A challenging mathematical puzzle
Вставка
- Опубліковано 29 сер 2024
- This problem went viral because teachers had trouble solving it. It even got covered in television news! Can you figure it out?
References
BlackPenRedPen Twitter
/ 1284719356799115264
BlackPenRedPen video (English)
• ABCDE*A=EEEEEE, Super ...
BlackPenRedPen video (Chinese)
• [Math for Fun] 小學四年級的數...
China Times News
• 中天新聞》超難數學！國小四年級考題 建中生也投降
Subscribe: www.youtube.co...
Send me suggestions by email (address in video). I consider all ideas though can't always reply!
Like many UA-camrs I use popular software to prepare my videos. You can search for animation software tutorials on UA-cam to learn how to make videos. Be prepared--animation is time consuming and software can be expensive!
Why are there comments before the video is published? Get early access and support the channel on Patreon
/ mindyourdecisions
If you buy from the links below I may receive a commission for sales. (As an Amazon Associate I earn from qualifying purchases.) This has no effect on the price for you.
Show your support! Get a mug, a t-shirt, and more at Teespring, the official site for Mind Your Decisions merchandise:
teespring.com/...
My Books (worldwide links)
mindyourdecisi...
My Books (US links)
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.1/5 stars on 97 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 3.7/5 stars on 9 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.2/5 stars on 19 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.3/5 stars on 21 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.6/5 stars on 14 reviews
amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 28 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.4/5 stars on 9 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.1/5 stars on 7 reviews
amzn.to/1NKbGlp
Connect with me
My Blog: mindyourdecisi...
Twitter: / preshtalwalkar
Newsletter (sent only for big news, like a new book release): eepurl.com/KvS0r
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter. - Наука та технологія
Dolphins be like: 0:09
JAJAJAJAJ
You made me laugh so weirdly that my ears are weird now.
LOL
Lol
This is hilarious
Thanks, in Taiwan a 45 minutes exam, we definitely have the time to draw a table and tying all the possibilities and find the answer for a single problem that cost 1 point in score. Very cool
It's become a template. Hard problem for [single digit number] year olds in [asian country].
Hard problem for 1 year olds in Japan. College level calculus required.
In the Asian countries if you fail to master calculus before your 10th birthday you bring eternal shame to your family
@@kevina5337 🤣
Next week’s problem was given to 6th Graders in Vietnam as a national math contest tie breaker.
Shawn Gollatz After That will be
Hard problem for 18 year olds in the USA.
*Basic Addition Required*
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
I’ve actually seen this question in my midterm when I was 12 being a Taiwanese student.
I got the point actually, but it’s definitely too hard for a 9yo student.(even in an Asia country)
IJMO 7th grade has a much easier version of this
@@aryamanjha9380 yeah there are easier problems for the younger kids
I want to touch the feet of the person, who has made such a wonderful question☺
This feels like a problem where you kinda brute force your way through lol.
Sometimes brute forcing is the simplest and fastest way.
Converting thousands of possibilities into a few cases and then trying to solve them is not brute force. Brute force is when you try every possibility... perhaps using a computer.
Hahhahahaa so true, I can relate
@@a.16.g you ruined the joke
yeah, I just wrote the algorithm into python, testing all possibilities and it took me like 3 minutes.
I solved it myself, but very differently. I saw that A x A had to be a square, and a square result more than 1 digit (since product has 1 extra digit) plus 'something' carried over that would result in matching digits. The carry over amount added would have to be less than the square root, otherwise it would be higher multiple. Try 16 (4x4): could be 16,17,18,19 (none with matching digits). Try 25 (5x5): could be 25, 26, 27, 28, 29 (no matching digits). Try 36 (6x6) could be: 36, 37, 38, 39, 40, 41 (still no matching digits). Try 49 (7x7) could be 49, 50, 51, 52, 53, 54, 55 (finally "55" has matching digits), so '7' (our trial square root) is a good number for 'A' and '5' is good for the matching 'E' digits. To test it divide the 6 matching digits of "5" by 7: 555555 ÷ 7 = 79365.
I first took the same approach as MYD but because there were still too many options I decided to go with the same approach as you! Interesting how you can solve this puzzle in 3 ways all differently labor (manual big devisions) intensive
I know this is an old comment, but could you perhaps tell me why the "something" carried over would have to be less than the square root? That's not quite clear to me.
I did it! This is the first time I did a question from your videos and got the right answer!! Thanks for sharing these puzzles!
Me too! 🙏🙏
Same, needed about 2-3min
Well done!
👏👏👏
Nice!!
I did it a different way - I knew EEE,EEE would be a multiple of 111,111 - thats is, EEE,EEE=E*3*7*11*13*37. A had to be 7, since A is a single digit and if A was 3, AB,CDE would be a multiple of 1,001 which would guarantee repeating digits. E had to be 5, as 5*odd number ends in 5. So AB,CDE = 555,555/7=79,365.
P.S. 75,924*7=444,444 is also a correct answer, if you count in base 12.
(In dozenal, 111,111 = 7*11*17*111, and 7*7=41, so E must be 4.)
P.S.2 Not all bases work, e.g. in base 60, the only single-digit prime divisor of 111,111 is 7, which 7*7=[Digit-Effent], which is smaller than the base. (That makes a 5-digit number instead of 6.)
I kept thinking something along the lines of 111,111 too... I just didn't know where to go from there like you did =-( Thanks for explaining that!
I took the same approach and came to the same conclusion. Not entirely sure what 9 year olds were meant to do though, being taught prime factor decomposition at age 9 isn't unreasonable (although in UK it's definitely later than that, around 11-12) but having to apply it without any prompting in a pseudo-algebraic context to a 6-digit number? It feels like the examiners either set this as the last question to give bright students something to do/bang their heads against or had a different method in mind.
Did the same thing but with my friend Brute Force, took 111,111 ; 222,222 and so on and divided those by all digits and pin pointed the right result.
@@alleyztak da mey Andrei dar ei vor BRain Force, not BRute Force ..
@@danmimis4576 Ce te aduce la rezultat mai repede contează.
I said it before and I say is now again:
Show me one of the 9 years old students, in Taiwan, that solved this problem.
Come on...
I'm 14 and it took me 15 minutes to did it, how could a 9 years old solve this in 20 minutes
@@Noname-67 Exactly!!!
At age 9 you can solve "problems" like:
Mom have 12 oranges, she gave 3 oranges to her son, and eat 1.
How many oranges left?
I asked mr. Presh (not for the first time) please show us one of these 9 years old who solved this problem... Give us his name... Show us the examination paper...
true, only if that kid takes olympic maths courses at a young age
אחי, למה אתה סקפטי? יש לטיאוונים חוש פי אלף יותר טוב במתמטיקה מישראלים.
@@Noname-67 if you are 14, 9 is onlyn five years younger doesnt that siggest a niner could solve itnin 20 then?
It's actually 6B60AC. Not EEEEEE
It's hexadecimal.
Lol, I also thought the first time the
Same thing
I didn't understand your thoughts.. Can you explain little bit more??
@@hasanemon4496 me too
@@chinnudanturi9730 He assumed EEEEEE is a hexadecimal number, not a decimal (0xEEEEEE)
But 0xA × 0xE = 0xC + 0x80 so that doesn't work.
3:38 I could tell by the way your voice changed that you were smiling while telling the answer! This is truly a very awesome solution and puzzle
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it..
That division method is great! I reasoned my way to A = 7 and E = 5, then multiplied it out to deduce the other digits. The reasoning took quite a bit longer than 4 minutes though! lol
But using logic to get A = 7 followed by E = 5 is much more nuanced than the computer science answer here.
How much longer than 4 minutes
@@m.talhamemon5831 it was a year ago, so I don't remember exactly 😅 a while, I imagine lol
@@idedimi was it worth solving these? Had it improved your reasoning. Or was your reasoning better from start
@@m.talhamemon5831 I love the mental exercise, so I think it was absolutely worth solving! I imagine it improved my reasoning skills, but I don't know by how much haha
There is an additional observation that allows for the removal of two of the test cases: if A < E, then EEEEEE / A will be a 6-digit number, while ABCDE is 5-digit number. That means A > E.
So, we can additionally discard (A, E) = (6, 8) and (3, 5).
Also, 444444 / 6 = 2 * 222222 / 6 = 2 * 37037 and will also have repeating digits, so we don't need to divide it at all.
i think there is one another way if you come down to those six ordered pairs,you can do the following:
see the right most digit it must be a^2+(y6.
similarly if e=5 then the last two places must be 55,that is a^2+y(where y
When you're forced to laugh at a bad joke : 0:09
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it..
You can eliminate a lot of options quickly by noting that A*A
I started out trying brute force for roughly 5 min then realized there were only a handful of possible A,E combos. Then determined 7 had to be A based on 7 being the first number in the dividend of 7 into 555.
That’s the fastest I’ve ever solved anything on your channel. Love it!!!
I was mostly the same. Realized only a handful of possible combos, made a table stating "if E, then A". Made quick work and determined only 3 combos of A and E were technically possible (4 and 6, 5 and 7, and 8 and 6, respectively). Then just verifying which worked after that.
Tough for 9 years old kids, but that teacher who took 20 minutes could use some retraining!
Took me roughly 6 minutes, and my technique was not optimal.
I'm 10 but I did not get😭
Plot twist: it was a geography teacher
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
I literally just chose a random number for EEEEEE and divided it by a random number for ABCDE and it was correct lol. It took me about 15 seconds
@@xfire3778 I'm pressing x for doubt as hard as I can. You have a 1/9 chance to guess the E and then you have a 1/9*8*7*6*5 chance to guess ABCDE. That's a 0.000007% chance at guessing correctly. That's giving you credit you didn't bother with the zero for obvious reasons.
I decomposed EEEEEE into E×3×7×11×13×37.
Then I decided to test 3 and 7 for A first.
EEEEEE is between 10000A² and 10000A(A+1), so if A is 3, then E is 1, but that does not work. Then I test 7, then E is 5. Then that works.
Here's what I did.
When I saw E x A=E, the first number that popped into my mind where it multiplies to a number and the ones digit is the same, was 5. So I took E=5.
5 when multiplied by odd numbers, yields a ones digit of 5.
So I took out a calculator, started dividing EEEEEE ( i.e. 555,555) by 3,5,7,9 to see which number yields an answer where all digits are different. It was 7. So A=7.
There you go.
All this literally took me just 30 seconds.
I like that solution. Thanks for sharing!
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
Someone who did the same as me! I don't know why but I got really happy because of that :))
Sure, division with akward numbers works, but this can be done 100% by only using logic, addition and a 9x9 multiplication table.
EE - some carryover < AxA < EE
In combination with the unique numbers condition
will give you E and A.
Then you can go D to B and use the known carry over from the less significant digit to quickly find D, C, B.
I followed the same steps up to the division part. Instead of division, I went through each multiplication. The long division was a superior method.
I'm pretty proud of my way of finding this too. What I did was take all possible values of EEEEEE (9 of them) and divide them by the 9 possible integers. This yielded a table of numbers. Out of that table it was a simple task to eliminate all values that had repeating digits, weren't whole, or were not 5 digits. That left only 6 values in the 7 column. Meaning A=7. Then out of those 6 values (15873, 31746, 47619, 63492, 79365, 95238) only one of them started with 7, so that was the answer.
I did the same... Using an excel sheet, it takes 1 minute to make the table, plus 1minute to eliminate the wrong possibilities 🤣
Foreign students in Turkey take like these questions and harder in the IQ test that is part of the Y.Ö.S. Exam for entering universities
Kinda like the SAT in the US but a lot harder and only maths and IQ questions (also geometry)
It has a pretty good questions though
Thats EXACTLY why I say it CAN'T be a problem to 9 years old (???!!!)
At age 9 years old you know something about multiple, division, adding, and substract numbers...
You can't solve such a problem at age 9.
No way.
(Read my comment)
The biggest failure in Turkish examination, in my opinion, is that it is a multiple choice questions with 30-100 sec per question depending on the exam... So the education system here forces students and teachers to go to pattern recognition rather than learning the actual subject.
I found most students way of thinking "if the question I get has X, then I should use Y formula to solve it" without giving any thought about why is it this way, or how did we get to that formula. To me, it seems like an exam for robots, not for humans.
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
@@tamirerez2547 Lol what 9 year olds r u talking to? You learn addition/subtraction/multiplication/division/primes composites/factors by 3rd grade (8 yrs old) in the USA, at least for California (bc education is done by state not by the federal gov)
So if even the public school standard have this by 8 then surely advanced kids can reasonably know this by 7 or 6, so it is not impossible for 9 years old to be advanced enough for this question. Would be hard, but not implausible
so bascially the creator of this problem took a number and divided by the other one and wanted students to figure it out.
this is like puzzle games where they just randomly scramble a complete set and they expect the person playing the game to figure out the solution lol
You need to certify that there is only a solution to the problem.
The thing is, you need the numbers to be unique which will be hard. But yeah I was thinking along the lines of this
Sudoku
not really, you just need to know what dividing by 7 does.
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
That happens to me every time I solve a time taking question, I feel like 'how can I take so long to solve such an easy solving!'
Solved it using AxA
We can use divisibility of 7; when a digit is repeated 6 times it is divisible by 7..
It will fasten the second step
more generally, EEEEEE = 3 × 7 × 11 × 13 × 37 × E
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it.....
I have played around with puzzles like this. I usually get through the first part, but never thought to use division to get a more direct way to the solution. I would bang my head against the table trying to figure out what could be the value of some other letter.
Took me about 10 minutes. After eliminating A as 0 through 5 quickly, I tried A=6 and E=4. Then I moved on to A=7 and E=5 and that worked. I used the squares of A as my guide for the possible pairsnof A and E.
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it..
I just straight up counted up A from 0 to 9 and saw the requirements it made on the other numbers and found contradictions at every point except for A=7 and E=5. Indeed, A=6 and E=4 ALMOST worked, that was definitely the closest one that didn't make it.
The division way is so much easier, didn't think of it
Went E*A = XE and A*A is within 9 of EE, which only left 2 possibilities for A,E =(6,4),(7,5)
(In hindsight, could lower the "within 9" to within 7 and only 7,5 would remain)
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
It's been a while since I was able to solve one just from the thumbnail!
Ya this one's alot easier than the "use a double surface integral to find the cube root of the area of this weird ass shape, no calculators allowed!" Lol
ua-cam.com/video/bmOZuadG4uU/v-deo.html
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
I took a slightly different approach… I used the multiplication grid, but then considered how a^2+[carryover]=11e… with this approach, I quickly calculated a and e, and the rest of it just fell into place…
I got it. Knowing this fact helps. In multiplication, the only way that you can get some number multiplied by a single digit number to end with the same digit is by the following:
1: Multiplying it by 1, but in this case that's not possible because you'd be left with a five-digit number.
2: Multiplying a number ending in zero with any number, but that's also not possible because A can't equal E, so they can't both be zero.
3: Multiplying an even number by six.
4: Multiplying a number ending in five by an odd number.
I first assumed that A was six, and thus case 3. Trying 2 for E and dividing it by six left a number too small, and trying 4 and 8 (since E can't be six because it has to be unique) left a number too large.
Thus, I determined that E had to be five (so case 4), but now I had to figure out what A was.
Taking 3BCD5 and multiplying it by 3 left a number too small, even if that number was as large as possible, and taking 9BCD5 and multiplying it by 9 left a number too large, even if that number was as small as possible, so I knew that A had to be seven. So taking 555,555/7 gives us the following: A=7, B=9, C=3, D=6, and E=5.
Well explained.🎓
Same
I'm the junior high school student from Taiwan.
This math competition in Chinese means
"REMARKABLE"
Now if any student wants to participate this competition,this problem is the basic question that they need to solve.
I have been sharing this video to my classmates and mathematical teacher,I hope this video will be popular in Taiwan.
In Taiwan,many of the parents want their children have acuity of logic,they will let children to solve the problem like this,and then there will be many competitive to let these children. According to Asian parents' amazing mental,they want their children have many prizes so they can feel more honor. Due to these factor,the question of the competition become more and more harder. Let's look forward to the questions of this competition 😂
我是一個來自臺灣的國中生
這個數學比賽以中文的意思來解讀即是
「卓越」（卓越盃數學競賽）
現在想要參加這項比賽的學生，這種程度的問題是他們一定要可以解決的題型
我現在把這部影片分享給我的同學和數學老師了，希望這部片能在臺灣發揚光大
在臺灣，家長們會希望自己的小孩擁有敏銳的邏輯，他們會讓小孩去解決這一類的問題，進而衍生出諸如此類的比賽
基於家長們神奇的心理，他們希望自己的小孩可以多多獲獎，使他們感覺的光榮
因此，競賽題目越來越變態
我們期待一下往後幾年這個競賽的題目 XD
Haven't watched yet but sure it involves Gougu theorem somehow.
Gougu has been tested positive for covid 19. So he has been in isolation this days.
Or the intersecting chords theorem.
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it..
"Gougu" in Mandarin Chinese (勾股) is read same as guess blindly in Cantonese (𨳊估)
so if someone says "Gougu theorem" instead of "Pythagorean theorem" in Hong Kong, that means he is guessing blindly 🤦♂️
Nick Wong ha ha
My approach:
EEEEEE = E * 111111 = E * 3 * 7 * 11 * 13 * 17 * 37, therefore A is either 3 or 7.
Then using the unit digit method that Presh mentioned, E must be 5. And here it's easy to see that A is 7.
Can’t believe people in my country actually had to solve this!
I think this problem is easier if you think how you would divide EEEEEE / A if A and E were numbers instead of letters.
To perform the division, you start by finding the first digit of the result. That is, you find a number X such that
A * X
I did that "mentally" (without paper or pen) in 3 or 4 mins... Used the same methodology as in the video.... But used the calculator for the large divisions
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it..
There is no need to complete the long divisions to rule out the cases; checking that the first digit in the division result is/is not the same as the quotient (i.e. "calculated A" = A?) quickly rules out all the cases except A = 7
From the thumbnail:
The thing to note is that from the first digit of the top line that A^2 + some possible carry = EE (=11×E), and from the last digit that A×E = E + 10×some other carry. (In both cases the carry is in the range 0 to 9 - the maximum carry is the carry from 9×A + the carry of 9×A.)
These severely limit what A and E can be.
Searching through the possibilities of A gives the corresponding E (the next multiple of 11 higher than A^2) and then a quick check to see if all conditions above hold.
A=1 obviously fails as the result is not ABCDE
A=2, A^2 + carry = 11 》 E=1, carry = 7, but 2x9 = 18, so max carry is 1; fail
A=3, A^2 + carry = 11 》 E=1, carry = 2, but 1×3 = 3 not 1; fail
A=4, A^2 + carry = 22 》 E=2, carry = 6, but 2×4 = 8, not 2; fail
A=5, A^2 + carry = 33 》 E=3, carry = 8, but 3×5 = 5 (+10) not 3; fail
A=6, A^2 + carry = 44 》 E=4, carry = 8, but 6x9 = 54 so max carry is 6; fail
A=7, A^2 + carry = 55 》 E=5, carry = 6, 5×7 = 5 (+30), possible
A=8, A^2 + carry = 66 》 E=6, carry = 2, but 6×8 = 8 (+40); fail
A=9, A^2 + carry = 88 》 E=8, carry = 7, but 8×9 = 2 (+70); fail
Instead of stopping at A=7, the first possible solution, all possible values of A were searched to check for possible multiple solutions - it is easy to do the next two checks.
So A=7, E=5 and it's a simple matter to fill in the remaining 3 digits from the right hand end.
Now to watch the video...
Nice, that's how I did it too
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
Mine as well, but compared to solution in video is vastly non-optimized
Every base has a trivial solution of A=E=0, B, C, D can be any number. That is excluded.
Solutions for Bases 2-2140. Base 885 has 2 solutions.
Base 4 Solution 1 A=3 B=2 C=0 D=3 E=2
Base 10 Solution 1 A=7 B=9 C=3 D=6 E=5
Base 12 Solution 1 A=7 B=5 C=9 D=2 E=4
Base 24 Solution 1 A=21 B=11 C=4 D=7 E=18
Base 48 Solution 1 A=37 B=4 C=31 D=43 E=28
Base 75 Solution 1 A=26 B=23 C=32 D=6 E=9
Base 80 Solution 1 A=49 B=47 C=77 D=28 E=30
Base 154 Solution 1 A=109 B=77 C=0 D=109 E=77
Base 159 Solution 1 A=130 B=74 C=32 D=98 E=106
Base 175 Solution 1 A=148 B=114 C=63 D=90 E=125
Base 217 Solution 1 A=156 B=112 C=0 D=156 E=112
Base 220 Solution 1 A=210 B=105 C=158 D=21 E=200
Base 248 Solution 1 A=63 B=59 C=75 D=12 E=16
Base 252 Solution 1 A=103 B=42 C=0 D=103 E=42
Base 324 Solution 1 A=229 B=296 C=133 D=228 E=162
Base 425 Solution 1 A=273 B=33 C=142 D=131 E=175
Base 464 Solution 1 A=385 B=133 C=186 D=199 E=319
Base 546 Solution 1 A=337 B=335 C=543 D=206 E=208
Base 615 Solution 1 A=124 B=119 C=144 D=20 E=25
Base 679 Solution 1 A=195 B=191 C=543 D=331 E=56
Base 705 Solution 1 A=546 B=674 C=554 D=496 E=423
Base 738 Solution 1 A=301 B=726 C=110 D=547 E=123
Base 832 Solution 1 A=465 B=635 C=617 D=707 E=260
Base 845 Solution 1 A=756 B=403 C=233 D=322 E=676
Base 885 Solution 1 A=511 B=431 C=726 D=215 E=295
Base 885 Solution 2 A=532 B=825 C=632 D=659 E=320
Base 966 Solution 1 A=806 B=226 C=149 D=434 E=672
Base 1284 Solution 1 A=215 B=209 C=245 D=30 E=36
Base 1294 Solution 1 A=915 B=905 C=1035 D=1174 E=647
Base 1300 Solution 1 A=456 B=638 C=1231 D=1209 E=160
Base 1377 Solution 1 A=892 B=1266 C=689 D=204 E=578
Base 1600 Solution 1 A=1281 B=80 C=1181 D=205 E=1025
Base 1745 Solution 1 A=1561 B=777 C=1425 D=113 E=1396
Base 1898 Solution 1 A=1665 B=359 C=1819 D=154 E=1460
Base 1980 Solution 1 A=511 B=1437 C=1569 D=1058 E=132
Base 2040 Solution 1 A=589 B=170 C=0 D=589 E=170
5 is the only option for E, for this multiplication to hold. So proceeding with odd numbers only gives you the solution. Once E Is found, rest is just calculate for remaining and multiply accordingly
Solution was tremendous. love from Bangladesh.wait for your!♥! react.
Waiting for* hobe! 😊
🇧🇩🇧🇩🇧🇩
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
Hello I'm also from Bangladesh
@Sadia subah I'm from Satkhira and you?
After looking at it for a minute, I started off with the assumption that the solution had to be 555555 as E had to be 5 and A would have to be an odd number to make AxE = N5. After that it was pretty easy.
Intuitively E is most likely 5, as all the odd multiples result in 5 as a one's digit. Also intuitively A is most likely 7, because the multiples of 7 are able to cover every digit from 1 to 9 without needing to go into double digits. And it just so happens that 7 squared is 49, which is a feasible scenario if E is 5 and therefore to get 55 by adding a single digit to 49 is possible. The rest is just plugging in numbers
The answer is relatively straightforward. It took me about 5 minutes, once you realise what E has to be. Think about which times table has the most consistent answers having the same unit as the number being multiplied - the 5x table. All odd numbers multiplied by 5 has a 5 in the unit column and, therefore, A has to be odd. It can't be 1 or 5 so it's simply a quick process of trial and error to get to the answer..
That doesn't mean E has to be 5. For instance it ALMOST worked for E=4 and A=6. 444444 divided by 6 is 74074, so it indeed ends in a 4, but it just fails because the first digit is a 7 instead of a 6. But that's not because of any clever modular arithmetic reasoning, that's straight up luck. Now it so happened that 79365 x 7 was the only nontrivial (A not equal to 0) solution, but your reasoning is still faulty, you just got lucky.
1:50 We can also eliminate (A, E) = (3, 5), (6, 8) in advance since (5 digits) × A = (6 digits) implies A>E.
From 2:47 to 3:23 it is usually not necessary to finish the division.
Example for 222222/6 = 37037 : as soon as we see that the first digit (of what would be 37037) is a 3, it is of no use to continue the division because this first digit should be equal to A, i.e. 6, and 3 ≠ 6.
Nobody:
Not a single soul:
Presh Talwalkar: EEEEEE
I would like to emphasize that I only upvoted your comment because it made me giggle.
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
Without looking at the video, I think I got a way to do it (and am commenting without looking, but I'll check in the end)
ABCDE x A = EEEEEE
My understanding for the first part is that E x A = YE (Where Y is a digit used in the same way as A, B, C, D, E)
There are also a few rules for the calculations that I will explain as I go on.
First, A and E cannot be equal to 0, since it will make the entire equation zero (A by multiplication of zero, and E by substituting it with 0)
The options we get are
A) 1 multiplied by anything (1xE = E and 1xA = A)
B) 6 multiplied by any even number (6xA = YA and 6xE = YE, where Y is a digit used like A or E)
C) 5 multiplied by any odd number (5xA = Y5 and 5xE = Y5, where Y is a digit used like A or E)
One of the things that eliminates E being 3 or 6 is a rule where the addition of the digits of a number if when their digits are added are a multiple of 3, the number itself is a multiple of 3, but the structure of the number (6 consecutive repetitive digits), means that if it was either 3 or 6, it would have a certain repetition of digits in ABCDE, and A cannot be equal to B, which cannot be equal to C, etc etc.
E cannot be 1 on the basis that if E was 1, then ABCDE=EEEEEE and that does not work.
From the three options, we are left with the one with 5 multiplied by any odd number, but from the point, we mentioned earlier, it cannot be 3, and by a similar logic, it cannot be 9 (the sum of digits must equal a multiple of 9). we are left with the two options of 7x5 and 5x5, but A cannot be equal to E, so the equation that fits our goal is 5x7. And in substituting them into the equation, A cannot be 5. If A is 5, from a well-known fact of multiplication, then the number it produces ends in either a 5 or 0, and we established why it cannot be 0, and it must end in 7 anyway from what we said in the beginning (AxE=YE)
So we have E = 5 and A = 7. From this, we can perform a division in order to find the rest of the digits. Divide 555,555 by 7 and we get the rest of our digits.
555555/7 = 79365,
A = 7, B = 9, C = 3, D = 6 and E = 5
It took me five minutes to write a program which found this answer. Sometimes brute force is best!
what is the language and code please give it
@@sb_infi This is Java, but only minor tweaking is required to put it into the language of your choice:
public class Abcde
{
public static void main (String[] args)
{
for (int a = 1; a < 10; a++) {
for (int b = 0; b < 10; b++) {
for (int c = 0; c < 10; c++) {
for (int d = 0; d < 10; d++) {
for (int e = 0; e < 10; e++) {
int number = a * 10000 + b * 1000 + c * 100 + d * 10 + e;
int product = number * a;
int target = e * 100000 + e * 10000 + e * 1000 + e * 100 + e * 10 + e;
if (product == target) {
System.out.println (number);
}
}
}
}
}
}
}
}
Generalized Approach for receiving all solutions (revised):
A ≠ 1 as the output is a 6-digit number.
Consider the multiple of E(unity) and A to achieve result of E; to be in the form A.E = 10a + E for positive integer a from which a = E(A-1)/10; which confirms E = 2 or E = 5; and Case I: for E = 2 (A-1) to be a multiple of 5; hence A - 1 = 5m for positive m integer from which A = 5m + 1; from which for m = 0, 1; A = 1 and 6; as A ≠ 1, for E = 2; A = 6.
Now consider the multiple of leftmost A with A to achieve result of EE (leftmost 2 digits of result) we need to write the multiple as 10E + E = A2 + k for any k positive integer. Substituting A = 6 we get 11E = 36 + k; as you are aware 1 ≤ k ≤ 9 and hence, 37 ≤ 11E ≤ 45; this will imply E = 4 which is contradicting with E = 2; now we conclude E ≠ 2, 4 and A ≠ 1, 6.
Case 2: For E = 5; substituting in 10E + E = A2 + k we get 55 = A2 + k; as you are aware 1 ≤ k ≤ 9 and hence, 1 + A2 ≤ 55 ≤ 9 + A2; right end inequality confirms 55 ≤ 9 + A2 and A2 ≥ 46 and A ≥ 7; similarly considering left end inequality 1 + A2 ≤ 55 and hence, A ≤ 7; now to satisfy both conditions A = 7 and E = 5. This will give result of ABCDE = 79365; for cross checking 79365 x 7 = 555555.
*Fun Fact: 1001 = 7*11*13*
*and 1001 * 111 * n = nnnnnn , n being a single digit number*
it's not worked if n=0
Yes for got to mention that but 000000 is also 0
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
There’s actually an easier way, construct EEEEEE from 111111 to 999999 to check which can start a long division with a perfect square. Only when EEEEEE equals to 555555 will start a long division with 7*7 which result in five different digits number.
Fresh morning and your question at 5. 30 wow .
79365
x 7
555555
Easy to search for a solution because 7 is the only number whose multiplication with single digits covers all 10 digits in the one’s place.
0, 7 , 4 , 1 , 8 , 5, 2, 9, 6, 3.
And we can add any 2 of those digits to get a 5 in one’s place.
I get the solution myself and its take a long time to figure out the answer.😅😅😅 And you have done it in four minutes.😍😍😍
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it..
I just ran into this problem on the Internet and tried my hand at it without having seen the full context, so I didn't know all the letters corresponded to unique digits. I approached the problem in a slightly different way. Looking at the multiplication, we know that A*E + c = E*11, where c is whatever is carried over from multiplying BCDE with A. From this, you can build a table with the possible tuples of (A,E,c). Only two of them pass the other obvious constraint, where A*E has to end in E. That is: (6, 4, 8), (7, 5, 6). With these options, you then use the same technique from the beginning to discover the other digits. In other words: B*E + d = 10c + E (where c is the carry from earlier, and d is the carry from CDE*A).
Definitely took me a while, and had to retrace my steps a few times, but probably no longer than if I would have had to do all those tail divisions without a calculator :)
1:30 If we just could do that multiplication chart in under 5 seconds just like in your video...
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it...
My take, after identifying the possible pairs, note that the 10's digit of A*A+carry from A*D=E. So the tens digit from A*A is either E, or E-1. This eliminates (6,2) and (9,5) because A is too big. It eliminates (3,5) and (6,8) because A is too small.
After that, your division method is faster than the modular arithmetic method I used to derive each digit. (7*5=35, so (A*D) mod 5=2 with A-7, means D-6 and similarly down the chain.
Great way to solve the problem. I just used a process of elimination. I then figured out that E=5, from there I figured out that A was either 7 or 9. First I tried to solve the rest of the letters using A=7. This process is definitely not the fastest way to do it, but I got there. Great problem!
What was your rationale to establish that E=5, not A=6?
So after some thought, I guessed that E is 5, because then A can be any odd number. But then you can work backwards from the left, where A squared is roughly 55 and from there it's straightforward. Nice little problem!
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it.....
Given that the product is in the form EEEEEE, it equals E * 111111, which then can be factored, from then on, it's just finding E and A, such that E*A ends with E.
That's pretty easy, E=5, and A is some odd number. Now it's just testing 4 cases for A (1, 3, 7, 9) and the answer is found (because of the first condition that the answer is 111111*E)
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
To start with, there are only two patterns for X * Y = (10 * Z) + X (where X-Z are unique one digit numbers). Either X = 5 and Y is odd. Or X is even and Y = 6. We can also exclude the cases where they match, but that's not necessarily relevant.
So we immediately know E = 5 xor A = 6.
We can now determine the other digit, because A^2 + N = (E * 10) + E and N must be a single digit number (the carryover from B * A).
For A = 6, E = 4 and N = 8. For E = 5, A = 7 and N = 6.
But if A = 6 and N = 8, it falls apart, because B * 6 < 60 so N can never = 8.
So we know E = 5 and A = 7. From there we can just work it out.
E * A = 35, so a 3 carries and D * A must end in a 2. That is 6 * 7 so D = 6
D * A = 42, so a 4 carries and C * A must end in a 1. That is 3 * 7 so C = 3
C * A = 21, so a 2 carries and B * A must end in a 3. That is 9 * 7 so B = 9
B * A = 63. That gives us the carrying 6 (previous identified as N). So it matches up. Giving us ABCDE = 79365.
(EDIT: Okay, the division thing is much quicker and somewhat more elegant, even if it is trial and error)
Makes look so easy
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
Method of finding a solution is fantastic !!!
we can solve by a system of equations
I got it in about 15 minutes, and I was so proud! It showed up in my Google Pocket as an article from Popular Mechanics, and the description made it seem impossible. (They used the word "maniacal.") I started asking what times E could end in E, and I got the most solutions with E = 5, far and away (all odd numbers). Then I asked what could A be if A had to be an odd number (A x 5 can't end in 0), A^2 + carryover had to give a two digit number beginning with E, and A had to give a carryover number for each digit. E=5 seemed an intuitive place to start, and that left only 7 & 9 as contenders for A. The rest of the digits fell into place by solving for the carryover number needed for the column to end in a 5.
I thought I was so clever until I saw this simple division method. The article also said teachers couldn't solve it, but this video says they could, with one of them taking as long as 20 minutes. So I'm no brainiac, but I guess I could substitute for an elementary school math teacher if they were hard up.
Okay, who else here initially thought ABCDE meant A x B x C x D x E ? 😂
😂
Don't need to brute force through that many options - either A=6 (and E is even) or E=5 (and A is odd). Then look at the LAST multiplication: A times A should give you a number with the tens digit being E, possibly after some carry, so A times A should give a tens digit of either E or (E-1).
For A=6 we now get A*A=36 so E=3 or E=4. E must be even in that case, so E=4.
For E=5 we only get A=7 because that's the only number whose square has a tens digits of 4 or 5.
Now only left to check the pairs (6,4) and (7,5), which is a lot less brute force, especially if you're doing it without a calculator.
But you can actually strike out the (6,4) option continuing the previous logic with one extra step: we looked at the last multiplication and acknowledged the possibility of a carry from the previous multiplication, but that carry cannot be any number - it must be lower than A. For the (6,4) pair we get an equation of 6*6 + carry = 44, which we can solve for a carry of 8 which is too big.
Me: Solved this within 2 minutes. Duh that was easy.
Presh: This problem is for 9 year olds.
Me: Gulp! I am 16.
Bruh
So guess correct, but I think intuitively I just knew that A x E = yE meant that E had to be 5. Because I also wanted to include carrying, so I would need a number that would tip into the next value place easily.
Then I had A x A = yx < EE, and then just step it backwards, If A = 7, what does B need to be to make 7x7 + y = 55.
So an intuitive guess got me there pretty quick.
I didn't get this table 1:45
How you find that
9*9 matrix of digits 1 to 9 and the units place result for each combination.
Generalized Approach for receiving all solutions:
A ≠ 1 as the output is a 6-digit number.
Consider the multiple of E(unity) and A to achieve result of E; to be in the form A.E = 10a + E for positive integer a from which a = E(A-1)/10; which confirms E = 2 or E = 5; and Case I: for E = 2 (A-1) to be a multiple of 5; hence A - 1 = 5m for positive m integer from which A = 5m + 1; from which for m = 0, 1; A = 1 and 6; as A ≠ 1, for E = 2; A = 6.
Now consider the multiple of leftmost A with A to achieve result of EE (leftmost 2 digits of result) we need to write the multiple as 10E + E = A2 + k for any k positive integer. Substituting A = 6 we get 11E = 36 + k; as you are aware 1 ≤ k ≤ 9 and hence, 37 ≤ 11E ≤ 45; this will imply E = 4 which is contradicting with E = 2; now we conclude E ≠ 2, 4 and A ≠ 1, 6.
Now Case 2: For E = 5; a = E(A-1)/10 and A - 1 = 2n for positive n integer; which confirms A = 2n + 1 and for all possible positive n values; A = 3, 5, 7, 9 substituting in 10E + E = A2 + k we get 55 = A2 + k; as you are aware 1 ≤ k ≤ 9 and hence, 1 + A2 ≤ 55 ≤ 9 + A2; right end inequality confirms 55 ≤ 9 + A2 and A2 ≥ 46 and A ≥ 7; similarly considering left end inequality 1 + A2 ≤ 55 and hence, A ≤ 7; now to satisfy both conditions A = 7 and E = 5. This will give result of ABCDE = 79365; for cross checking 79365 x 7 = 555555.
You just gave me
The answer and Im still confused 😂
Dude it’s not that hard
Just takes time
Got the right solution, but my method focused more on the A × A at the beginning, though the E × A did play a role. My method:
A five digit number became a six digit number. Therefore, the EE in the ten-thousands digit is the solution of of A × A + x, which I'll represent with variable S, with mind paid that a possible solution involves x = 0. Since no variable is greater than 9, the largest possible value for A × A is 81, which also proves the largest possible value for x is 8. Knowing the number can be represented as EE means S is a multiple of 11. At this point came a little brute force. I took a random A value, squared it, and set the range of possible values for S by setting x to both 0 and 8, then looked for a multiple of 11 within this range. If I found a multiple, I would note the values for E and x. The summary was:
A = 1, S = [1, 9]: No multiple, No Good
A = 2, S = [4, 12]; E = 1, x = 7
A = 3, S = [9, 17]; E = 1, x = 2
A = 4, S = [16, 24]; E = 2, x = 6
A = 5, S = [25, 33]; E = 3, x = 8
A = 6, S = [36, 44]; E = 4, x = 8
A = 7, S = [49, 57]; E = 5, x = 6
A = 8, S = [64, 72]; E = 6, x = 2
A = 9, S = [81, 89]; E = 8, x = 7
With this list, dividing eight times is an option, but I chose to reduce the number of possibilities first by testing the E × A at the beginning. With mind that the E in the ones digit only represents the product's baccarat value and there is no means to ascertain its true value yet, I'll stylize that number as € and recognize that € = E. Now back to brute forcing using the values derived above.
If A = 2 and E = 1, then € = 2; No Good
If A = 3 and E = 1, then € = 3; No Good
If A = 4 and E = 2, then € = 8; No Good
If A = 5 and E = 3, then € = 5; No Good
If A = 6 and E = 4, then € = 4; Possible
If A = 7 and E = 5, then € = 5; Possible
If A = 8 and E = 6, then € = 8; No Good
If A = 9 and E = 8, then € = 2; No Good
The list is now down to two possibilities. Dividing from here is more viable, but I noticed something else before I could start.
Recall that if A = 6, then x = 8. In order to generate that x value, B × 6 + y > 79 while noting the range for y was identical to the range for x. With largest values of B = 9 and y = 8, the largest x value possible is 6. Therefore, A = 6 is also not possible.
Thus, A = 7 and E = 5. Then I went division to get B = 9, C = 3, and D = 6.
Took me less than 2 mins to solve this..
Oh wow...a lifetime award for you, perhaps? Or a chair in the Royal Society for Maths? CHootia
@@pseudorealityisreal its ok bro if you couldn't solve thjs... And you dont need to use slangs. I just commented what i did,
@@paramgoswami7224 As expected, an assumption I couldn't solve it. Do you have any other assumptions?
@@pseudorealityisreal if you did solve it congrats, but why does it matter what i comment?.
If you didnt like the comment, simply ignore it, and move on
@@paramgoswami7224 Actually, you are right. I should have just moved on rather than commenting. I was very irritated with similar comments posted on this channel's videos, and your comment was the one where I chose to vent my annoyance. Anyway, sorry for the previous verbal abuse, I agree it was totally uncalled for.
You COULD solve this problem using sophisticated mathematics...
Or you COULD just guess and check.
This probably has a maximum of 100 possible combinations of the product and "A". A can be one of 10 digits, and E can be one of 10 digits.
Obviously you can break that down because we know A isn't 1 or 0 (based on the EEEEEE product). So that's 80 possible combinations maximum.
We can then figure out that since E x A must end in E, that the only combinations for A and E are 6,2 , 6,4, 3,5 , 7,5 , 9,5 and 6,8. We literally have 6 possibilities to check now:
222222/6, 444444/6, 555555/3, 555555/7, 555555/9, and 888888/6. The only possibility that gives a result with all 6 unique numbers is 555555/7 = 79365.
And look! 7 is the A at the beginning too!
A=7
B=9
C=3
D=6
E=5
Now to watch the video...
E
Math : polynomial
Factor higher degree polynomial
ua-cam.com/video/vU7-06A3KXM/v-deo.html
One time see it....
After recieving 6 possible pairs it is possible to eliminate 4 of them:
A cannot be less then E, lets take (A, E) as (6, 8). 6*6 (when multipling A by A) = 36 (plus smth from multiplication from previous digit). No way you can get 8 by this as firts digit of 6th digit final result. So (6, 8) and (3, 5) are out.
Also 6 by 6 result 36 and smth more, so E cannot be 2, E will be at least 3, so (6, 2) also out.
9 * 9 will be 81 plus smth, so you cannot get 5. This leave us with only 2 variants to check.
how is a not less than e?
A=0, E=A, :.B, C, D{IR}
From what was given 7 and 5 seemed like strong candidates for A and E, especially since 7 squared is close to 50 (to begin the product).
You can avoid all but two long divisions by eliminating a few options based on A x A + digit = E E.
You quickly eliminate A = 1-4, which gives 1, 4, 9, 16 + a digit, so an E value of (0 or) 1 or maybe 2 none of which work.
A = 5 gives 25+digit, and neither 2 nor 3 works for E.
A = 6 gives 36+..., so 2 is too low and 8 is too high (no long division necessary), 4 is the only number to check!
A = 7 gives 49+digit, E must be 5 and is correct
A = 8 has no digits for E
A = 9 gives 81+... while E must be 5 so is too low (again no long division necessary).
Alternate way to reach a solution: We take the basics of the problem, namely that an unknown five-digit number multiplied by the first digt of that number becomes a six-digit number where all the digits are the same. This already puts a nice limit on the possibilities. How can we limit this even further? Simply, by taking the lowest possible value of the first number and the highest possible value, multiplying both by the second number, then checking if there is any number in that range that has all digits be the same. If not, then we can discard that option entirely without needing to do anything else.
The first digit obviously cannot be zero, for reasons that really don't deserve explanation.
If the first digit was a 1, then multiplying it by 1 wouldn't actually do anything.
If the first digit is a 2, then the unknown first number could be anywhere from 20000 to 29999 -- or rather 30000 for the sake of easy calculations. Also, we're going to ignore that all the digits must be different for a moment, and solely focus on the range of these numbers. 20000x2 is 40000, while 30000x2 = 60000. Still a five-digit number, so it can't be the right one.
Next up, 3. Here, the unknown first number ranges from 30000 to 40000. 30000x3=90000, while 40000x3=120000. This one is a possible candidate: 111111 lies within this range. Dividing this by 3 actually does get us a five-digit number, namely 37037. The issue, of course, is that this breaks several rules of the problem: E is both 1 and 7 at the same time, and AD and BE both have the same value, namely 3 and 7 respectively. Either way, this can't be right either.
4 then? This gets us a range from 160000 to 200000. No all-same six-digit number, so no dice.
5 is up next. This time it's 250000 to 300000, so still no.
6? 36000 to 420000, same issue.
7? This gets us 70000x7=49000 to 80000x7=560000. Another candidate, 555555 lies within this range, if just barely. That's convenient though, because it means that the number we want to check is very close to the upper bound. To be specific, it's 4445 less than 560000, or 635 less than 80000 after dividing both sides by 7. 80000-635=79365, which fulfills all of the conditions of the problem. And there you have it!
I solved it by noting that A*A + some carry from penultimate multiplication (A*B) must give EE. A= 1, 2 are eliminated because we won't get a 6 digit number, for A = 3, E has to be 1 which isn't possible. A= 4, 5, 6 are eliminated because no carry over added to them squared will give a repeating 2 digit number...for example, for A= 4, the possibilities are only 16, 17, 18 and 19. But with 7, the square is 49 and a carry over of 6 can take us to the 55. Now I simply divided 555555 by 7 to check if it is divisible and all digits are different. We have a match here...
We can continue with 8 and 9. For 8 the closest repeating digit number is 66 and for 9 it is 88.. we 666666 isn't divisible by 8, 888888 not by 9. So we have our unique solution.
Here's how I did it. In the original equation A x A plus whatever number was carried over from A x B equals a two digit number whose digits are the same. This immediately eliminates A equals one and A equals two. So, A=3 (and we carried a 2) if E=1; A = 4 (and we carried a 6) if E=2; A=5 (and we carried an 8) if E=3; A=6 (and we carried an 8) if E=4; A=7 and (we carried a 6) if E=5; A =8 and (we carried a 2) if E=6; and finally A=9 (and we carried a 7) if E equals 8. From there, it's just a matter of seeing which of those sequences will work by substituting in numbers--and only the correct solution does. Probably not the most "mathematical" way to solve it, but it works. ;)
The alignment of the bottom E's de-emphasized the sixth digit. Now that I've spent so much time trying to solve for five E's, I can't be bothered for six.
A different way to solve this:
We need A times A + (a carry over digit) such that it's a multiple of 11 (11,22,33,...) This will be true because the carry overs can never exceed the digit 8. Now, We also need A times E to be such that the units digit is E. So, for example if you choose A to be 5, then A times A + 8 (In this case) is 33. or E is 3. but 5 times 3 is 15 so this does not work. only A,E = 7,5 and A,E=6,4 work. but In case of A=6 You will also need the carry over from B times A to be 8, which is impossible as it cannot exceed 5 (even 10 times 6 is 60). So now, the only option left is A,E = 7,5, which works out to the given solution of ABCDE = 79365.
Figuring that E = 5 was pretty easy. The rest I did like this: A must be odd to multiply with 5 and give the last digit 5. you can eliminate 1 and 3 quickly and are left with 7 or 9. So 7x9 = 35 so your first carry over is 3. B x 7 must end in 2 (2+3 = E) , so B is 6 and so on.
Here is the way I used :
I first checked how we could get AxA=EE, keeping in mind that the multiplication AxB coming next could add at most n=9 (if not only 8) to the result of AxA.
I then checked if it could match with AxE=?E
Then I ended up with this :
(A=1 is impossible for the reason explained in the video)
A=2 -> A²=4 so the result A²+n could only lead to EE=11 (with n=7) so E=1. Then 2x1=2 -> Impossible
A=3 -> A²=9 -> EE=11 (n=2) -> E=1 -> 3x1=3 -> Impossible
A=4 -> A²=16 -> EE=22 (n=6) -> E=2 -> 4x2=8 -> Impossible
A=5 -> A²=25 -> EE=33 (n=8) -> E=3 -> 5x3=15 -> Impossible
A=6 -> A²=36 -> EE=44 (n=8) -> E=4 -> 6x4=24 -> OK
A=7 -> A²=49 -> EE=55 (n=6) -> E=5 -> 7x5=35 -> OK
A=8 -> A²=64 -> EE=66 (n=2) -> E=6 -> 8x6=48 -> Impossible
A=9 -> A²=81 -> EE=88 (n=7) -> E=8 -> 9x8=72 -> Impossible
The only solutions were (A;E)=(6;4) with n=8 or (7;5) with n=6.
But if A=6, AxB could then at most equal 6x9=54, so n could never reach 8 as required.
So the only solution was A=7 and E=5.
Then it became much easier to find the other digits :
AxE=35 so you will have to add 3 to the unit of AxD to reach E=5. So AxD should have a unit of 2 -> As shown on the multiplication table : D=6
AxD=42 -> AxC should have a unit of 1 -> C=3
AxC=21 -> AxB should have a unit of 3 -> B=9
So the result is ABCDE=79365.
Multiplication table of 7 :
7x0=0
7x1=7
7x2=14
7x3=21
7x4=28
7x5=35
7x6=42
7x7=49
7x8=56
7x9=63
Once you reach the “A = 6 or E = 5” step, you can prove that A cannot equal 6 because 6BCDE x 6 must be between 360000 (60000 x 6) and 420000 (70000 x 6), and there is no number with six of the same digit in that range.
Likewise, with E = 5 established, you can prove that A cannot equal 3 because 3BCD5 x 3 must be lower than 120000 (40000 x 3), and A cannot equal 9 because 9BCD5 x 9 must be at least 810000 (90000 x 9), so 555555 cannot be the resulting product in either case.
From the possible solutions A=6 and E=2, 4 or 8. Based on multiplication of the most significant digit A x A:
6 x 6 + [0-8] = EE
So 36 + [0-8] can only be 44, but then you would have to come with an 84 from the 6 x B + [0-8] which is impossible.
Similarly from solutions A = 3, 7 or 9 and E = 5 you would have:
A x A + [0-8] = 55
The only possibility is 7 x 7 = 49 to reach the 55, so then A = 7 and E = 5 is the only option on the table.
A bit faster way.
Start by the last multiplied digits AxA on lefts side so we have
AxA +n = EE where n is an integer from 0 to 9 from the preceding multiplication BxA,
The only integers possible
A=6 -> 6x6+8=44, and
A=7--> 7x7+6=55
Thus (A, E)= (6,4) or (7,5)
Now we callculate EEEEEE/A = ABCDE
444444/6= 74074 does not start with A=6, so excluded
555555/7= 79365 which starts by 7(=A) and ends with 5(=E)
Thus , (A,B)= (7,5) and
ABCDE=79365
after watching your video about the problem "send more money" I wrote a program to find the solution of "word1 + word2 = word3" where every letter of the input words is a different digit, all I had to do is change the + into a * and i found the solution of this problem :)
I actually have another way, see if this is better
Since 1 is impossible, we will rule it out
2x2=4 +7=11 --> not divisible by 2
3x3=9 +2=11 --> 37037 (x--repeated no.)
4x4=16 +6=22 --> not divisible by 4
5x5=25 +8=33 --> not divisible by 5
6x6=36 +9=44 --> 74074 (x--repeated no.)
7x7=49 +6=55 --> 79365 (correct answer)
8x8=64 +2=66 --> not divisible by 8
9x9=81 +7=88 --> not divisible by 9
Before seeing the video I tried to solve and got 79365.
I just thought that no matter the rest, the A had to be a number that squared, could turn into a number with two equal digits when added with a single digit number, so I tried 7 bc 7x7=49, and 49+6=55. So that meant that if A was 7, B had to be 9 so that when 9x7=63, the six would add to the 49 to get 55. Then I just put the numbers I had given into the calculator, 79005x7=553035, then I just thought, 6x7=42 so D is 6 to get to 553455, and to complete, 3x7=21.
Seemed obvious that E was going to be 5, that's pretty much how you end up with the same digit in each place for the product. Looking at the end, I was going to need to get AxA to be close to 55, so that lead to A being 7. From there, work the problem figuring out what you carry and what you needed the next letter to be. I.E. AxE using my values was going to be 35, so we need something that multiplied by 7 that ended in 2 making D = 6. 7x6=42 add the carried 3 and you have another 5, and you carry a 4 this time.
Lather rinse and repeat and you get the B and C values of 9 and 3 respectively.
Took about a minute.
@MindYourDecisions I solved it using almost the same method as you did, except that I took a shortcut and only had to calculate the **first iteration** of most of the long division problems. We know the first digit of the final product is A so we can eliminate any division-by-A quotient that does **start** with A. Thus 22/6=4, 44/6=7, 8/6=1, 5/3=1, 55/9=6 all fail leaving only 55/7=7 so I only had to complete the entire long division problem for one candidate.
I HAVE A EVEN EASIER WAY
First it's a 6 digit no. And all the digits are same in the end A * A should be eaqual to EE+ remainder from the last one and incase the A is 2 then max remainder will be 1 and if it's 3 max reminder will be 2 and so on so I worked them out and found out 2 cases
Case1 - E =6,A=7,B=9
Case 2 - E= 6,A=8 , B= 2
Try both the first one satisfys the question and the values of rest comes
(I am a little bad at explaining but that's the main thing)
As someone who, as a 9-year-old, counted exponentiation of base five for fun at the back of my maths notebook by manually typing "5 * 5 * 5 * 5 * ..." for each subsequent product since I obviously wasn't taught the notation for that at the time until my teacher noticed my notes, I was immediately drawn to a "E is most likely 5 and A 7 so that it A*A can be filled to 5" solution. I am quite confident that I would've solved this back in the day.
At the time, I chose the base of 5 specifically because of the pattern it seemed to create. Made it much easier to count the next product in my head when the last few digits were always 125 or 625 one after another.