible A can be 1 or 2 For this reasons the maximum value of A+B+C = 19 But it can't be less than 10 So in the second column we have to add 1 So: If A=1 1+1+B+C = 10 + B Or if A=2 1+2+B+C = 20 + B The second one is always false infact to reach a number bigger than 20, B+C have to be bigger than 17 and it is impossible because the maximum value of B+C is 9+8 = 17. So 1+1+B+C = 10 +B C=8 AND Analyzing the right column we have A+B+C = 10 + C A+B= 10 1+ B = 10 B = 9 ..11 ..99 +88= 198 Right. Thanks for listening, and sorry if I did some mistakes in writing.

[@argothiel]

OK, I don't see my solution in the comments (using the fact that all digits are distinct and positive (in range 1 to 9)): 11*a + 11*b + 11*c = 100*a + 10*b + c 89*a = 10*c + b [moving all a to one side, and b,c to the other] But 10*c + b

[@hart-of-gold]

My method was this: Break down the units and tens.so first A+B=10, from the units same as the video solution.Then for the tens, 10A+B=A+B+C+1 leads to C=9A-1,Since C

[@thomasr.jackson2940]

Re the difficulty of this puzzle, and the puzzles in general, I appreciate a variety of difficulties and puzzle types. I especially enjoy those that lend themselves to multiple, or unusual, methods for solving. This one took longer to write out than to solve, but as long as there is a variety I am not complaining. Simple pleasures.

[@davidsousaRJ]

I solve it the following way: looking vertically at the right-hand column in the sum, we see A+B+C is a number that ends with C. Therefore, A+B=10 and A+B+C = 1C (it could not be 2C because the maximum values for A and B are (9,9) and 9+9 = 18 < 20). At the left column, 1 will raise up from the right column, and we will have 1 + A + B + C adding up to a number that ends with B. Therefore, 1 + A + C = 10, and 1 + A + B + C = 1B. But the number shown below the left is AB, therefore A = 1. From A+B =10, B = 9 and from 1 + A + C = 10 we get C = 8.

[@yohopostop]

I used your method, but here's another way: This equation is the same as 11(A+B+C)=ABC, therefore ABC is divisible by 11. Hence by divisibility rules in base 10, A+C=B. This means 22(A+C)=11(10A+C). Rearranging, C=8A. The only solution to this in one digit integers is A=1, C=8. And B=A+C=9

[@nasrullahhusnan2289]

Consider the right most colum: A+B+C=C It means that A+B is a two-digit number with 0 as the last digit. A+B=10 as the sum of two single digit number can"t be greater than equal to 20. Now consider the middle colum. A+B+C=B instead of C as in the right most column. It is because 1 must be added when suming the middle column. Hence C=B+1. Hence A=1, B=10-A=9 anc C=0. Thus

[@nasrullahhusnan2289]

Sorry clicking send untentionaly while checking. Correction first: Instead of C=B+1 it should be C+1=B. And A=1. Hence from A+B=10, B=9 and C+1=B or C=8. Hence AA+BB+CC=11+99+88=198 or ABC.

[@plentyofpaper]

My steps: 1s column: A+B=10 10s column: B=C+1 100s column: A=1 (A+B=10, -1

[@DanWS100]

I worked it somewhat backwards from the solution given. I realized that A + B = 10 to give C in the ones digit. That AA + BB + CC could equal 264 at most so A was either 1 or 2 in the hundreds digit. Since the tens digit answer is B and we know that 1 is carried over from the ones addition, it follows that B=C+1 (I pondered the longest on this part). That leaves only 2 solutions (A=2, B=8 and C=7) or (A=1, B=9 and c=8 the correct answer).

[@candiduscorvus]

This was pretty easy actually. Just working it out in my head I came to the solution. All you have to realize is that the sum of A and B is 10, which leads you to the conclusion that AA must be 11 since ABC cannot be larger than 100 and the A must be a 1. That means that BB must be 99, which if you then add them together leaves you with 110, and in order to make a 9 out of the tens spot you need an 8, so CC must be 88. 11 + 99 + 88 = 198. Don't even need a pen and paper or a calculator for it.

[@candiduscorvus]

I should have said that the A in ABC cannot be greater than the 100's. That would be more accurate.

[@candiduscorvus]

+Matthew Ciaramitaro because if you try to solve for A being 2 then AA is 22, BB is 88, and CC can only be 11 since you can only add 1 to 8 in order to get 9. You can't go over 9 for the tens place. And 22, 88, and 11 do not get the correct solution. See?

[@candiduscorvus]

+Matthew Ciaramitaro My "solution" was worked out in my head, like my initial comment reads. I wasn't providing a formula for you to follow and then write a paper about. Anyway I initially could see no way that A could be 2 and lead to a solution. It just kind of came to me the way I realize two shades of blue are different.

[@definitelyoldfag6167]

!!SPOILER!!: * (small note on my notation: with aa I mean 10a + a, and NOT a times a) I started with the fact that a + b + c = 10d + c (where d is some arbitrary integer 0

[@sayanghosh6996]

A = 1,B=9,C=8 we can just do it by algebra the problem gives : 11(A+B+C) = 100A+10B+C which on simplifying gives: B = 89A - 10C now we just need single digit numbers for A B and C by seeing only we can say that C will be 8 so that the term becomes 80 and then 89-80 gives 9 (a single digit term = B) so A = 1 as we are using 89 itself so (B=9) == 89*(A=1) - 10*(C=8) or, (B=9) == 89 - 80 == 9 it satisfies the eqn. so it is the answer

[@thaolinh92]

the sum ends in C and they are single digits therefore a+b=10 11(a+b+c) becomes 11c+110 100a+10b+c becomes 90a+c+100 subtract the two, we have c=9a-1, therefore a=1 and c=8, b=10-1=9