I just found your channel and wanted to let you know that I absolutely love it. It has been over 30 years since I studied electrical engineering, so I am a math geek. And while I use math regularly, I don't deal with anything particularly complex. Your channel has been a wonderful reminder of how fun math can be. Thank you so much for sharing.
It's a neat trick that the result of any 3-digit number being multiplied, successively, by 13, 11 and 7 is a 6-digit number presenting the original 3 digits repeated. The reason why this happens is fairly obvious, given this video.
That is exactly what I noticed. Then knowing from experience that 1001 = 7 × 11 × 13, the prime factorization comes out as 7⁷ × 11⁷ × 13⁷. The whole process took less than a minute.
@kinshuksinghania4289 Very nice. It's also helpful to note that the number is just a bit over 10^21 or 1000^7. Lastly, with your way of using Pascal's Triangle the origin of the of three sets of double zeros is obvious.
01:15 "If you like it, like it at the end." Many others *_start_* their videos by asking for a 'like,' which is backwards if I have never seen their channel before. Your videos never disappoint, so a 'like' at the start would be confirmed at the end anyway. But your approach was respectful of the viewers in a way that I have not seen before -- in the watching of many videos.
This is smart. All along, I found that magnitude checks can help things make sense. For example, 101³≈100³=(10²)³=10⁶, so it should have 6 numbers tailing the 1. 7×11×13≈7×12×12=7×144≈7×140=7²×2×10 =49×2×10=980≈1001, so even if we're just guessing prime factorisations, it's a reasonable start.
So cool. The relationship between powers of 11, 101, 1001, etc., and Pascal's Triangle, was completely new for me! (And I've got a degree in math; lol.) Thank you so much for this awesome trick! :)
On first glance, it has 1, 7, 21, 35, 35, 21, 7, 1, so it obviously is of the form 1(x 0s)1^7. Gap of two zeros between the 1 and the 7, so it's 1001^7. 1001 is equal to 7*11*13 Ans: 7^7 * 11^7 * 13*7
Dropping a comment here for another video I saw of yours so you see it, you really explained the video of finding inverse of a matrix really well, it was a topic I found really hard and could not understand much. Thank you so much for your efforts and will surely recommend your channel to others. Much love from India.❤❤
Very nice problem find! Took me a minute or two to recognize the number wasn't palindromic - but then the binomial expansion only took a few seconds to identify and verify. Spotting 11 as a divisor of 1001 was trivial as 11 divides (10 + 01 = 11); and the rest was easy. Now to check out the video, and how you solved it. P.S. Welcome back.
I'm going to disagree about memorizing (with intent) random facts such as 1001 = 7*11*13. One inevitably collects a few of these over time spent solving puzzles, but the deliberate memorizing not so very useful. More valuable, IMHO, is to learn a wide variety of divisibility tests - or, even better, how to construct the divisibility test for any (smallish) prime in just seconds.
Been binging your videos since YT decided to recommend them to me. You're an awesome teacher dude, and the videos are super well structured. Keep up the good work! PD: Would love to see more Number Theory / prime numbers content 😃
The zeroes between the ones seem to be padding every number. If you take 101^5=10510100501, add a single zero before it and take every pair from it, you get 01, 05, 10, 10, 05, 01 which is exactly the fifth layer of the Pascal triangle without the carry from adding messing up the result. 11^5=161051
How do you add links to UA-cam comments? In your comment the term "Pascal triangle" is highlighted as a link. Edit: Nevermind, must be some automatic highlighting by the YT app.
If I’m remembering correctly the guts round has 3 problems at a time, and you have to submit your 3 answers to get the next 3 problems. So it helps to go fast/use tricks so you can finish and get to more problems, but at the same time you can’t fix/correct any previous problems
It is easy to test for factor of 11 by alternately subtracting & adding digits. You will get number modulo 11. You can do the same for 7 by using the following "weights" in the calculation: (units) 1, (tens) 3, (hundreds) 2. The weights go 1, 3, 2, -1, -3, -2, 1,... Test for 3 is simply sum the digits. Test for 13 uses weights 1, -3, -4, -1, 3, 4, 1,.... These weights can be calculated using modulo math, starting with units weight = 1 & multiplying by 10 & applying modulo, repeating for each digit until looping occurs. But your number is too easy as all the prime factors are "small". In reality you need to test for all primes up to the square root of the number. & if none found, declare it a prime.
Beautiful example of experience-based intuitive mathematics. How come there are people that do not see the esthetics of maths? I think that even without particulat maths skills and relieved from the intimidating myths of how terribly difficult maths is, one should be able to meditate to e.g. this example just to cherish the pure beauty of numbers. Creds again to you!
nice i also recognizing the mirrored number thing and thought it might be divisible by 11 and confirmed using the divide by 11 trick. wasnt sure what to do after. never knew about the 11^x, 101^x, 1001^x,... thing representing pascal's triangle before. neat
Yeah, I think I saw another video on a divisibility test for 1001. It's similar to the divisibility test for 11 since 1000 is equivalent to -1 (mod 1001). The repetition is what would quickly show this is equivalent to 0. And the divisibility test might even make sense after testing 3 since, while 11 is easy to test for, 7 and 13 are not.
@PrimeNewtons I saw a video on checking for divisibility by 7. I think it was Matt Parker on Standup Maths (but it could have been Numberphile). It was not straightforward, and I can't remember how it worked. I really enjoyed your video, and I learnt something that I think I will remember.
@@PrimeNewtons not that quick😀. Imagine the number is written with commas 1,007,021,035,035,021,007,001 Now sum all the odd trios 001+021+035+007=64 Sum all the even trios 007+035+021+1=64 Take the difference of these results 64-64=0 If this number is divisible by 7 then the original number is divisible by 7 If this number is divisible by 11 then the original number is divisible by 11 If this number is divisible by 13 then the original number is divisible by 13
Very nice explanation. One thing you did not mention early on was the observation of the pattern of the numbers in the middle - which was also easily seen. You held it under your belt until the key had already been found.
You can also solve it with the palindromic method. You would have to divide out 10^10.5. Adding everything up you would end up with 10^10.5 * (10^1.5 + 10^-1.5)^7 = 1000 + 1. This is easier to see with the substitution of x=10^1.5.
I haven't watched the full video yet. First thing I would try is, find the digital root of this number. If it's divisible by 3 or 9, or some power of 3, then the number is divisible by that power of 3. Ok, that didn't work. The next thing I would try is, the alternating digital root (taking each digit with an alternating +/- sign). That will tell me if the number is a multiple of 11. And, lo and behold, by repeating this process 7 times, I found that this number has a factor of 11^7. This is a good start to beeak it down. Using modular arithmetic, there are other easy ways (similar to digital roots) to check for divisors of other small primes, Since this is like an Olympiad problem, it is probably going to be divisible by some small primes. Not 2 or 3 or 5, but maybe 7, 13 or something. I did figure it out, but I won't spoil it for you.
PS: I did not notice that this number is a palindrome in base 10, until I actually watched the video. Not sure how I missed that, but, your solution is very clever.
My factorization program written in C# factored this number quite fast Factorization has seven 7 , seven 11 , seven 13 We could observe that this number equals 1001^7 (Binomial expansion and Pascal's triangle for coefficients of this expansion can be useful)
Could probably be more rigourous when ignoring zeroes... Otherwise could make a mistake with 10070021003500350021007001 or 100702103503502107001. Probably best to just expand out (1+x)^7, and then match to determine x = 1000.
I wonder if grouping factors would help 1*(10^0+10^21)+7*(10^3+10^18)+21*(10^6+10^15)+35*(10^9+10^12) 1(1+10^21)+7*10^3*(1+10^15)+ 21*10^6*(1+10^9)+35*10^9*(1+10^3) from this I observe 10^3+1 is a common factor because if we note 1000=A then the rest of the expressions are: 1+A^3=(1+A)(1-A+A^2) 1+A^5=(1+A)(1-A+A^2-A^3+A^4) 1+A^7=(1+A)(1-A+A^2-A^3+A^4...) Dividing by 1001=7*11*13 we get quotient 1006015020015006001 Dividing by 1001 again quotient 1005010010005001 Dividing by 1001 quotient 1004006004001 Dividing by 1001 quotient 1003003001 ........................................... 1002001 repeat 1001 so the original number is 1001^7=(7*11*13)^7
Very interesting, thanky you! The question would be, if the mentioned long number wasn´t a "Pascal´s triangle number", what approach would you take to make a quick guess, on how big the lowest prime factorial must be and which of the frequently memorized prime numbers (1-100) would be most likely to start with?
Before: I quickly divided in my head up to 7 and got to 100000000000....1 through my division process. No way is that length of division going to occur mentally. I guessed this number must be divisible by 101, or 1001, or something similair like 10001, etc, just from looking. Since i also suspect it is divisible by 7 (and 11), 101 is not divisible by 7, so i was pretty sure it is some multiple of 1001, which is divisible by 7 (and 11) So I did 1001^2,3,4,5... (with a calculator, ie: cheating) and found the number in question. I guess I could have written out 1001^7 with a piece of paper and then its "fair"? Idk, but surely calculators wouldn't be allowed for this type of problem. I am so exicited to hear your approach!!! Edit: I am grinning as I hear your explanation. Wonderful!!!
This is a 22 digit number - large enough to scare people from trying to get started. Last digit indicates that it is not a multiple of 2 or 5. Sum of the digits indicates that it is not a multiple of 3. Alternate digits add to 19, so 11 has to be a factor of this number. On dividing by 11, the same rule works on the quotient. So, the number (or at least some factor) is some power of 11. This worked 7 times, so 11^7 is a factor but 11^8 is not. 7th power indicates, that lets try prime roots at least for single digit primes. Square root, cube root and 5th root are not integers but 7th root is 1001 which is multiple of 11 (same alternate digit sum equal rule). 1001 / 11 = 91 = 7 x 13. So, the 22 digit number = (7 x 11 x 13)^7. This number represents a very bad RSA public key modulus that banks should never use for encryption of high value transactions 😁
Any good prime factorization problem MUST avoid multiples of small primes like 2, 3, 5 and 11 due to easy divisibility rules that expose the solution a bit too easily, as in this example.
Dear sir I have a very interesting puzzle for you I am Surya from India here is the puzzle The problem involves finding an even number:- Joh has an even number and asks his friend to guess the number using the following (clues 1) the sum of the reciprocal of the number of possibilities of writing the number as the sum of two different integers and the actual number is 13 /2 (clue 2) the repetition of the number is not allowed and the order of the pair doesn't matter. What is Joh's number?? Si r I hope you see my message Thank you 😊
I'm a retired prof and have a stupid question? What are you using to get the blackboard so clean? 🙂 I could never get one that clean. Of course, later we went to the white board with erasable markers and then to video projectors. Nice to see someone use chalk. I've just started (I'm always late to the party) working through all The Euler Project problems to prevent Alzheimers or at least put it off as long as I can. Spent all those hours learning applied mathematics, might as well use it. I also love to feed such problems to Wolfram Alpha to see if it will break.
Here I began with exploring the magnitude of the given number by splitting by thousands: 1|007|021|035|035|021|007|001, which can be written as 1.10^21 + 7.10^18 + 21.10^15 + 35.10^12 + 35.10^9 + 21.10^6 + 7.10^3 + 1, which corresponds with the seventh row of the Pascal triangle (for 10^3 and 1), therefore the whole number is (10^3 + 1)^7 = 1001^7. Then the easy factoring of 1001 = 7 . 11 . 13. Lucky guess at the beginning, but it paid off :-)
The number of zeroes correct? I think there's an extra 0 between 21 and 7 at the end, otherwise the rule of the number of zeroes doesn't apply regularly(?)
صباح الخير أستاذ الاحظ أن العدد ليس مماثلا من اليمين واليسار فكيف يمكن أن نطبق الفرضية التي استعملت 10072 0 21035 0 35021 00 21001 هنا صفر واحد عل الشمال وصفرين على اليمين ، أتمنى أن ترد علي وشكرا
If your calculator cannot handle such a long number, try to split it into a higher and a lower part at a certain digit, e.g. 1'007'021'035'035'021'007'001 = 1'007'021'015'000'000'000'000 + 20'035'021'007'001, and test if both complements are divisible by the same. The above, for instance, are divisible by 7×11×13=1001. Anyway, apply a test or just try out. It will work or not. 1 007 021 015 000 000 000 000 ÷ (7×11×13) = 1 006 015 000 000 000 20 035 021 007 001 ÷ (7×11×13) = 20 015 006 001 1 006 015 000 000 000 + 20 015 006 001 = 1 006 035 015 006 001 (Hope I didnt present any typing errors, I had some of them) You better split it closer to the middle. My calculator, for example, can handle 15 digits. Just saying. Didn't watch the vid, will do it later.
We get very very similar results even when not working in decimal. In every base b above 2, 11² = 121 (1b+1)² = 1b² + 2b + 1 (this fails in binary because 2 is not a digit and we have to carry over And this can be extended to bigger powers, and with putting zeros in between. 11⁴ translates to (b+1)⁴ which is b⁴ + 4b³ + 6b² + 4b + 1, and if your base b is greater than 6, this can be written as the digits 14641. Now, the fifth row of Pascal's triangle is 1,5,10,10,5,1, so in decimal this will carry over and ruin the forms. To preserve the pattern we could look at higher bases, eg. in hexadecimal, 11⁵ = 15AA51, where A is the hex 'digit' for the number we would call ten (it's a good idea to confirm 17⁵ for yourselves) The other thing we can do is pad with zeros, kinda like what we had in the problem in the video. Then the carry-over just hides zeros, which our eyes are often happy to ignore, so back to decimal, 101⁵ = 1|05|10|10|05|01, where I added the vertical lines to help visualize. ofc we can do both. The 6th row is 1, 6, 15, 20, 15, 6, 1 In hex, this is 1, 6, F, 14, F, 6, 1 So in hex, we can find 1001⁶ = 100600F01400F006001 The thing I really like with swapping bases is what happens to fractions. In decimal we know that 1/9 = 0.111..., 1/99 = 0.010101..., 1/999 = 0.001001... Turns out this generalizes too! I'll write q as the last digit in base b, representing (b-1). Then the number qq is just b²-1. Using infinite geometric series, we can show that (note that b is an integer starting from 2 and up) 1/(b²-1) = b⁻² + b⁻⁴ + b⁻⁶ + ... So in base b, this will be written as 1/qq = 0.010101... 1/qqq = 1/(b³-1) = 0.001001... etc. Every repeating decimal (but not only in decimal lol) is just one of these 1/qqqq's, times the integer that appears in the repeating sequence, shifted with some power of b⁻¹, with additional first non repeating digits. Also also, you know how to check if something is divisible by 9 you just add the digits? This is a result of 10ⁿ modulo 9 always being 1, and can be applied to any basis. Is a number divisible by q?/what is remainder mod q? Sum the digits in base b. Whatever you get, that is, or has the same remainder. Did you get zero? Great, it's divisible by q
You really show beautiful questions and beautiful and clear solutions. It's just a shame that the screen is too dark with a somewhat claustrophobic atmosphere. My advice, and I hope you'll see this as constructive criticism, is to replace the board with a clean white board. Write on the board with a black marker, and you can also use other colored markers to highlight important things. In red, green, and blue... Thanks for interesting videos.
These mathematics tricks do not contribute much to the science of math nor are they used in the engineering field. Such questions often appear in math competitions but I think they get way to much attention, also on UA-cam, as there are few real life applications for them.
It shows a broad understanding of mathematics, on the part of those who can solve them. Personally, I had little exposure to Pascal's triangle or number theory along the way.
There isn't much practical use for regular exercise except to keep your body from deteriorating into formless mush. Think of these math problems as exercise for your brain.
Could you please provide a reference to this information? We know that the so-called Pythagoras theorem was actually known in Ancient Egypt and India long before Pythagoras was born.
@@slavinojunepri7648 Chu-Shi-Kie was a Chinese mathematician, and "A.D. 1303" is the date on the manuscript illustration of the triangle. I found it on p. 720 in my textbook "Precalculus" by David Cohen, 4th Ed. Googling Chu's full name gives various spellings.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Before watching the video: «That's impossible»
After the video: «That was surprisingly easy!»
I just found your channel and wanted to let you know that I absolutely love it. It has been over 30 years since I studied electrical engineering, so I am a math geek. And while I use math regularly, I don't deal with anything particularly complex. Your channel has been a wonderful reminder of how fun math can be.
Thank you so much for sharing.
how do you find such problems
They can be found under a rock on a rainy day
He said he found it in a competition
Some students or subscribers also e-mail him questions
There are we bsites that keep track of these math competitions and provide links. It's a hobby to keep up with them and find interesting problems
It's a neat trick that the result of any 3-digit number being multiplied, successively, by 13, 11 and 7 is a 6-digit number presenting the original 3 digits repeated. The reason why this happens is fairly obvious, given this video.
Really good to see you back with these great videos again, sir..God bless.
Welcome back, I am a 76 year old physicist/engineer and have never seen one of your videos that was a waste of time. Gold in every vein!
1, 7, 21, 35, 35, 21, 7, 1 → binomial coefficients of (a+b)⁷
So if we expanded (1+1000)⁷ we would get the exact same number
That is exactly what I noticed. Then knowing from experience that 1001 = 7 × 11 × 13, the prime factorization comes out as 7⁷ × 11⁷ × 13⁷. The whole process took less than a minute.
@@zanti4132 exactly
Woah
@kinshuksinghania4289 Very nice. It's also helpful to note that the number is just a bit over 10^21 or 1000^7. Lastly, with your way of using Pascal's Triangle the origin of the of three sets of double zeros is obvious.
Thank you, sir! That makes it really clear what's going on here with this number. The whole video could be summarized in this one sentence. 👏
01:15 "If you like it, like it at the end." Many others *_start_* their videos by asking for a 'like,' which is backwards if I have never seen their channel before. Your videos never disappoint, so a 'like' at the start would be confirmed at the end anyway. But your approach was respectful of the viewers in a way that I have not seen before -- in the watching of many videos.
I also liked "share it with your enemies, they might need it" 😆
This is smart. All along, I found that magnitude checks can help things make sense.
For example, 101³≈100³=(10²)³=10⁶, so it should have 6 numbers tailing the 1.
7×11×13≈7×12×12=7×144≈7×140=7²×2×10
=49×2×10=980≈1001, so even if we're just guessing prime factorisations, it's a reasonable start.
you are a pearl of a human being. bless you ❤.
cheers from austria.
So cool. The relationship between powers of 11, 101, 1001, etc., and Pascal's Triangle, was completely new for me! (And I've got a degree in math; lol.) Thank you so much for this awesome trick! :)
Same for me. Degree in maths but never heard of this trick. 😂
On first glance, it has 1, 7, 21, 35, 35, 21, 7, 1, so it obviously is of the form 1(x 0s)1^7.
Gap of two zeros between the 1 and the 7, so it's 1001^7.
1001 is equal to 7*11*13
Ans: 7^7 * 11^7 * 13*7
Dropping a comment here for another video I saw of yours so you see it, you really explained the video of finding inverse of a matrix really well, it was a topic I found really hard and could not understand much. Thank you so much for your efforts and will surely recommend your channel to others. Much love from India.❤❤
I appreciate the feedback
Very nice problem find!
Took me a minute or two to recognize the number wasn't palindromic - but then the binomial expansion only took a few seconds to identify and verify. Spotting 11 as a divisor of 1001 was trivial as 11 divides (10 + 01 = 11); and the rest was easy.
Now to check out the video, and how you solved it.
P.S. Welcome back.
I'm going to disagree about memorizing (with intent) random facts such as 1001 = 7*11*13. One inevitably collects a few of these over time spent solving puzzles, but the deliberate memorizing not so very useful.
More valuable, IMHO, is to learn a wide variety of divisibility tests - or, even better, how to construct the divisibility test for any (smallish) prime in just seconds.
Been binging your videos since YT decided to recommend them to me. You're an awesome teacher dude, and the videos are super well structured. Keep up the good work!
PD: Would love to see more Number Theory / prime numbers content 😃
The zeroes between the ones seem to be padding every number. If you take 101^5=10510100501, add a single zero before it and take every pair from it, you get 01, 05, 10, 10, 05, 01 which is exactly the fifth layer of the Pascal triangle without the carry from adding messing up the result. 11^5=161051
How do you add links to UA-cam comments? In your comment the term "Pascal triangle" is highlighted as a link.
Edit: Nevermind, must be some automatic highlighting by the YT app.
If I’m remembering correctly the guts round has 3 problems at a time, and you have to submit your 3 answers to get the next 3 problems. So it helps to go fast/use tricks so you can finish and get to more problems, but at the same time you can’t fix/correct any previous problems
It is easy to test for factor of 11 by alternately subtracting & adding digits. You will get number modulo 11. You can do the same for 7 by using the following "weights" in the calculation: (units) 1, (tens) 3, (hundreds) 2. The weights go 1, 3, 2, -1, -3, -2, 1,... Test for 3 is simply sum the digits. Test for 13 uses weights 1, -3, -4, -1, 3, 4, 1,.... These weights can be calculated using modulo math, starting with units weight = 1 & multiplying by 10 & applying modulo, repeating for each digit until looping occurs.
But your number is too easy as all the prime factors are "small". In reality you need to test for all primes up to the square root of the number. & if none found, declare it a prime.
A good hint of factoring such a kind of number is thinking about the binomial expansion
Excellent point! And that's how the Pascal's triangle came about.
Excellent intuitional solution. You are superb teacher.
the furthest i got was that there's no even factors.
Beautiful example of experience-based intuitive mathematics. How come there are people that do not see the esthetics of maths? I think that even without particulat maths skills and relieved from the intimidating myths of how terribly difficult maths is, one should be able to meditate to e.g. this example just to cherish the pure beauty of numbers. Creds again to you!
Bro you are the GOAT (guy of all time)!! Amazing 👏👏👏
nice
i also recognizing the mirrored number thing and thought it might be divisible by 11 and confirmed using the divide by 11 trick. wasnt sure what to do after. never knew about the 11^x, 101^x, 1001^x,... thing representing pascal's triangle before. neat
Greetings! Never Stop Teaching!
There is a divisibility test for 7,11,and/or 13 which would have quickly said the number was divisible by all members of the trio
Quickly?
Yeah, I think I saw another video on a divisibility test for 1001. It's similar to the divisibility test for 11 since 1000 is equivalent to -1 (mod 1001). The repetition is what would quickly show this is equivalent to 0. And the divisibility test might even make sense after testing 3 since, while 11 is easy to test for, 7 and 13 are not.
@PrimeNewtons I saw a video on checking for divisibility by 7. I think it was Matt Parker on Standup Maths (but it could have been Numberphile). It was not straightforward, and I can't remember how it worked.
I really enjoyed your video, and I learnt something that I think I will remember.
@@PrimeNewtons not that quick😀. Imagine the number is written with commas 1,007,021,035,035,021,007,001
Now sum all the odd trios 001+021+035+007=64
Sum all the even trios 007+035+021+1=64
Take the difference of these results
64-64=0
If this number is divisible by 7 then the original number is divisible by 7
If this number is divisible by 11 then the original number is divisible by 11
If this number is divisible by 13 then the original number is divisible by 13
@@robertpearce8394based on powers of 10 mod 7. It's doable but not intuitive.
Fantastic.Thank you very much.
The legend is back!
Absolutely wonderful!
Thank You for This Video.
I hope you're well
Very nice explanation. One thing you did not mention early on was the observation of the pattern of the numbers in the middle - which was also easily seen. You held it under your belt until the key had already been found.
Hello Prime Newtons! I’m one of your subscribers and a big fan. Please I need your help with a Mathematical problem. How do I reach you sir?
I guess, he provides his email in the video
Welcome back. I liked the video so I left a like, just like you said.
This is really cool. Thank you so much.
OUTSTANDING!!!!
Maths is not a science: it is a kind of art
Extremely Good Question...!!!
You can also solve it with the palindromic method. You would have to divide out 10^10.5. Adding everything up you would end up with 10^10.5 * (10^1.5 + 10^-1.5)^7 = 1000 + 1. This is easier to see with the substitution of x=10^1.5.
Welcome back!
We were missing your session.
Thank you mster😊
that's amazing. I love it.
I haven't watched the full video yet. First thing I would try is, find the digital root of this number. If it's divisible by 3 or 9, or some power of 3, then the number is divisible by that power of 3.
Ok, that didn't work. The next thing I would try is, the alternating digital root (taking each digit with an alternating +/- sign). That will tell me if the number is a multiple of 11. And, lo and behold, by repeating this process 7 times, I found that this number has a factor of 11^7. This is a good start to beeak it down.
Using modular arithmetic, there are other easy ways (similar to digital roots) to check for divisors of other small primes, Since this is like an Olympiad problem, it is probably going to be divisible by some small primes. Not 2 or 3 or 5, but maybe 7, 13 or something. I did figure it out, but I won't spoil it for you.
PS: I did not notice that this number is a palindrome in base 10, until I actually watched the video. Not sure how I missed that, but, your solution is very clever.
My factorization program written in C# factored this number quite fast
Factorization has seven 7 , seven 11 , seven 13
We could observe that this number equals 1001^7
(Binomial expansion and Pascal's triangle for coefficients of this expansion can be useful)
Superb!!!!!😊
Pascal’s triangle and binomial expansion jump on me what I saw the title.
I intuitively said that number is (1000 + 1)^7. Got stuck there for a while…
This is really amazing.
I didn't know the factorization of 1001, but it's a fairly easy check that it 1001 = 990+11 hence 91*11 and I did know the factorization of 91.
Could probably be more rigourous when ignoring zeroes... Otherwise could make a mistake with 10070021003500350021007001 or 100702103503502107001.
Probably best to just expand out (1+x)^7, and then match to determine x = 1000.
Nice approach
I solved by applying divisibility rule by 11. But this solution is way more cool
I wonder if grouping factors would help 1*(10^0+10^21)+7*(10^3+10^18)+21*(10^6+10^15)+35*(10^9+10^12)
1(1+10^21)+7*10^3*(1+10^15)+ 21*10^6*(1+10^9)+35*10^9*(1+10^3)
from this I observe 10^3+1 is a common factor because if we note 1000=A then the rest of the expressions are:
1+A^3=(1+A)(1-A+A^2)
1+A^5=(1+A)(1-A+A^2-A^3+A^4)
1+A^7=(1+A)(1-A+A^2-A^3+A^4...)
Dividing by 1001=7*11*13 we get quotient 1006015020015006001
Dividing by 1001 again quotient 1005010010005001
Dividing by 1001 quotient 1004006004001
Dividing by 1001 quotient 1003003001
........................................... 1002001
repeat 1001
so the original number is 1001^7=(7*11*13)^7
Very interesting, thanky you! The question would be, if the mentioned long number wasn´t a "Pascal´s triangle number", what approach would you take to make a quick guess, on how big the lowest prime factorial must be and which of the frequently memorized prime numbers (1-100) would be most likely to start with?
Before:
I quickly divided in my head up to 7 and got to 100000000000....1 through my division process. No way is that length of division going to occur mentally. I guessed this number must be divisible by 101, or 1001, or something similair like 10001, etc, just from looking. Since i also suspect it is divisible by 7 (and 11), 101 is not divisible by 7, so i was pretty sure it is some multiple of 1001, which is divisible by 7 (and 11)
So I did 1001^2,3,4,5... (with a calculator, ie: cheating) and found the number in question.
I guess I could have written out 1001^7 with a piece of paper and then its "fair"? Idk, but surely calculators wouldn't be allowed for this type of problem. I am so exicited to hear your approach!!!
Edit: I am grinning as I hear your explanation. Wonderful!!!
You also can do so for 202, 3003, 40004 etc.
this is superb
Almost a quarter million subs. I'm one of them.
This is a 22 digit number - large enough to scare people from trying to get started. Last digit indicates that it is not a multiple of 2 or 5. Sum of the digits indicates that it is not a multiple of 3.
Alternate digits add to 19, so 11 has to be a factor of this number. On dividing by 11, the same rule works on the quotient. So, the number (or at least some factor) is some power of 11.
This worked 7 times, so 11^7 is a factor but 11^8 is not.
7th power indicates, that lets try prime roots at least for single digit primes. Square root, cube root and 5th root are not integers but 7th root is 1001 which is multiple of 11 (same alternate digit sum equal rule).
1001 / 11 = 91 = 7 x 13.
So, the 22 digit number = (7 x 11 x 13)^7. This number represents a very bad RSA public key modulus that banks should never use for encryption of high value transactions 😁
Any good prime factorization problem MUST avoid multiples of small primes like 2, 3, 5 and 11 due to easy divisibility rules that expose the solution a bit too easily, as in this example.
Good video like always! Just some other question though: going to enter some math olympiad next year, any advice?
This was a fun one!
Dear sir I have a very interesting puzzle for you I am Surya from India here is the puzzle The problem involves finding an even number:- Joh has an even number and asks his friend to guess the number using the following (clues 1) the sum of the reciprocal of the number of possibilities of writing the number as the sum of two different integers and the actual number is 13 /2 (clue 2) the repetition of the number is not allowed and the order of the pair doesn't matter. What is Joh's number?? Si r I hope you see my message Thank you 😊
No! Im going to like the video at the beginning!
I'm a retired prof and have a stupid question? What are you using to get the blackboard so clean? 🙂 I could never get one that clean. Of course, later we went to the white board with
erasable markers and then to video projectors. Nice to see someone use chalk.
I've just started (I'm always late to the party) working through all The Euler Project problems to prevent Alzheimers or at least put it off as long as I can. Spent all those hours learning
applied mathematics, might as well use it.
I also love to feed such problems to Wolfram Alpha to see if it will break.
Ah yes. The question of life itself
Here I began with exploring the magnitude of the given number by splitting by thousands:
1|007|021|035|035|021|007|001, which can be written as
1.10^21 + 7.10^18 + 21.10^15 + 35.10^12 + 35.10^9 + 21.10^6 + 7.10^3 + 1,
which corresponds with the seventh row of the Pascal triangle (for 10^3 and 1), therefore the whole number is
(10^3 + 1)^7 = 1001^7. Then the easy factoring of 1001 = 7 . 11 . 13. Lucky guess at the beginning, but it paid off :-)
The number of zeroes correct? I think there's an extra 0 between 21 and 7 at the end, otherwise the rule of the number of zeroes doesn't apply regularly(?)
Great !!!!!!!🎖
Wow, that's some trick! But who, who does not already know it, is going to figure it out?
صباح الخير أستاذ
الاحظ أن العدد ليس مماثلا من اليمين واليسار فكيف يمكن أن نطبق الفرضية التي استعملت
10072 0 21035 0 35021 00 21001
هنا صفر واحد عل الشمال وصفرين على اليمين ، أتمنى أن ترد علي وشكرا
Wow! Great!
Nice!
If your calculator cannot handle such a long number, try to split it into a higher and a lower part at a certain digit, e.g.
1'007'021'035'035'021'007'001 =
1'007'021'015'000'000'000'000
+ 20'035'021'007'001,
and test if both complements are divisible by the same. The above, for instance, are divisible by 7×11×13=1001.
Anyway, apply a test or just try out. It will work or not.
1 007 021 015 000 000 000 000
÷ (7×11×13) =
1 006 015 000 000 000
20 035 021 007 001
÷ (7×11×13) =
20 015 006 001
1 006 015 000 000 000
+ 20 015 006 001
= 1 006 035 015 006 001
(Hope I didnt present any typing errors, I had some of them)
You better split it closer to the middle. My calculator, for example, can handle 15 digits.
Just saying. Didn't watch the vid, will do it later.
I HAVE AN INQUIRY. HOW THINK OF THIS METHODE
We get very very similar results even when not working in decimal.
In every base b above 2, 11² = 121
(1b+1)² = 1b² + 2b + 1
(this fails in binary because 2 is not a digit and we have to carry over
And this can be extended to bigger powers, and with putting zeros in between.
11⁴ translates to (b+1)⁴ which is b⁴ + 4b³ + 6b² + 4b + 1, and if your base b is greater than 6, this can be written as the digits 14641.
Now, the fifth row of Pascal's triangle is 1,5,10,10,5,1, so in decimal this will carry over and ruin the forms.
To preserve the pattern we could look at higher bases, eg. in hexadecimal, 11⁵ = 15AA51, where A is the hex 'digit' for the number we would call ten (it's a good idea to confirm 17⁵ for yourselves)
The other thing we can do is pad with zeros, kinda like what we had in the problem in the video. Then the carry-over just hides zeros, which our eyes are often happy to ignore, so back to decimal, 101⁵ = 1|05|10|10|05|01, where I added the vertical lines to help visualize.
ofc we can do both. The 6th row is
1, 6, 15, 20, 15, 6, 1
In hex, this is
1, 6, F, 14, F, 6, 1
So in hex, we can find
1001⁶ = 100600F01400F006001
The thing I really like with swapping bases is what happens to fractions.
In decimal we know that
1/9 = 0.111..., 1/99 = 0.010101..., 1/999 = 0.001001...
Turns out this generalizes too!
I'll write q as the last digit in base b, representing (b-1).
Then the number qq is just b²-1. Using infinite geometric series, we can show that (note that b is an integer starting from 2 and up)
1/(b²-1) = b⁻² + b⁻⁴ + b⁻⁶ + ...
So in base b, this will be written as
1/qq = 0.010101...
1/qqq = 1/(b³-1) = 0.001001...
etc.
Every repeating decimal (but not only in decimal lol) is just one of these 1/qqqq's, times the integer that appears in the repeating sequence, shifted with some power of b⁻¹, with additional first non repeating digits.
Also also, you know how to check if something is divisible by 9 you just add the digits? This is a result of 10ⁿ modulo 9 always being 1, and can be applied to any basis.
Is a number divisible by q?/what is remainder mod q? Sum the digits in base b. Whatever you get, that is, or has the same remainder. Did you get zero? Great, it's divisible by q
(7×11×13)^7=7^7x11^7x13^7=1001^7=1007021035035021007001 final answer
That's what he got too!
Nice.
1007 ... 7001, mmm, let's try the 7th root.
Let's get into the video
amazing
7^7, 11^7, and 13^7
You can write a book for us.
I did it the long way, i waa misguided by the double zero at the right 21.
Very close 😊
ATTEMPT:
Wait, that's just pascal's triangle. That's just (1000+1)^7 = 1001^7 = 7^7 * 11^7 * 13^7
Done
7C0=1
7C1=007
7C2=021
7C3=035
7C4=035
7C5=021
7C6=007
7C7=001
N=sum[n=0,7](7Cn•1000^n)
(a+b)^x=sum[n=0,x](xCn•(a^n•b^(x-n)))
a=1000
b=1
x=7
N=1001^7
1001=1050-49=7(150-7)=7•143
143=11•13
1,007,021,035,035,021,007,001=7^7•11^7•13^7
I'm also back.
1001 = 1111 - 110 = 11 × 101 - 11 × 10 = 11 × 91 ...
wow
You really show beautiful questions and beautiful and clear solutions.
It's just a shame that the screen is too dark with a somewhat claustrophobic atmosphere.
My advice, and I hope you'll see this as constructive criticism, is to replace the board with a clean white board.
Write on the board with a black marker, and you can also use other colored markers to highlight important things. In red, green, and blue...
Thanks for interesting videos.
These mathematics tricks do not contribute much to the science of math nor are they used in the engineering field. Such questions often appear in math competitions but I think they get way to much attention, also on UA-cam, as there are few real life applications for them.
It shows a broad understanding of mathematics, on the part of those who can solve them. Personally, I had little exposure to Pascal's triangle or number theory along the way.
Burpees for the brain.
There isn't much practical use for regular exercise except to keep your body from deteriorating into formless mush. Think of these math problems as exercise for your brain.
Is there a mistake in the question? It appears that there's an extra "0" before the last "7".🤷🏿♂️
Too many blah-blah-blah in the beginning. It is nothing about math. No way, bye.
"Pascal's Triangle" is really Chu's Triangle figured out centuries before Pascal. You can Google Chu-Shi-Kie; the spellings vary.
Could you please provide a reference to this information?
We know that the so-called Pythagoras theorem was actually known in Ancient Egypt and India long before Pythagoras was born.
@@slavinojunepri7648 Chu-Shi-Kie was a Chinese mathematician, and "A.D. 1303" is the date on the manuscript illustration of the triangle. I found it on p. 720 in my textbook "Precalculus" by David Cohen, 4th Ed. Googling Chu's full name gives various spellings.
@@slavinojunepri7648 Oops. My reply disappeared. I hate when that happens.
Google "Chu-Shi-Kie". The spellings will vary. Chinese, circa 1300.
@slavinojunpri7648 I learned the info in one of my Precalc texts. You can Google "Chu-Shi-Kie".
@@slavinojunepri7648 My reply keeps disappearing, but I edited my comment.