Can you solve this?

[@MindYourDecisions]

Thanks to everyone that pointed out a different method. The result number is SSS for some value S, and it must be a multiple of 3. So you can just check 111, 222, ..., 999 divided by 3 and see which number works out where the middle digit is S. The answer is 444/3 = 148. I had employed this strategy in a similar puzzle ABCDE x A = EEEEEE but I forgot it for this video. Good to practice to stay sharp! kzbin.info/www/bejne/nZKnmqyNpNyVi8U

[@warlock415]

not only that, but since you know the result is a three-digit number and it's not 999, then circle is 0, 1 or 2.

[@tobyfitzpatrick3914]

​@@warlock415 Can't be 0. That would make it a 2 digit number.

[@ambika69]

damn... took me 5 minutes to figure that out, then I click on the video, and i'm ninja's by the pinned comment.

[@arakelyanavetik]

@@MindYourDecisions Actually since SSS÷3 must be at least 3 digit number, then SSS÷3 is at least 100 hence SSS is at least 333, so 111 and 222 are not needed to divide by 3.

[@Aksel37]

Or else the trio 0,0,0 also work 😅

[@arakelyanavetik]

Its faster to check what results give 444, 555, 777 and 888 dividing by 3, the rest AAA are easily ruled out by using little effort...

[@TheChamp1971]

Exactly!!..... It only took me a few seconds to come up with 148.

[@londongaz2]

Yep this was my method!

[@Buckminster.Fullerene]

Exactly, the result of 3 additions of the same number is clearly divisible by 3 then you got at most 10 tries to figure out which one's middle digit matches the result.

[@daveincognito]

I used brute force to solve for each of the numbers, but knowing that the solution was a multiple of three was a big part of finding them.

[@Giulienas.coconuts]

OH that's smart!! 😭 I shouldve thought of that I didn't do it like Presh though. ____________ I did it like if triangle × 3 = 1 square or square => if square is something then triangle is something for all ways. If you are curious, the ways are: ____________ 1. If triangle × 3 = 1 square •triangle = 4 => square = 2 •triangle = 5 => square = 5 •triangle = 6 => square = 9 •triangle = 7 => square = 1 •triangle = 8 => square = 4 •triangle = 9 => square = 7 2. If triangle × 3 = square •triangle = 1 => square = 3 •triangle = 2 => square = 6 •triangle = 3 => square = 9 Then I did if square = smth then triangle = smth (example: if square = 5 then triangle is = 5) and checked if it was right (also I did both ways for 6) Then finding circle was pretty simple. From what I had done I knew triangle (which is 4) times 3 is 12, so we add the 1 to circle × 3's result So circle × 3 = 4 - 1 circle × 3 = 3 => circle = 1

[@Vienticus]

I focused on the fact that the carryover from the ones column would be 0, 1, or 2.

[@justatest90]

More generally, you can solve in about 10-20 seconds when you realize the 🔵CAN'T carry over, and the other two MUST carry over (or they'd all be circles). So 🟨* 3 + 1 or 2 (remainder from the 🔺) must have a 🟨 in the 1's digit and also carry. 0, 1, 2, and 3 are immediately out. 4 works (4*3 + 2 = 14) and at that point the solution is self-evident (🔺* 3 = 24 means it's 8, and the circles are 1 to make 🔵* 3 + 1 = 4). But for completeness: 5, 6, 7, and 8 don't follow our rule (5 would have a 6 or 7 in the 1s place, 6: 9 or 0, 7: 2 or 3, 8: 5 or 6). And 9 doesn't work even though it passes the test (9 * 3 + 2 = 29) because it would mean 🟨 is 9, and you'd need 🔵* 3 + 2 to equal 9, and no whole digit does that.

[@Sunday797sns]

Let's do this without the c, s, and the t,

[@farhanrejwan]

thinking about the other digits a bit more, you'll eventually figure out that the first carry needs to be true and thus t has to be 7, 8 or 9.

[@tHEuKER]

You need it to be non-zero or otherwise the result of 3x square won't end with square.

[@braxtonbunch]

Me too. Solved much quicker than the long, drawn out method in the video.

[@fritz46]

I admire your math, but in this case using common sense is way faster, and you can easily do it in your head without needing to write down formulas. Just test all nine possible values for the triangle, and you'll easily find out that only one, 8, doesn't cause a contradiction in the 2nd column. The only math you need is to to remember the carry-over!

[@loiscampos4553]

That was the method i used in the end, i end with the same equation of the video but just stuck in middle of the process, and then try this and it was way faster and simple.

[@andrasferencz7948]

Yes, that's how I done it in a minute in my head, and I don't consider myself a mastermind. Oh, and you don't need to test 0 and 5 for the triangle...

[@glassjester]

That's what I did. Took like a minute.

[@roginutah]

Same here. give or take.

[@annaclaragoncalves2235]

Yeah, easy, did it based on the thumbnail alone in less than 5

[@cannot-handle-handles]

A systematic approach, such as the one you use, is probably the best way in general. In this specific case, I immediately saw that the sum must be a multiple of 111 by a factor a priori between 1 and 9, so the summands had to be multiples of 37, by a factor of at least 4. It worked for 4 and no other factor, so I was done.

[@joopysalem16]

Exactly what I did

[@sxboson]

148 easy with making green triangles from 0 to 9. At 8+8+8 we get 24, then 4+4+4+2 we have 14. So last 1+1+1+1=4.

[@Giulienas.coconuts]

Yess!! That's how I did it too :))

[@GojoArc0]

Same answer brother

[@Sohaaanv-xt2qf]

Fun fact: Whoever made this (I think unintentionally) made this to be powers of two... 2⁰ 2² 2³ Like, 2^0(circle), 2^2(square), 2^3(cube/triangle) Dunno guys, just my assumption.

[@araneda665]

⏺️Circulo=1 ⏹️Cuadrado=4 🔼Triángulo=8

[@aWecaDo0]

Yes

[@MautozTech]

0, 7, 4 also works

No it does not

[@mateuszmrzygod1676]

You can also revere engineer it, by dividing number sss (111, 222 ect.) By 3 and checking which has the same number as number in the 10s as in the final number, you can only get 4

[@josephinemahon5082]

0= 2 square=5 triangle is 9.9x3=27 carry 2 3 squares = 15 +2=17 carry 3 circles =6 6+1=7 777

[@briant7265]

I did it the grade-school way. The circle must be 1, 2, or 3, or you end up with a 4-digit result. If the circle is 3, the square is 9, you carry 2 from the second column and fail. If the circle is 2, the square is 6, 7, or 8. 6 will cause a carry from the middle column, fail. 7 will carry 2 from the middle column, fail. 8 will give a middle column result of 4, 5, or 6 (not 8), fail. Circle must be 1. That means square is 3, 4, or 5. 3 fails. 5 only carries one, fail. 4 works. Triangle × 3 must end in 4 and carry 2. Thus 8.

[@williamverhoef4349]

It's far easier to start with the last digit starting with 0 and working upwards to 9. it gives you the answer in a bout 15 seconds when you get to 8: 8+8+8 = 24. write down the 4, carry the 2. So, we have as the answer: **4 So, with 8 as the last digit, the middle digit would have to be 4 and then 4+4+4+2 (carried) = 14. Write down the 4 and carry the 1. So, we have the answer as *44. Then 3 times the first digit + 1 = 4, so the first digit is 1. So, we have the answer = 444, the last digit of the number added = 8, the middle digit = 4, and the first digit = 1. So, 148 +148+148 = 444

[@briant7265]

@williamverhoef4349 Nice approach. My favorite from other comments is the approach of abc×3=bbb, then looking at 111/3, 222/3, 333/3.... My solution goes to a common puzzle format that relies more on deductive reasoning than strict math.

[@williamverhoef4349]

@@briant7265 I started doing it your way the first time round, figuring out what the first digit can't be and then looking at the middle digit. But I then decided that it is better to start on the last digit, which is what you do when you actually add up numbers. I found it felt more natural, and therefore easier to solve it this way. I didn't think of the "abc×3=bbb" approach. I find it interesting but also somewhat unnecessarily complex.

[@cmuller1441]

Or you can just try the 10 possible sss: 000=3*000 111=3*037 222=3*074 333=3*111 444=3*148 OK 555=3*185 666=3*222 777=3*259 888=3*296 999=3*333 (I didn't use any calculator. For 333 666 999 it's obvious... For the others just add 37)

[@BramLastname]

Interestingly enough if you change the sss to ttt the problem still works

[@paparmar]

@ 3:10, I took a different approach. I referenced t = 27s - 100c. This indicates that 27s and 100c must be very close together (within 9 of each other, with 27s being the larger). The first multiple of 27 greater than 100 * 0 is 27 so no good. It’s 108 for 100 * 1, so that would work. For 100 * 2, it’s 216, so also no good (and we've covered all candidates). So c must be 1 and s must be 4, yielding t = 108 - 100 = 8.

[@Yusso]

As stated in the comments you can do xxx/3 and the result must have x in the middle and that works only for x = 4.

[@fbj9648]

Much faster

[@BendoesyourHW]

I did it with modular arithmetic, starting in the middle (let yellow square be y) 3y+r is congruent to y mod 10. By inspection r cannot be 0 (or y=5, which is not a multiple of 3 so gives non integer results) or r ≠ 1 (or 2y+1 is a multiple of 10, but one is even the other isn’t), so 2y+2 is congruent to 0 mod 10. Solve for y = 4 (y

[@josenunes8473]

I made it the buldozer way: I tried every single value of the triangle that allowed a solution. For example, if triangle = 1, the square should be 3, but 3+3+3=9, so 1 is not a value for the triangle. I kept doing this until I found that 8 is the first possible answer: 8+8+8=24, so the square is 4. 4+4+4+2, the 2 being the carry from the sum of the triangles, gives us 14, so the square is indeed 4, and with this it is easy to see that the circle is 1, so 148 should be the answer

[@Nikioko]

Start in the middle: Which one-digit number multiplied by 3 plus a carry-over makes a number ending with that digit? That are only 0+0+0+0=0, 4+4+4+2=14, 5+5+5+0=5, and 9+9+9+2=27. We can immediately eliminate 0 and 5, because then square would be equal to triangle, and we assume that different symbols are different numbers (and for triangle=5 we would get a carry-over of 1, anyway.) And 000+000+000=000 is quite boring. If square=9 was correct, then we would need 3×triangle=29. But since 29 isn't divisible by 3, this can't be the answer. That leaves square=4 and thus 3×triangle=24. 24 is divisible by 3, and triangle=8 The last step is to get 3×circle+1=4, which is the case for circle=1. So, we finally get 148+148+148=444.

[@zehradiyab3439]

I work with the circle digit first 1

[@PekoraMamiCouncilHK]

No You should have 3c≤9 Then go on with 0≤c≤3 s=3c or s=3c+1 or s=3c+2 because the addendum from 10th degits might be 0,1,2

[@chawuplays8012]

While it’s true you can do trial and error, I did it using algebra and logic because its more fun for me this way. The simple satisfaction of getting it right first try cannot be replaced by anything.

[@CarolCanGame]

What i do is to simplify the puzzle. Call those shape abc, we got 3 x abc = 111b or abc = 37b Now just brute force b from 1-9 to find the answer (you can also eliminate b = 0, 3, 6, 9 first). if the brute force number is the same as the middle number from abc, thats the answer, and so i got 4, 37 x 4 = 148, and thats the entire answer lol

[@vekyxeivo5595]

0:07 Easy 20 seconds solutions: 3(100a+10b+1c)=111b (divided 3) So 100a+10b+c=37b --> 100a+c=27b 27b means b can be 1,2,3,4,5,6,7,8,9. And we need 3 digit numbers. B-->27,54,81,108,135,162,189,216,243 --> So B cant be 1, 2 or 3. --> 100a means a can be only 1 or 2. If 1, than b is 4 and c is 8, cuz you can not add any number for good equal. If a=2 c need to be 16 or 43 what is impossible, cuz that two numbers is two digits. So only 1 solution is a=1, b=4 and c=8. That was only 20 seconds.

[@diegomartinez7180]

I solved it differently: Lets keep the variables as called by Presh ("c", "s" and "t"). The typical sum algorithm creates a "carry" for the next column to the left if the result of the sum is 10 or higher, so, being the highest possible number a 9 and we have a total of 3 repeating numbers, 9x3=27, so we know that the "carry" for the next column (tens, hundreds, etc.) can be at most a 2, so it must be either 0, 1 or 2. Next step, find "s", so we focus on the "tens" column. We need a number such that "t" multiplied by 3 then added 0, 1 or 2, gives the same number on the "tens" units (removing the "carry"). Hence, we can evaluate the numbers as follows: 1: 1x3=3 and for the "tens" to sum 11 we would need a "carry" of 8, not possible. 2: 2x3=6 and for the "tens" to sum 12 we would need a "carry" of 6, not possible. 3: 3x3=9 and for the "tens" to sum 13 we would need a "carry" of 4, not possible. *4: 4x3=12 and for the "tens" to sum 14 we would need a "carry" of 2, POSSIBLE. *5: 5x3=15 and for the "tens" to sum 15 we would need a "carry" of 0, POSSIBLE. 6: 6x3=18 and for the "tens" to sum 26 we would need a "carry" of 8, not possible. 7: 7x3=21 and for the "tens" to sum 27 we would need a "carry" of 6, not possible. 8: 8x3=24 and for the "tens" to sum 28 we would need a "carry" of 4. not possible. *9: 9x3=27 and for the "tens" to sum 29 we would need a "carry" of 2, POSSIBLE. *0: 0x3=0 and for the "tens" to sum 0 we would need a "carry" of 0, POSSIBLE. So we now know "s" is either 4, 5, 9 or 0. Now we continue with the evaluation of possible results for "s": 0: The problem states that "c", "s" and "t" are different digits, and for "s" to be 0, we would need "t"x3 to be 0, 10 or 20. "t"=0 would mean "t"="s" which is a given false statement. 10 and 20 are not divisible by 3, hence "s" cannot be 0. 9: "s" can only be 9 if "t"x3 is equal to {9, 19 or 29} with a "carry" of 2, so it cannot be 9 ("carry" of 0) or 19 ("carry" of 1). But 29 ("carry" of 2) is also not divisible by 3, so "s" cannot be 9. 5: "s" can only be 5 if "t"x3 is equal to {5, 15 or 25} with a "carry" of 0, so it cannot be 15 ("carry" of 1) or 25 ("carry" of 2). But 5 ("carry" of 0) is also not divisible by 3, so "s" cannot be 5. *4: So, we have only one option left, "s"=4. For that to be right we would need "t"x3 to be equal to {4, 14 or 24} with a "carry" of 2, so it cannot be 4 ("carry" of 0) or 14 ("carry" of 1). But 24 ("carry" of 2) IS DIVISIBLE by 3, so "s"=4, and "t"x3=24, hence "t"=8. Now that we know "s"=4 and "t"=8, we need to find "c". For that we follow the sum algorithm: "c"48 "c"48 +"c"48 ----------- 444 So, 8+8+8=24, we take the 4 and "carry" 2, 4+4+4+2=14, we take 4 and "carry" 1. Now we need a number such that "c"+"c"+"c"+1=4, so ("c"x3)+1=4, so "c"x3=3, hence "c"=1. Final result: "c"=1, "s"=4, "t"=8, the number is 148. If we remove the statement where "c", "s" and "t" must be different digits, we can also evaluate "c"="s"="t"=0, and it would also work.

[@BrotherAlpha]

I solved it while looking at the thumbnail. Three squares have to add up to a double-digit number where the unit digit is one or two less than Square. The first digit that could possible work is 4... and 4 is the correct answer. Yes, there are faster ways to get to the correct answers, but when the first thing I think of after getting up less than 20 minutes ago works it is hardly a Mastermind question.

[@verkuilb]

Your video arrives at the correct answer-but that method is WAY too inefficient for the situation. Trial and error by dividing triple integers by 3 is a far better approach.

[@justatest90]

Yeah, I like when his videos show multiple solutions. The posted solution is great for when other, more intuitive approaches don't work (or you lack the proper 'intuition' - which we all do sometimes!). Like, it's good to include that approach. But there are way easier approaches that solve in seconds of consideration, and I wish he'd have presented some of those.

[@glassjester]

Even faster than dividing - just tried each digit mentally for the triangle; took about a minute to arrive at 8. From there, 4 is a given for the square, and 1 is very obvious for the circle.

[@ThatsMrSomeGuyToYou]

@@glassjester That is not faster. You know that the [square square square] is 111, 222, 333, 444, 555, 666, 777, 888, or 999. 111 and 222 won't give a 3-digit result, if you divide by 3, so they are excluded. 333, 666, and 999 won't give a 3 digit-result which includes 3 different numbers, so they are excluded. So with just that bit of intuition, you have a maximum of 4 calculations to make, before you arrive at the correct answer: 444/3, 555/3, 777/3 and 888/3. 4 total calculations, if you want to make sure that there is only 1 correct answer. To be fair, I used a calculator, but I had the whole answer in less than 10 seconds.

[@glassjester]

@@ThatsMrSomeGuyToYou Good distinction. But I did it without a calculator (or paper). Done mentally, my method is faster, than if I tried mentally dividing.

[@ThatsMrSomeGuyToYou]

@ I think that may depend on the person, but I can't say for sure. Some people are fast at calculating in their heads. I'm not, mind you.

[@arnautarnautsen2564]

3 times a 3-digit number equals a multiple of 111, which is itself 37X3. This means that the number must be a multiple of 37 with three different digits. The numbers 148, 185, 259 and 296 solve the problem with much less trouble than you took.

[@MrSummitville]

Umm no. There is only one solution.

[@markoboksic2238]

Only answer is 148, your other answers dont take in consideration that result have to be 3 digits that are in place of 10

I think he only listed multiples of 37 with 3 diff digits, as to find the solution from those, not suggested theyre all correct answer.

[@MrSummitville]

He wrote, " ... The numbers 148, 185, 259 and 296 solve the *problem* ... ". Wrong. *ONLY* 148 solves the problem. We only know what he wrote, not what he meant. Words have meanings. You should not assume that you know what he meant, because you don't know.

I love how you cherrypicked *ONLY* last sentence out of context and argue "words have meanings" (plural), and also how you chose to skip the part and meaning behind the words "I think" from my comment and replaced it with...whatever the hell you replaced it with. Winning attitude, man, winning attitude.

[@wesleysuen4140]

All of the following can be learnt in split seconds: 1. It’s impossible to have no carries from the units digit column. 2. From the tens digit column, we know that square must be 4 or 9. 3. From the hundreds digit column, we know that square can’t be 9. Then everything falls into the right places.

[@Brickryan]

Wow that was easier than I expected. I got c=1, s=4, t=8 (I haven’t seen the solution yet) I started with t=9 to get that 2 to carry then tried it with t=8 Edit: welp now I realize I probably should’ve proven it better

[@저녘놀]

abc × 3 = bbb cause bbb = 111 × b = 37 × 3 × b, so abc = 37 × b 37 × 3 = 111 37 × 4 = 148 37 × 5 = 185 37 × 6 = 222 37 × 7 = 259 37 × 8 = 296 37 × 9 = 333 only 37 × 4 = 148 is allowed cause b in the both side is equal. a = 1, b = 4, c = 8

[@vekyxeivo5595]

I like your problem solving. I used different.

[@harisfakiridis6416]

You have a number abc that is equal to 100a+10b+c if you divide this by the sum of their digit it give you a quotient of 30 and a residue of 16 find all the possible numbers abc It was one of my favorite problems try solve it if you want !!!

[@Cigam_HFden]

As others have said, you can solve it by comparing the middle number. I put it in a spreadsheet and with m as the middle number for the formula: m=floor(n/10)-floor(floor(n/10)/10)*10 and m=n/37 You can graph it and will only get the correct result with m=4 and n=148

[@aveenmahabal]

just go 444/3 555/3/ 777/3 888/3. only 444/3=148. the sq digits match. solution by trial and error within 20 secs on your calculator.

[@Qwadra481]

Exactly what i did 😔

[@timothysteadham331]

You forgot 333.

[@TheAssaisa1]

you forgot 000 which is also a sol'n

[@TheAssaisa1]

wait nevermind, I just noticed they all have to be different digits

[@小綸林-t3x]

555÷3=185😮

[@ritwikghosh7743]

Considering yellow square couldn't be less than 4, just took me 10 seconds with trial and error😅

[@RJRJ]

Circle can only be 1. 2+ results in thousand column. Square needs a carryover so hundreds (square) will be 4. 24/3 = 8 so 1 4 8.

[@07Pietruszka1957]

I tried to find a non-algebraic method of solving, and I noticed that it is easy to find a solution by substituting the next numbers for the triangle, the next substitutions generate themselves. e.g. triangle=1 is square=3 but then we reach a contradiction. ... e.g. triangle=5 is square=5 because 3*triangle=15, but then we reach a contradiction again, ... and so on until triangle=8, square=4 because 3*triangle=24, then 3*4=12 but we have to add 2 so we have 14. To get 4 again we have to add 3 to 1 so the circle is 1. Answer: triangle=8, square=4, circle=1

[@DumToasty]

Before clicking on the video I proceeded to solve it by setting an equation: 3(100b+10y+g)=111y, then simplify to 100b+g=27y, since all variables are between 0 and 9 inclusive we must have a y such that it is within at most 9 off from a multiple of 100, the only (non-zero) solution would be y=4 (27*4-108) with b=1 which is 8 off, therefor g=8 to get 100+8=108. Additionally the obvious solution with such questions is that you set everything to 0.

[@stevespenceroz]

I did it by trial and error starting with the green and when i got to 8 it worked. Giving yellow of 4. Meaning blue is 1. However...there is another solution - if you allow non-integers. Try yellow is 5 and the middle column works. So that means green has to be 1/3 of 5 (ie 5/3) and blue has to be 1/3 of 5-1 (ie 4/3).

[@titirenee]

I did it a very efficient way The 4 square column implies that the squares + a possible remainer makes 10 or 20 Without remainer it has to be 5 but then you can't divide 4 by 3 so the circles fail (you have to make 5 with a remainer of 1 so you have to make 4) With a remainer of 2 it can be 4, then 3 is divisible by 3 and 24 also so it is a solution With a remainer of 2 it can be nine but 8 isn't divisible by 3 so circle fails

[@evanj3535]

I wrote an equation. Let X be the circle, S be the square, and T be the triangle. 3(100C + 10S + T) = 111S 300C + 30S + 3T = 111S 300C + 3T = 81S 100C + T = 27S There is not enough information to solve it if the variables can be any number, but these have to be one-digit numbers. I did trial and error to find C = 1, T = 8, and S = 4.

[@todd727300]

Circle can't be larger than 3, but simple trial and error shows 3 and 2 can't work. So, circle is 1. The only carryovers that exist are 0,1,2. Trying those carryovers on circle shows that the carryover must be 1 and that makes square 4. Same logic then to solve for triangle. The square column must sum to 14, and since square is 4, the carryover from triangle must be 2. We now know the triangles sum to 24, which divided by 3, results in 8 for triangle.

[@nitailevi8001]

I checked each same digit three digit number divided by three for which has the middle digit as the digit in the answer. 444 was the only one where the middle digit was 4.

[@kestrelwings]

The left column is different than the right column. That means that with the middle column, you carry 1 or 2 over. C must be 1 or 2, because 3 would give you a 4 digit number. If C = 1, then S = 4 or 5. I tried 4 and that worked with T = 8.

[@cyfur7858]

I solved it in around 2 minutes by simply varying the value of the triangle. 1,2,3,4,5,6,7 fail, but eight works ensuring that the square equals four and the circle equals one.

[@KhaasPotohar]

I did it in my head and got 2 8 16, which is also a correct answer but strangely 2X of OP's answer.

[@darthbernstein9889]

Claim to think I had a more elegant solution to this problem. Since 3x blue still has to be a single digit number blue would have to be 1,2 or 3, now you can imagie what would yellow be in each case, 3 doesnt work because if yellow were 9 you would end up with a 4 digit number in the sum. hence blue can only be 1 or 2. If blue is 1 then yellow has to be 4 as it cant be 3 for same sum issues if blue is 2 then yellow has to be 8 as 3x6 is 18 and 3x7 is 21 always rising the 100 digit by one too much. now your left to figure out the green yellow single digit that works for either 14x or 28x to reach either 444 or 888 since 3x8 is 24 you would need 3x to be 48 for the second option to work which is impossible with x being a single digit, so 148 is the only solution left

[@gelbkehlchen]

Solution: 3*5 = 15 = 5+1 in the sense would work for the middle column, but this 1 in the sense for the 1st column then does not work for the 1st column. 3*9 = 27 = 7 +2 in the sense of the 3rd column = 9 would work for the middle column, but this 2 in the sense of the 2nd column then does not work for the 1st column. The only thing that works for the 2nd column is 3*4 = 12 = 2 +2 in the sense of the 3rd column. So the square is 4. In order for 2 in the sense to arise in the 3rd column, the triangle must be = 8, because 3*8 = 24. Since 14 arises in the 2nd column with 3*4 = 12 + 2 in the sense of the 3rd column, a 1 arises in the sense in the 2nd column and the 3 circles must add up to 4-1 = 3. So the circle = 1. The whole puzzle is: 148 148 +148 ---- 444

[@hundredpercentjuice]

Weird. I just started in the middle and ruled out 0 and 5 since that couldn’t work for triangles. That means 3*triangle has to be greater than 10. But a quick check shows it has to be greater than 20. Then you can check where the ones digit of 3*X+2=X This locks square in as 4, triangle as 8. Then square as 1 is trivial.

[@aquawoelfly]

Going down the middle. Adding 0 (because the first three maybor may not carry over) 5+5+5=15 means 5 is in the middle space but no other number gets 5 in the first column (5 is not divisible by 3) Adding one to the middle collumn no number adds up to itself in the middle space (ie carried over 1+5+5+5=16 and therefore wont work) Adding two means that carried over 2+4+4+4=14 carried over 2+9+9+9= 29 Wince no 3 numbers =9 and carries over a 2 the solution is 148+148+148=444

[@ExileXCross]

Easier, way less tedious Since {a, b, c} ≤ 9 Given 3 {c} = 3, 6, 9, 12, 15, 18, 21, 24, 27 We can infer that 4 (12) and 8 (24) is the only possible pair that will result in the same digit, naturally leaving one as the last digit.

[@markc2643]

I got the answer in my head in 5 seconds by starting with the squares. I thought of 4 immediately being 4x3=12 and adding 2 for a carry over. Then figured what number would give me 4 with a carry over of 2. 8x3=24 The circle was obviously 1 at that point. I often find using guessing on problems like this is quicker than forming equations to get an answer.

[@mathmannix]

I started by trying each value for Square, knowing that 3*Triangle ended in Square, which gives 9 different values of Triangle for the 9 different values (1-9) of Square (like if Square is 7, Triangle is 9, because 3*9=27.) (Square can't be 0 because that would mean Triangle was also 0.) And then added up the three Squares with the carry from the three Triangles to see if I got Square. I stopped when I got to 4. Then I tried it a different way, looking at the thirds of each possible Square-Square-Square sum (111 through 999) to see which made sense (three different digits, with the middle digit Square.)

[@devondevon4366]

148 Answer Let the first digit =a, the second b, and the third c, then 3c + 30 b+ 300a= 111b (100b + 10b+ b) 3c + 300a =81 b c + 100a = 27 b (divide both sides by 3) Linear diophantine equation with three variables which can be solved in a number of ways. a=1 b=4 and c =8 is a solution 8 + 100 = 27*4 108 =108

[@ニミウ]

I just do 37*n (where n is 1-9) and check the 10's digit of the results if it's matched with n or not. Ex. if n=8, 37*8=296, 8≠9 so 8 is not correct. Quick check only give 4 as a solution (37*4=148) so the final answer is 148+148+148=444 as an only unique solution.

[@ニミウ]

Note: this is equivalent to dividing 111,222,...,999 with 3 and check them.

[@phungcanhngo]

Thank you professor for awesome solution.

[@ghosttwo2]

Did it in my head through trial and error. A bit of trial shows that the third column needs to carry over a two, so triangle has to be 7/8/9. 8 is the only one that works, and gives square = 4 in the process. After that, it's 3n+1 = 4 to get circle.

[@nasrullahhusnan2289]

To simplify let triangle, square, and circle be x, y, and z respetively. Carry over of the third column may be: 0 if z=3 or less, 1 if z={4,5,6), and 2 if z=7 or greater. How do you know that the third column needs to carry ove a two when in fact the carry over may be 0,1,2

[@ghosttwo2]

@@nasrullahhusnan2289 I think I had to test the options that carry over a 1, but since the least significant digit has to be in the middle too, there's no 3n+1 mod 10 = n. 3 can't be reached with a single digit, so 2 is the winner. Knowing that the carry over has to be two makes it possible to brute force the middle: 1->5, 2->8, 3->11, 4->14 (check), 5->17, 6->20, 7->23, 8->26, 9->29(check). Square is four or nine, but the original 7/8/9 on the right only give 1/4/7. Thus they have to be something-4-8. Middle is four, so the left digit has to be 3n+1=4. Trivial.

[@markfernee3842]

You can do it in your head with a little bit of deduction. Considering the possible carry overs of 0, 1, 2, the squares can only be 4, 5, 9. You can rule out 5, due to the triangles. You can rule out 9 because 7 is prime and so cannot be a factor of 3. Once you have the squares, the other digits immediately follow.

[@SmoMo_]

I did it like this.. Looking at the 1s column, the square can’t be zero, unless the triangle was also zero. Similarly it can’t be 5, unless the triangle was also a 5. The middle column therefore must involve a carry from the 1s column. If the carry was a one, then the parity of the three squares being added couldn’t be the same as the parity of the square below. Therefore there is a carry of two from the 1s column. So either: 3 x square = (10-1) + square Or 3 x square = (10-2) + square Of 9 and 8, only 8 is even, so square must be 4 Therefore triangle is 24/3 =8 Finally 3 circles plus the digit one from 14 is equal to 4

[@brokenursa9986]

I ruled that the number represented by 🟥🟥🟥must be greater than 333 because 111 and 222 divided by 3 would give only two-digit numbers, and it must by less than 1000 because then it wouldn't have only 3 digits. From there, rule out 333, 666, and 999 for all of them providing factors with 3 identical digits when divided by 3. That left 444, 555, 777, and 888. I plug and chug, and immediately got my answer with 444: 🔴=1, 🟥=4, and 🔺=8

[@gamefacierglitches]

I solved this using the carried numbers. 3T = 10A + S 3S + A = 10B + S 3C + B = S Where A and B must be 0, 1, or 2. B = S - 3C A = 10B - 2S A = 10(S - 3C) - 2S A = 8S - 30C The highest 8S can yield is 81, so 30C must be below 81 thus C = {0,1,2}. If C = 0, S = DNE If C = 1, S = 4 If C = 2, S = DNE Therefore: A = 2 C = 1 S = 4 3T = 10A + S = 10(2) + (4) = 24 T = 8 So we have the final results: - Circle = 1 - Square = 4 - Triangle = 8 3 * 148 = 444

[@simonharris4873]

The square can be one of 6 values. 1, 2, 4, 5, 7, 8. If it were 3, 6, or 9, then dividing SSS by 3 would lead to another number where the digits are all the same. Of the 6 possibilities, only one leases to a number with S in the middle; S = 4. 444 / 3 is 148. And that's the answer.

[@marvhollingworth663]

Here's how I did it: It was obvious that square had to be higher than circle & triangle, so square could not be 0 or 1. It also couldn't be 2 as 222 is not a multiple of 3. So I tried square as 4. If square was 4, circle had to be 1 or the total would be too high. 140 x 3 = 420. 444 - 420 = 24. 24 / 3 = 8. Which gave me 8 for triangle & showed me that 1 for circle & 4 for square worked.

[@Сергей_Петров_85]

It can be faster. Just analize the second digit which multiplied by 3 gives 10 + x + [1,2], or 20+x+[1,2] and is the last digit of another 3y product. 1 - no (too low) 2 - no (too low) 3 => 9 = 3+6 != 3 + [1,2] 4 => 12 + 2 = candidate! Then the last digit = 8 (8 * 3 = 24) and the first digit = 1 (1 * 3 = 3 + one more "1" from 4*3 = 4) => 148 * 3 = 444 check another digits 5 => 15, but the last digit must be the same 5 6 => 18 => can't be, because it needs +8 for get 6 at the end 7 => 21 => need +6 to get 7 at the end 8 => 24 => need +4 to get 8 at the end 9 => 27 +2 = candidate! But in that case the last digit must give us x*3 = 20 + 9. There is no digit to respond this. ======================= So there is the only solution: 148 * 3 = 444 Just analysis without algebra -- that is the power of math!

[@FoxMcCloudV2]

The circle can only be 1, 2, or 3, and the square can't be divisible by 3 (because if the square is divisible by 3 then all three shapes represent the same number, which is a contradiction). Now we examine the eligible multiples of 111 to see which one fits the form of the problem: 111 = 037 + 037 + 037 (does not fit) 222 = 074 + 074 + 074 (does not fit) 555 = 185 + 185 + 185 (does not fit) 777 = 259 + 259 + 259 (does not fit) 888 = 296 + 296 + 296 (does not fit) 444 = 148 + 148 + 148 (fits) So the circle represents 1, the square 4, and the triangle 8.

[@diegelbeseegurke2116]

I know this is not optimal, but: 3×(100a+30b+3c) = 111b => 100a+c = 27b Then I broke it down into the parts you would use to solve it manually: 3c = b+10x 3b+x = b+10y 3a+y = b Then I used 3a+y = b and assumed numbers based on a,b,c,x, and y being single digit whole numbers: y=?1 => 3a+1 = b => 3×1 = 4 or 3×2 = 7 7=/=b because 3×7+x =/= 7+10 Which means: 3×4+x = 4+10 =>x = 2 3c = 4+10×2 => 3c = 24 => c = 8 if a=1 then b=4 and c=8 100+8 = 27×4 => 108 = 108 ✅️ This can now also be written as: 148+148+148 = 444 ✅️

[@AutomatedTeller]

just divide 111, 222, 333, etc by 3 and see which one has the middle digit that's right.

[@Xoruam]

I worked through it this way: 1. Square (S) CANNOT be equal to 0, otherwise the result would be 000, which is impossible, unless ALL digits are equal to 0. 2. Since the square has to be higher than 0, and the result of both the circle AND the triangle being added ends in a square, then tripled square has to have a sum that's larger than 9. Otherwise, Triangle (T) and Circle (C) would have to be the same digit. 3. That means that square has to be higher than 3. 4. As a test, let's see what would happen if S=4: 4+4+4=12 1 is carried over, and if C=1, that would make it 14+14+14=42 5. The first digit is correct, but now I need the sum of the third digit to carry over the 2, while having 4 as the last digit: In other words, the sum of T+T+T has to be equal to 24. 6. 24/3=8 148+148+148=444

[@dentonyoung4314]

Very easy to solve. The sum is a 3 digit number of the form ZZZ. Clearly Z has to be at least 3, at most 9, and can't be a multiple of 3. So divide other 4 possibilities by 3 to see which one has a middle digit of Z.

[@JDrapic]

Not too bad this one. Since each is a single digit, circle (x) cannot be greater than 3, and if it is 3, square (y) must also be no greater than 3, since carrying any digit over will result in square being 10 or higher. Because we're adding 3 digits, the highest possible carry-over is 2. This means each possible circle is now narrowed down to 3 cases each: 3x=y; 3x+1=y; 3x+2=y. Early on we can eliminate x=3. Any carry-over would result in y being greater than 9. Since 9x3=27, we can eliminate x=3. We are left with 6 possible combinations for x and y. For x=2, we start with no carry-over. 3x=6 -> y=6 -> 3y=18>10. This case is eliminated. Let's try carrying 1. 3x+1=7 -> y=7 ->3y=21>20. This case is eliminated. Maybe we can carry a 2. 3x+2=8 -> y=8 -> 3y=24. So far so good, but now we need to have 3z=39 3y=9. From here, we would need to carry a 4 for the y column to work out. This is eliminated. How about carrying a 1? 3x+1=4 -> y=4 -> 3y=12. We would need z to carry a 2 for this to work. Happily 24 splits equally into thirds, giving us 8. We have a good case! x=1, y=4, z=8. Let's see if the third works. 3x+2=5 -> y=5 -> 3y=15. We would need to carry 10 for this to work out, so it is not a valid case. For fun, we can try x=0, but none of those cases are possible either.

[@siavashakrami3444]

How should we define the duration it takes for a person to solve a problem? Is it (A) from the moment they start to read and understand the problem, (B) from the moment they figure out the correct approach, or (C) from the moment they start applying the correct approach toward solving it? I see people claiming this could be solved in a few seconds, but I'm sure they are measuring the (C) time, and not A+B+C time.

[@gavindeane3670]

I'm sure they're measuring the (C), but then that's what Presh is showing in the video - he's already figured out his approach, and the video is demonstrating that approach. The point is that there is a much simpler and quicker approach, which takes much less (C) time. Presh has acknowledged that now.

[@justinstephenson9360]

I started with the observation that as the result of 3(cst) was a 3 digit number the maximum value for c is 3 as any higher integer means 3c>=10 which would result in a 4 digit number. Then I continued as per the video until I got to 300c+3t=81s (eq1). I then considered the right hand column where we can conclude that because the largest single digit in 9 and 3x9=27 if we treated the right hand column as independent sum we have 3 possible solutions 3t=s, 3t=10+s and 3t=20+s. As c,s & t must all be integers we can test each solution by substituting into Eq1 if 3t=s =>300c+s=81s=>300c=80s=>s=30/8c which does not yield an integer for s & c if 3t=10+s=>300c+10+s=81s=>300c+10=80s=>30c+1=8c again not possible as the only possible values of c (0,1,2,3) in that equation do not result in a number divisible by 8. if 3t=20+s =>300c+20+s=81s=>300c+20=80s=> 30c+2=8s and if c=1 we get 30+2=8s=.s-4 But if 3t=20+s and s=4 then 3t=24 therefore t=8 Therefore c=1, s=4 and t=8 so the sum is 148+148+148=444!

[@Neodynium.the_permanent_magnet]

Thanks as usual, but a bit disappointed that we didn't explore more methods.

[@quantumcaffeine]

Here's a quicker algebraic approach that avoids most of the trial and error. Look at the middle column, and you get 3s + carry = 10x + s, so 2s + carry = 10x. Since all the other terms are even, the carry must be 0 or 2. However, a zero carry means s = 3t, which is impossible (a sum of 333 means s = t = c = 1, and similarly for 666 and 999). Thus the carry is 2 and we have 2(s + 1) = 10x, or s + 1 = 5x. Thus s = 4 or 9, but we've already ruled out 9 so s = 4 and the sum is 444. From there, c = 1 and t = 8 follow quickly. Not as fast as just trying 111, 222 etc., but hopefully more interesting, and do-able in your head.

[@brunogrieco5146]

I just ignored the circles and went brute force for t varying from 1..9 (0 cannot be the case otherwise t and s would be the same digit) this yielded 3,6,9,12,15,18,21,24,27. taking the last digit, multiplying by 3 and checking quickly revealed that t=8 and s=4, from there, calculating circle was a piece of cake.

[@sgarstin]

Once you arrive at 100c + t = 27s common sense is the quickest method: 100c can only produce multiples of 100 ie. 100,200...900 t is 0-9 so the only result of 27s that works must have the second digit as zero - only 108 satisfies this - hence c=1,s=4 and t=8

[@prit707gameacc8]

How about this mental math: Consider the number being added to itself twice is 'abc'. Now, the answer necessitates 3c should have b in the one's digit and the carry over from 3c plus 3b should have b in the one's digit. So, we go like : if c = 9 then b has to be 7 but if b = 7 then 21+2(carried over from 27) doesn't have 7 in the one's digit. Might sound complicated but do it once in your head and you can run all the digits like this within a minute. Another eg: C= 7, then b = 1 but 3b + 2 doesn't have 1 in the one's digit. Hence not the answer.

[@timwatterson8060]

My solution was a bit of trial and error, 3cst=sss becomes s*111/3=cst to s*37=cst s, can only be 9 possiblities run each one and if the solution follows the correct silhouette and has the "s" used as the middle digit i have a winner.

[@jeremiahlyleseditor437]

Great Solution. Those types of problems always get me.

[@damonf6564]

Anything under 333 when divided by 3 is two digits. 333 comes to 111 when divided by 3. Working upwards, 444 we get 148, three different numbers and the middle number matches the digit that is the total in triplicate.

[@achandagumbs6565]

How I solved it: By pure logic we can deduce only 10 combinations for the total given the square symbol is prevalent 3 times. This solutions are 000, 111, 222, 333, 444, 555, 666, 777, 888, and 999. Through this, we can also deduce that the three circle, square, and triangle numbers are the same. Therefore, we can divide each solution by 3. If any of them don't have the second number equivalent to the numbers in the solution we can eliminate them as an option. This leaves 000 and 444. Either of these can be a solution with circle = 0, triangle = 0, and square = 0, or circle = , square = 4, and triangle = 8. We can get rid of solution 000 since each shape must be unique. Interesting to note how many different ways there are to solve this problem.

[@spafon7799]

I solved it this way: Define the 100s, 10s, and 1s digits as c, b, and a, respectively. Look at the 10 cases for a. A cannot be zero because the 1's digit of the sum is b, and the problem indicates a,b, and c are different integers. The other 9 cases of a each lead to a specific value for b, which is the 10s digit of the three numbers. For a= 1,2,3,4,5,6,7,8,9 b=3,6,9,2,5,8,1,4,7 respectively. Now that you know b, you can add up the 10s digit of the three numbers along with the number carried over to the 10s digit from the sum 3a. This gives for the 10s digit of the sum of the 3 numbers b=9,8,7,7,6,5,5,4,3. This b must be equal to the b already determined previously. The only case in which b agrees between the two is for b=4, the value that goes with a=8. Now that we have a and b, we find that the sum in the 100s column is 3c+1=b. Thus 3c+1=4, hence c=1. We now have the solution 148.

[@toray424]

Start with the circle, only 1,2,3 can fit. if 3, square cannot be 9. if 2, square will be either 6,7,8, which doesn't work, so circle can only be 1, then square can be 3,4,5, the only option is 4, lastly triangle will be 8

[@CalliopePony]

I feel like the method shown in this video is the most complex and time-consuming option. We know the sum 🟨🟨🟨 is a three digit number where every digit is the same. There are only nine such numbers (111, 222, etc). It's quick and easy to divide each of those possibilities by 3 until you find a three digit number that fits. Alternatively, there are only ten digits that the ones place can be. Using guess-and-check for the green triangle quickly eliminates possibilities until you find the number that works.

[@williamito7650]

now that was easy, guess I am a mastermind according to the thumbnail.

[@monroeclewis1973]

Ramanajan could have solved this in a flash, 5 seconds or less. I, a retired attorney, with neurons in decline, took 10 laborious minutes, floundering around randomly, then drawing an elimination grid. After that it was easy to test values for the triangle to see what impact they would have on the rest. Quickly settled on 8 which implied the square could only be 4. Then easy to see the circle must be 1. No algebra needed. Using algebra is like dancing Swan Lake to get from one side of the stage to the other.

[@Vfdking]

I used this method, I counted all multiples of 1 to 9 and saw if the second term would contain the same number as the first digit of the multiple.

[@brianwade4179]

Once you have 100c + t = 27s, just look for a multiple of 27 that is a little greater than 100 and see if you can make it work. Trying s=4 gives c=1 and t=8. AND THAT'S OUR ANSWER.

[@billhill897]

Easy. Took 5 minutes tops. Blue=1 yellow = 4 green = 8 for 3 times 148 = 444. Tried 9 in the first place, that didn’t work. 3x8 =24 so 4 in second works since 3x4=12 plus 2 puts four in second place in the answer. All the yellow squares are 4 and the rest is obvious. No fancy algebra needed.

[@taflo1981]

There's an easy solution without case distinctions and that only uses extremely simple calculations (not even any division of multi-digit numbers). For the sake of shorter notation, I'll denote the value of the circle symbol by C, the square by S, and the triangle by T. (1) Observe that the last digits of 0*3, 1*3,... 9*3 are distinct (this also follows directly from the fact that 10 and 3 are coprime). Thus, since C, S, T are supposed to be pairwise distinct, there has to be a carry-over in the ones and in the tens, and those two carry-overs have to be distinct. (2) Looking at the tens, we see that 2S plus the carry-over from the ones (which we already know not to be zero) needs to be divisible by 10, which implies that 2S ends in the digit 8 and the carry-over from the ones is 2. As the carry-over from the tens needs to be different from that from the ones, we know that the carry-over from the tens is 1. This already gives us S=4, in order for 2S to end in the digit 8 *and* the carry-over from the tens to be 1. This also gives us C = (S-1)/3 = 1, and T=8 (the only digit for which 3T ends in the digit 4).

[@FrostedFaunOfficial]

I solved this by using logic for the system :) 3t must equal s or a double digit ending in s. Meaning s must be 3 at a minimum, but if it’s 3 then the second digit (sss) equals 9 so invalid, but if s is any higher it’ll reset by goin above 9 and resulting in a low number. So instead t must be a high number who will add to sss. Then you can just fill it in, 9+9+9 is 27, assuming s equals 7 we’d have 7+7+7+2 = 23 which is invalid. So now Try t as 8 and we get t*3 = 24. Assume s is 4, s*3+ 2=14 valid. 3c+1 must equal 4 so c is 1. Bam answer is 148

[@bart6650]

Circle must be 20 and have 4 as last so 24 works so z=8 so x=1

[@Hambrack]

I had a different solution myself. I started with the squares. S+S+S plus possibly whatever it gets from the T+T+T (so either 0, 1, or 2) equals something that ends in S. So I went down the list. With 0 extra, only 5 works (3×5=15), but there's no way for C+C+C+1 to be 5. So I started going down the multiplication table and adding 1 or 2. 1, 2, 3 didn't work, but 4 did if I added 2. And 24 is divisible by 3, and 4×3=12, so 3×1+1=4, and that's how I got my breakthrough. Then I went on with my grocery shopping.

[@gumbycat5226]

I haven't watched the video but I solved this in about 40 seconds, in my head. The middle row came down to 8 or 4 - it ended up 4 when I tried combinations for the first row. The last didn't require any thought.

[@timsmith8489]

I started with the s column. Let x be the carry from the t column (x=0,1, or 2). We have to have 3s+x = s + 10 or s + 20. That gives s = 5-x/2 or 10-x/2. Three cases to consider: If x = 0, s = 5 (we eliminate s = 10 because s must be a single digit). But x = 0 also means s = 3t, which does not work when s = 5. We can scratch the x = 0 case. If x = 1 then s is not an integer so scratch that case. That leaves the x=2 case which gives s = 4 or 9. We can eliminate 9 because s = 9 means 3t = 29, which would give t > 9. So all that is left is s = 4. That gives 3t = 24, or t = 8. With s = 4, the carry from the s column is 1. The c column then tells us that 3c + 1 = 4, or c = 1. We thus arrive at c = 1, s = 4, and t = 8.

[@rivkah9929]

Solved this in my head in a minute. The middle number had to be 4 with a carry over of 1, so the triangle had to have a carry over of 2, so 8*3

[@OctavioAlvarez]

I just checked the 10 possible square values. Easy to eliminate options because of the needed triangle carry. Fast.

[@krjhg17fddt]

100c+t has 0 in the center, that means 27s has 0 in the center as well, the only solution for that is 27*4=108

[@philz3ro823]

It is very easy to solve, just find out the numbers on the Tens column and basically you solve everything. basically the square can only be replace by 4 so you get all the answer, this can be done even with large numbers

[@russell2952]

This is more a logic puzzle than a math puzzle. I made a quick table of 3n, 3n+1, and 3n+2 and saw that the square had to be either 4, 5, or 9 and it's pretty obvious it can't be 5 or 9 because there's no digit where 3n+1=5 or 3n+2=9. And from there it's trivial to get the triangle and circle.

[@jmcminn1076]

I solved it in my head in like 30 seconds. Once I determined that the yellow square had to be 4s with a carried 2, the rest just fell into place.

[@ebbitty132]

Less technical, but felt shorter than the video. I quickly ran 9 digits multiplied by 3 and calculated the difference to get the last digit back to the original parent digit. (example 4 * 3 =12 but it takes 2 to get last digit back to original 4) If the difference was greater than 2 it is eliminated since at most you can only carry 2. Leaves you with s = 4, 5, or 9. 5 is eliminated because the only way t can become 5 is if it too is 5. Between 4 and 9 only 4 has a tuple that produces a 4 in the last digit place. When you realize the s digit place will carry a 1 it leaves 148 as solution.

[@kix4christ130]

I just brute forced what digits could work in as the triangle. I put a number in as the triangle, which gives me what the square should be, and see if the second column works with that number. This only works with 8, and then it is trivial to find the circle after you have the triangle and square.

[@JohanAulin]

I had 100x+z = 27y, and knew x,y,z were all [0,9], so I wrote out the 27 table up to 27 * 9, and knew it must be possible to subtract 100x from it and for the remainder to be single digit. Only 27 * 4 = 108 fulfills that, thus I knew y = 4, z = 8, and x = 1.