Solving AB*A*B=BBB

[@bprpmathbasics]

Try this puzzle next: kzbin.info/www/bejne/hKCZnY2Hq7OCi5Y

[@balthazarbeutelwolf9097]

I started from the result BBB. That is B*111, and 111=3*37. This means that AB must be divisble by 37, giving us only two options 37 and 74, and only 3,4 7 as options for A and B. But 111 is also divisible by 3, and neither 37 nor 74 is, so either A or B has to be divisble by 3. That leaves us with AB=37, which we admittedly still need to check, but we had it down to one option only.

[@0hdhjfbdi]

Smae

[@Cau_No]

​​​@@0hdhjfbdiI shortcutted the answer from AB * A = 111 = 37 * 3 , seeing that is already the only fitting option. (it's the prime factors) How do you think 74 * 1.5 would be feasible, when there are no fractions nor sums involved?

[@pietergeerkens6324]

Yep! Perhaps I was specially primed for this challenge, by recently working on several challenges around multiples of 111, but this problem took me only 10 or 12 seconds to solve with this line.

[@houstonceng]

I worked it backwards so BBB/B=111. A single digit number multiplied by a double digit number to give 111 is 3x37. That matches A=3 and B=7.

[@哀恩欸芙屁]

37 x 3 x 7 = 777. The first question ever solved within 5 seconds 🤩🤩🤩 I notice that BBB = 111 x B = 37 x 3 x B (by prime factoring 111) And then I guess A = 3, B = 7 immediately I plug in A = 3 and B = 7, and the whole thing works, yay 🤗✌🏻

[@tinybabybread]

From the thumbnail, my guess was 37 × 3 × 7. My reasoning is that AB × A _HAS_ to equal 111, because multiplying 111 by the variable B is the only way for the product to end up as "BBB" 111 doesn't have many factors, and 37 pretty much immediately came to mind that suited the parameters. I got to my answer quickly but it still involved guess work and logic, so time to watch the solution! edit: awesome. love to see the straightforward coherence

[@Frederic7594]

Same reasoning, adding the fact that 111 is semiprime with nontrivial prime factors equal to 37 and 3. Therefore AB is forced to be 37 (two digits) and B to be 7 (one digit), because 111 semiprime, no other factor. A matches with itself, there is a solution. Finally, finding A=3 and B=7 iwas quite obvious.

[@martinpenwald9475]

Yeah, same here, I know that 37×3=111, which give the obvious solution that A is 3 and B is 7.

[@ОлегПолканов-д1н]

If smth * by B = BBB, then this smth is 111, since B is a 1 digit Then we have AB * A = 111. A 1 And 111 is divisible by 3, leaving 37 So 37 * 3 * 7 = 777

[@pbolchover]

Dividing by B, you get AB x A = 111. Now consider the fact that (AB) lies between 10A and 10(A+1), and therefore AB x A lies between 10A^2 and 10(A+1)^2. We can therefore find A as being the largest integer whose square is less than 11.1 (ie 3). And then dividing 111 by 3, we get AB = 37, so B=7

[@tristanridley1601]

Cool to see someone solved it differently.

[@tristanridley1601]

My solution was the same but a different order slightly. I thought about the number BBB, saw that it was 111, 222, etc, some multiple of 111. I looked for prime factors and found 3*37. So AB * A * B = B * 3 * 37. I only cancelled out B at that point. Since I had a one digit number and a two digit number starting with the same number, I had the solution. Maybe not as clean as the oure algebra way, but maybe easier to do entirely in your head.

[@DanielBearden]

Ah, but the prime factorization is arguably a 'guess and check' ;)

[@brianwade4179]

Def a deus ex machina move :(

[@triniasta]

i like when you make videos about problems that arent like impossible to do in your head

[@stokmlnes-flame2025]

im guessing you like the number 7 if you like the number 17 for no reason at all

[@skc4188]

I can relate to that. 😅

[@louisrobitaille5810]

I like 17 but not 7. 17 is prime, 7 is overused.

[@anonymouscheesepie3768]

to make this easier, write it out in algebraic terms: ab(10a + b) = 100b + 10b + b ab(10a + b) = 111b b's cancel a(10a + b) = 111 111 prime factorization: 111 = 37 * 3 a and b are integers contained within {1,2,...9} so the only possible system of values for a and b is that a = 3, and 10a + b = 37. it cannot be the other way around bc a =/= 37, as it is contained within {1,...9} so a = 3 and 10a + b = 37 10(3) + b = 37 b = 7 thus, a = 3 and b = 7. case closed without guess and check!

[@muskyoxes]

Since 37 is prime, there's immediately no other choice but 37 * 3 * b = bbb

[@phoenixarian8513]

I've seen another puzzle on one of the non-textbooks when I was primary school child. It says AABC*DEF= (vertical multiplication adding in 5,5,4 digits) =BBBBBB and the answer is 3367*198=666666. It's the same logic they tried to crack it from B*111111. Guess you would be interested in this puzzle and I don't mind if you make a video out of it.

[@compiling]

Simplifying, gives AB * A = 111 = 3 * 37. 37 is prime, so that's the only way to factor it into a product of 1 and 2 digit numbers. Therefore A = 3, and B = 7.

[@JoshuaTagliarino]

I solved it algebraically: (10a + b) * (ab) = 111b 10*a^2*b +ab^2 = 111b ab(10a + b) = 111b 10a + b = 111/a The only number that is natural, less than 10, and a divisor of 111 is 3, so *a* must be 3: 30 + b = 37 and that gives out b = 7.

[@graf_paper]

We know that BBB is dividable by 37 because all numbers of the form BBB are dividable by 37. That pretty much solves the problem.

[@navghtivs]

Remembering 111=3*37 and it can be very quick to guess the answer

[@Silvar55x]

You can just divide both "sides" by B. Then you'll be left with AB × A = 111. Factoring 111 = 37 × 3 (only way, since these are primes), so A=3, B=7.

[@dlevi67]

All three-rep digit numbers are multiples of 37, which is prime and two digits, so AB can only be 37. Which means AB = 37, yielding A= 3, B= 7, BBB = 777

[@rogerkearns8094]

Knowing a thing about the number 37 led me immediately to the solution.

[@jamestschirhart269]

for 111/A = AB. I thought of the largest square under 11 since A divided into 11 needs to equal A, therefore A*A = 9. giving a remainder of 21 and 21 divided by 3 gives 7 resulting in 37 which is AB

[@МаксимАндреев-щ7б]

BBB=B•111=B•3•37 -> AB is divisible by 37 -> AB = 37 or 74 1) AB = 37 -> 37•3•7=111•7=777 2) AB = 74 -> 74•7•4≠444 Answer: A = 3, B = 7

[@Dravignor]

Knowing Linear Algebra for this problem makes it a breeze

[@ghostfacevillah]

Any other fairly chill algebra problems like this?

[@not_vinkami]

Simply because 111=37×3 and 37 is prime, we can easily figure that *AB* must be 37, so A=3, B=7

[@goseigentwitch3105]

(10A + B)(A)(B) = 111B factor out B (10A + B)(A) = 111 assuming we only want integers, 111 = 37 x 3 so we have 3 and 7 gg

[@zachansen8293]

(a*10+b)(a)(b)=(b*100)+(b*10)+b and we know that B*A*B=(ends in)B divide both by B B*A ends in 1. Very limited choices for that when A=/=B. not 1*1 not 9*9. 3*7 works.

[@robertholder]

algebra with a dash of number theory

[@Fox3303-l1t]

Thx math man

[@cyruschang1904]

AB x A x B = BBB AB x A = 111 A0 x A + B x A = 110 + 1 A = 3, B = 7

[@shogun6943]

Nice puzzle

[@bprpmathbasics]

Thank you 🤗

[@RoderickEtheria]

111=3×37. So this reduces the options of large number to 37 or 74. 21×37 vs. 74×28, clearly 3×7×37 is the only one that could work, given 4×8=32. So now to check, 21×37=630+147=777. It's true I could have just left out the step by dividing B from both top and bottom first.

[@msolec2000]

I thought you were going to get to 10A²+BA-111=0 and solve the quadratic in A. A = (-B +/- sqrt (B² + 4440))/20. That would be tricky.

[@interdimensionalsailboat]

Is simple it ends as AxA at the top if you cancel out the B's meaning thats S tier on top. Devided by B is tricky, however its only 2 tiers below S so it probably a large fraction of S. I do think that the scale drops off faster at the lower ends. But that seems irrelevant since A- is not real and devided by B would almost certainly not be S anymore. Its A.

[@brianwade4179]

You pulled a rabbit out of your hat when you magically decided to factor 111. Not cool. One can solve this analytically without knowing the factors of 111. AB x A = 111 ends in 1 so we know both A and B are odd. Further, because the first two digits of 111 are "11" we know A^2 must be less than 12. The only A that satisfies both conditions is A=3. From this we get (30+B)x3=111. Solving gives B=7.

[@nikolakosanovic9931]

AB*A*B=BBB AB*A=111 The only way is for a to be 3 and b to be 7

[@doge9737]

nice one.

[@bprpmathbasics]

Thanks!

[@tashkiira7838]

Before watching the video, just glancing at the thumbnail, mental process. BBB is a multiple of 111. 111=37*3. if A=3 and B=7, AB*A*B=37*3*7=777=BBB. While an argument could be made that any ordered pair (A, 0) could work if you write two leading 0s, 0 is not in the valid possible set of numbers B could be in, so there is no other solution.

[@ahborn6961]

My thought process is different: only 1×b=b so a×b have to end with 1. And from 1 to 9 the only case of multiple that fit the criteria is 3×7 (9x9 is invalid bc a=/=b). And 73 didn't fit the result so it's 37.

[@sgtsodium6472]

Hmmm… you said no “guess and check” in the title. At 2:50 I feel like ‘knowing’ 3x37=111 is essentially guessing. I get that someone who does lots of math would know 111 is the product of 2 primes, but… idk. Seems like a cheap way to skip solving half the problem. I was hoping you’d show how to solve this problem algebraically, even if it’s harder to do it that way. Once you’ve solved for B in terms of A, is there not a way to use algebra to derive the values of A and B without preemptively setting A=3 based on 111 being the product of 3 and 37? 111 could also be the product of 1 and 111, and the only reason that doesn’t work is because of the constraints of the problem (if A = 1, B = 101 by the equation we create at the top, B is not in the set of acceptable answers.) Knowing that in advance and solving based on that knowledge is still guessing, imho. You’ve mentally eliminated the only other possible solution, rather than use algebra to solve for the only possible solution. Granted, I dropped out of college, maybe for good reason. Lol My understanding of math is probably weaker than the average commenter.

It's very well known that if the sum of digits in a number is divisible by three, the number itself is also divisible by three. We were taught that when I was like 12 or so in school. So you don't need to *know* that 111=3*37, it's obvious that it must be divisible by three since 1+1+1=3. The factor 37 is found by division and we can tell that it's prime since it's not in any of the multiplication tables we had to memorize in school. :)

[@sgtsodium6472]

I guess so. I don’t remember being taught that rule either, but I’d still think knowing the ‘proof’ behind that rule would be good too. To be clear, I’m not saying that in the real world you shouldn’t use tricks and rules of thumb like that to solve problems - you should - I just thought the “guess and check” clause in the title was interesting. Not wrong, per se, just an interesting claim. I was just expecting a more comprehensive breakdown of how to solve the problem, because otherwise the finite possible series of values for A and B (especially when neither can be equally to one another OR equal to zero) basically means the problem is pretty easily solved without doing any math. Haha

[@lucatherine4089]

(10 A + B) A B = 111 B ⇒ (10 A + B) A = 111 ① 10 + 101 = 111 ⇐ A = 1 ✔, B = 101 ✖, ② (30 + 7) 3 = 111 ⇐ A = 3 ✔, B = 7 ✔, 👈 ③ (70 + B) 7 = (15+6/7) 7 ⇒ NIL, ④ (90 + B) 9 = (12+⅓) 9 ⇒ NIL, 37 × 3 × 7 = 777 checked to be true, A = 3, B = 7.

[@howareyou4400]

Something times B is BBB, that means something is 111 from there it's second grade

[@ra1nman_mashups]

A = 3, B = 7, AB has to be a multiple of 37 since BBB is a multiple of 111

[@eliechaya9690]

Excuse me but you guessed that you had to divide 111 by 3…..

[@Prowler1000]

It's not guessing, it's recognizing that if the digits of a number sum to a multiple of 3, then the original number is divisible by 3

[@ElevatorFan1428]

37 * 3 * 7 = 777 A = 3 B = 7

[@tontonbeber4555]

Took me half a second to find the solution

[@CompletelyCr]

Solved by inspection. And that is why my maths teachers didn't like me.

[@Prypak]

37*3*7

[@andrewrobison266]

How about A=37/12, B=31/6?

[@fifiwoof1969]

Hardly single digit integers.

[@mayankthakur5341]

10*1*0=000 technically

[@alkrab764]

Dang it, we cannot use zero.

[@Madgearz]

AB ×A ×B =BBB (10A +B)A×B =100B +10B +B 10A²×B +A×B² =111B 10A²×B +A×B² -111B =B(10A² +A×B -111) =0 B=0 10A² +A×B -111 =0 10A² +A×B =111 0≥A≥9 1≥B≥9 A∈Z A²∈Z 10A²∈10Z 0≥10A²≥111 10A²

[@沈博智-x5y]

i did it a convoluted way consider the units place BAB = AB^2 If AB^2 = B => AB = 1 (since B is non zero) Consider the whole product (10A + B)(AB) = 111B (10A + B)(A) = 111 (for some reason, i did not see A had to be 3 from here yet) => 10A^2 + AB = 111 => 10A^2 + 1 = 111 (since we found earlier AB = 1) => 10A^2 = 110 => A^2 = 11 => A is not an integer. So it must mean AB^2 = 10C + B (xxx) Considering the product with the tens place (AB)(10A) + 10C = 110B => BA^2 + C = 11B => C = 11B - BA^2 substitute this into (xxx) => AB^2 = 10(11B - BA^2) + B => AB^2 = 110B - 10BA^2 + B => AB = 110 - 10A^2 + 1 (B is non-zero) => AB = 111 - 10A^2 => B = 111/A - 10A (A is non zero) Since B is a single digit non zero positive integer and A is also a single digit non zero positive integer And using the fact that integers are closed under subtraction It must mean 111/A is an integer. so A divides 111. This means A = 3 so B = 111/3 - 10(3) = 37 - 30 = 7

[@沈博智-x5y]

So convoluted Because I should have got the result immediately from (10A+B)(A) = 111 (considering the whole product already) => 10A^2 + AB = 111 AB = 111 - 10A^2 => B = 111/A - 10A

[@terrypold]

This took seconds.

[@robertstorlind2302]

Video not sponsored by Boeing 😊

[@m.h.6470]

Solution: (10a + b) * ab = 111b |:b 10a² + ab = 111 |-111 10a² + ba - 111 = 0 a = (-b ±√(b² + 4440)) / 20 We need a square number of an integer, that is between 4441 (=4440 + 1²) and 4521 (=4440 + 9²) √4441 ≅ 66.64 √4521 ≅ 67.24 So the only value for b² + 4440 can be 67² = 4489, therefore b² = 49 and b = ±7 a = (-(±7) ± 67) / 20 a = ±3 Since the original Term requires positive single digit values, we end up with a = 3 b = 7 and the original equation becomes 37 * 3 * 7 = 777

[@_Odanert_]

b=0

[@zachansen8293]

both are stated as being 1-9

[@jeremyashford2145]

Poorly presented puzzle. I read it as ABxAxB=BBB, which is A squared x B squared = B cubed. A=2 B=4 The capital letters and the boxes made noise. Clean it up. And don't ever present this to anyone who expects marks for it.

[@blankspace178]

A^2*B^2 = B^3 3^2*7^2 = 7^3 9*49 = 343 441 /= 343 *These solutions do not match!!! His argument about placeholder values is simply wrong, this problem is written in simple arithmetic format. It's just wrong.*

[@ericgoldman7533]

You clearly do not understand the notation. The red boxes are there to indicate that A and B are base-10 digits, not regular algebraic variables. The top number is not A times B, it's a 2-digit number where A is the tens-digit and B is the ones-digit. The bottom number (i.e. the product) is not B cubed; it is a three digit number where all three digits are the same (i.e. B). A and B being digits

[@saltd1953]

B = 0 A = anything 🗿

[@dhahm6884]

Look again at the start of the video! A,B must be element from {1, 2, 3, 4, 5, 6, 7, 8, 9} You did not solve this riddle.

[@saltd1953]

@@dhahm6884 indeed