Japanese Math Olympiad Question | You should know this Trick!
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- Опубліковано 20 вер 2024
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If you're notating x to mean multiply, don't use x as a variable. Or if you're keeping x as a variable, use the * to notate multiplication.
Yes, I would have picked a different letter, both for that reason and because x typically represents an unknown variable, not a known constant. Other than that, though, this is a good solution.
I agree 1000%. Using x to mean multiply and to stand for a number at the same time can be very confusing or ambiguous. For example, 5xt can mean either "5 times t" or
"5 times x times t"
That's why x is generally not used in algebra to mean multiply. I also agree that x, y, z, t and other letters typically used for unknown or changing values should not be used for constants to avoid confusion.
or do what most people do and write the variable x curly
@@kendlemintjed7571 most use conjugate or dot to mean times.
For example 1 + 2(3) = 7
or 1 + 2x or 1 + 2.x
@@lyrimetacurl0 I've seen that but not nearly as often as a curly x
i usually just write my × very small instead of a dot
I used a trick: usually the result is very simple. So if it's a square root, than probably the number under the square root could be sth like 4,9,16...
Let's do a quick test and get the approximate value of the division: the numerator has 16 digits and the denominator has 15..so it will be close to 11/1,2 and it is something a bit less than 10. Could it be 9? Let's check the nominator if it is divisible over 9. It is. Let's try to multiply the divisor by 9. It worked!
Thanks for your insights..
It's so obvious but hidden at the same time.
Thanks again...
W.o.W
Could you please explain how you inferred that it'll be 11/12 but less than 10 from the number of digits?
It’s even easier than that. Realize that the numerator is all ones, eights, and zeros except for the final digit of nine, and add the digits together to see if it’s a multiple of nine. Which it is.
I did it less than 30 seconds (in my head) and got 3: 16 digits on top 15 on bottom - 1111/ 123 = about 9 and sq of 9 = 3
As an Engineer myself I Agree with your simple solution.
This is far and away the best solution here
Although I agree this is the fastest and the most parsimonious way, this is an aproximate Solution.
Plus, there is a certain beauty in seeing how the numbers are transformed along the way.
Fast is good, But is not always better.
I watched the video at fast speed and got the answer in 28 seconds. Try harder next time.
Well done only if you are an engineer. But if you are an engineer you are an idiot not to use a calculator.
The way I did it was, check the last digits of the nominator and the denominator (9 and 1 respectively) and determine what the trailing digit of the quotient is, in our case it's 9 (the last digit of a product is determined only by the last digits of the numbers being multiplied together), but given that the top line has 16 digits, the bottom only has 15, and the left most two digits of the top is smaller than the left most two digits of the bottom, there is a good chance that the quotient is 9.
To confirm that, either do a long division, or multiply the bottom by 9 to see if we get the top, or add the top and bottom together and see if the sum is exactly 10x the bottom, I chose the latter since adding is much faster than multiplying.
The result was that the sum was indeed 10x the nominator, so the whole fraction reduces to 9, leaving it's square root as 3
Set x =11111111
Observe that the denominator is x²
adding x to the numerator gives 1111111100000000 = 10^8 * x = (9x + 1) * x = 9*x² + x , so the numerator = 9x²
so the fraction is 9 and the square root of that is 3.
The key elements are "seeing" step 2 and 3 of which 2 is the hardest.
As others have pointed out, the alternative way is suspecting that the fraction equals 9, by estimation, then checking it is.
Yes, the denominator is the only trick one has to know. The 123....321 rule should have been known a priori and the rest becomes straightforward. Pretty flat for an Olympiad question
Took me a couple of seconds to determine it is 3... but only because I was looking at repeating number square patterns during the past week.
I recognized the top to be the pattern for 3 (1..08..9 the number of 1s or 8 indicates the number of repeating 3s); and the bottom to be the pattern for 1 (123..321 with the highest number at the inflection point indicating the number of repeating 1s)...
3333 3333 / 1111 1111 = 3
Interesting solve. A few steps could be abbreviated and better to avoid using x to denote multiplication when also using it as a variable.
Some people can't follow if he starts to skip steps, the ones with more practice can be like "Factor the nine already" but we lost nothing. On the other hand, someone with less practice could get lost. I guess the channel is for everybody is not like this is a problem on Lie Algebras or the version of functional analysis that we use in physics.
Should always write x as 𝓧
Yeah - no need for all those steps. Just look at the numbers - answer top division is clearly just a little under 10. And with a denominator ending in 1 and numerator ending in 9, 9 is the obvious answer to the division problem. Far more math work than needed to actually solve.
The early steps are ok, but after around 5:33 it becomes rather ponderous in simplifying the algebra.
Can use * or force parentheses
Why so complicated? 16 digits over 15 digits, and after the addition / subtraction the last digits of numerator and denominator are 9 and 1, respectively. So the last (and only) digit of the quotient must be 9. To test just multiply the denominator by 9 and it checks. Sqrt(9)=3. Done.
True
why should it be 9? from what I understand, 16 digits over 15 digits only imply that the result is between 1 and 100. does the last digit even matter? not saying your wrong, I just genuinely dont understand what you mean.
if anything, you would simplify it to something like 111/12 which is pretty close, but this is a math olympiad question so I assume they're not looking for approximations.
Actually the numerator 1, 111, 111, 088, 888, 889 = 33, 333, 333^2.
And as for the denominator 123, 456, 787, 654, 321 = 11, 111, 111^2.
So after being sqaure-rooted, the number becomes 33, 333, 333 / 11, 111, 111 = 3. And this is the answer!
Beautiful!❤
You are wright, but without calc you couldnot know a value of numerator and denominator! Our beautifull lady was show the nice trick how to do this! Thanks her!
@@zvonimirkujundzic6867 also the trick shown in the video requires that you identify the denominator as a number at the power of 2. Without a calculator is not simple to guess in any case.
Very interesting observation... :)
Wisdom gained after the event no doubt.
Brutforced this one.
1. sqrt(a)/sqrt(b) = sqrt(a/b)
2. Answer should be simple so a/b should be integer and complete square
3. 1111111088888889 can be divided by 9 because 7*1+7*8+9 = 8*9
4. Take a minute to carefully divide it. 1111111088888889 / 9 = 9 * 123456787654321
5. sqrt(9*123456787654321 / 123456787654321) is trivial.
Answer is 3
But your solution is more elegant
More systematic
Очень нравятся Ваши ролики, развлекают лучше фантастики. На скорости 1,5 просто супер! 😍🙋♀️
The denominator is an obvious pattern, that is achieved by repeating the digit 1 and squaring that number. Assuming the question has a clean answer, the numerator is likely a multiple of the denominator. Sum of numerator and denominator is denominator x 10. Numerator is denominator x 9. Simplify to square root of 9
Answer: ±3 (the answer you're looking for is probably just positive 3 though)
that's why we use the dot * for multiplication and not the x
Amazing step by step explanation!
Thanks! Your welcome
In China, we use the following method to do this kind of questions（ We never tell anybody this secret. Today I have to make a decision that goes against my ancestors😇）
.
11/1.2=9.16
so answer is 3😂
Thank You for your insights..
God Bless YOU and your Ancestors too :)
Your ancestors must be followers of Aryabhatta.❤
😂😂😂 哈哈！牛逼
😂😂
In the test you have to show the long solution to get high score so you can’t just use approximation 😂
I note that the numerator is really 11111111*sqrt(9), from which the result follows. So it won't work in base 12. There you get 11111111*sqrt(e) where e is the digit eleven. The square root of eleven is irrational. It works for bases n^2+1 where n is an integer. For example, sqrt(1111111033333334) in base 5 is 22222222 in base 5.
Very cool problem and solution!
Thank you very much!!
I kinda just lucked out, I felt the numbers were too specific so it was probably a round number, I tried subtracting them got nothing, then I tried adding them (1,111,111,088,888,888 + 123,456,787,654,321) and got 1,234,567,876,543,210, which as you can see gives the bottom number multiplied by 10, so if x + y = 10y, then that means x = 9y, the top number is the bottom number multiplied by 9, so the division gives 9, so the answer is 3, the square root of 9
Could have gone straight to long division, writing out the 123456787654321 times table by repeated addition.
As soon as you hit 123456787654321 x 9, the answer pops right out.
Might even be quicker too, but it assumes that the answer is going to be simple.
All you have to do is recognize the numerator is divisible by nine, do the long division (which isn’t hard), and realize that the quotient is also the denominator. Cancel and take the root and the answer is three.
No thanks, i'm good. Have a nice day. Goodbye.
There is a much more elegant solution. As you showed 11^2=121 and 111^2=12321 so 123456787654321=11111111^2. Now 33^2=1089 and 333^2=110889, now using this logic 33333333^2=1111111088888889 (which is the top part of that fraction). Therefore, under the square root, you get (33333333/11111111)^2=3^2. Take the root and you get your 3. But I can see I'm not the only one who has commented that solution.
The way you solved it was needlessly complicated and long.
Pretty fancy way to write three!
That was crazy. All that to get 3. Niiiiiice
Brilliant, can I say that some of the very obvious steps need not be written twice. You could easily make this video half the length it is. In conclusion fantastic proof anyway.
Would be fun to see it solved when the denominator had been 123456299503506-3 :) Of course without calculator.
I never really watched yt that much until I found your channel
a/b where a+b =10b
Easy.
(Just add both numerator and denominator)
Directly add 1 to get perfect square of numerator which can be written as 1111111 × 9999999 / and denominator as 1111111^2
Which can be divided by 9 therefore getting sqrt value of 9 which is 3
Excellent ! 😊👍🏻
Surprisingly my iPhone gave me the good result with 10-14 precision !
I did not know it could handle such long figures …
I am usure Excel can ! 😊
The denominator is obviously 11111111^2, the rest easily follows as this type of Q usually has rational answer
Dear Friend: Excellent Video !!!
Your work was incredible..!!!
I enjoyed the journey and numerical gymnastics you presented :)
I love the solution, very formal and step by step.
I did it a different way, and I'm glad i tried before watching your method, as i wouldn't have noticed this trick if not for my stubbornness:
I decided to factorize the numerator (having recognized that the denominator is 11111111^2). Applying a quick divisibility check to the numerator reveals 9 as a factor (digit sum is divisible by 9).
Once you start long dividing the numerator by 9, you'll see the beautiful number 123456787654321 emerge as the quotient. It's quite pretty.
So those giant factors cancel leaving root 9 equals three as the answer.
Do the addition and the substraction, then multiply numerator and denominator by 9. The result inside the square root is: (1111111088888889)*9/1111111088888889. Simplify. The result inside the square root is 9. So, the solution is: +-3
I did it extremely fast 33^2=1089 , 333^2=110889 11^2=121 ,111^2=12321 , followed by pattern ans will be the sqrt of 3333333^2/1111111^2= 3
I found the sqrt of 1111111088888889 in the same way as with the divisor, by starting the sequence with sqrt(1089), which must be a bit larger than 32 = sqrt(1024), so I tried 33, it worked. I noticed that 3 is sqrt(09) where 09 is the first iteration of the number and 1089 the second, so I had the sequence, where for 1111111088888889, which is the 8th iteration, the sqrt must be 33333333. Dividing by 11111111 gives 3.
Wow... impressive but pretty hard way to get the answer and you had to realize that 123456787654321 was 11111111 squared. For me once I realized that, it seemed that the top number was also likely some repeating number and 3 seemed like a good choice with 33 squared giving 1089 and 333 giving 110889 so the numerator should be 33333333 squared. Get rid of the squared signs and square root reduce the fraction and get 3.
Extraordinary result, greetings
What kind of math knowledge is needed to solve this. You're either lucky or you're not. Ridiculous.
Real number properties and number theory.
Tricks and a little bit of a mathematical sense. Like the pattern with squares of 11, factoring equations. You don't really need much math knowledge.
Add numerator(n) + denominator(d), you get 10 times denominator =>
n+d=10d
n=9d
n/d =9
(n/d)^0.5 =3
Здорово. Сначала выгоядело как галиматья, но такое изящное решение. Восторг
Got it in 10 seconds. I already knew the property of repeating 1s and 3s when squaring, so that led me to get 3 as an answer
я методом подбора определил, что числитель делится на 3 и вынес 3 за скобки, затем ещё раз сделал то же самое, в скобках осталось число равное знаменателю. Определить делится число на 3 можно посчитав сумму всех его цифр. Если она делится на 3, то и всё число делится на 3. Но это не честный вариант, хотя и без калькулятора, и два раза пришлось делить столбиком.
This is a very interesting problem. Thank you very much for introducing it to us. However, the video is toooooo long. I believe it can be made to be less than 3 mins.
The top one is the multiplication pattern 3 × 3, 33 × 33... and the botton one is 1 × 1, 11 × 11... The answer is three.
I solved it instantly. The Answer was same as given in the question
You can also realize that 1111111088888888 = 1111111100000000 - 1 - 11111111
Then 1111111088888888 + 1 = 11111111E+8 - 11111111, or 11111111 x (1E+8 - 1)
Dividing by 11111111^2, you get (1E+8 -1) / 11111111, or 99999999/11111111 = 9
😮😮😮 Wonderful maths is so beautiful
Sum of digits in nominator is divisible by 9, that is why the nominator is divisible by 9. If you divide it - you will get denominator.
Beautifully done!
The result is absolute value of 3
Absolute value of 3 is 3 FYI
@@ZAWARUD00 how about -3? It is not 3, but will be correct in case of absolute value of 3
@@hiryu70 the absolute value of -3 equals the absolute value of 3, but they are not the same.
Maybe use a dot for multiplication if you are using an x variable? Just a suggestion
111111108888889 (numerator) can be written as 1111111 x 9999999 and denominator can e written as 1111111² and after cancelling out everything we end with √9 = 3
Clever. 😊
Wow❤
❤️❤️❤️
Проверить разложение числителя на множители, делится на 3. Разделить на 3, проверить разложение, понять что делится еще на 3. Разделить на 3. Сократить числитель и знаменатель.
Very nice
Thanks
This kind of Math Olympiad Questions singles out kids who spend too much time inside, not the bright ones.
I like how you say hello.
it is 3 or -3. Great approach anyway. Regards.
@@ThatYapper x doesn't matter. (any number)^2 woudl be positive. as soon as we have positive numbers in square root we shoud consider negative outcome
@@nikolayparygin610 you are absolutely right, I mixed up the fact that we only solve for positive value of y when in √y situation, I'll delete my previous comment oopsie
it is just 3 tho
The top is divisible by 9, which is clear because all the digits add up to a number divisible by 9. Once the 9 is factored out, the numerator is just 123456787654321, which cancels out the denominator. So sqrt(9) = 3.
We can directly estimate the value to match it with given options
Well presented, but in reality this problem only tests whether one knows that 111...^2=... No real reason to learn, remember, or teach this rule (unless you aspire to be a number theorist).
Surprisingly the answer was a simple 3
That just made my day 😂
Damn why i am thinking his math is week
He took very long way to solve such simple question
This is nuts. Although if I'd just taken the first 3 digits I'd have gotten sqrt(111/12) which is around sqrt(9) = 3. So I could have gotten a very good estimate.
Everytime she say "over" its always remind me of wall-e saying "eve", love ur vid
The Doctor of Arithmetics!!! Ar.D.
Nice math quest!
Nice! One small complaint, though ... the answer should be +/- 3
Sorry, but you may have mistakenly remembered that √9=±3. √9=+3 and -√9=-3.
By definition, √9=√(3)²=|3|=3 and √9=√(-3)²=|-3|=3.
A definition does not require proof. It is what it is because it is decided. "√" is supposed to represent the positive root of the squared number inside the √, either positive or negative. Therefore, when extracting a √, the absolute value symbol is added to avoid mistakes even if you do not realize during the calculation that the inside of the √ is a negative square.
If √ was a symbol representing both positive and negative roots, even simple calculations would cause confusion. For example, √9-√4=3-2=1, but if √9=±3 and √4=±2, then 3-(-2)=3+2=5, (-3)-2=-5, and (-3)-(-2)=-1 would also be solutions. This is very strange.
Joli petit exemple de calcul 👍. C'est aussi un bon exemple d'étude de rédaction. Comment faire pour rédiger d'une manière plus courte et beaucoup moins fastidieuse ? On va utiliser un petit peu D'ABSTRACTION et quelques LETTRES et prendre un peu de RECUL :
Pour bien comparer les deux rédactions, comme lui, je ne vais pas écrire les commentaires. Ma rédaction commence donc ici. On va éviter de recopier la racine et on va couper en morceaux.
quotient sous la racine = N/D
Calculons D seul
X=11111111 (8 fois le 1)
111x111 = 12321 etc. donc D = X.X = X^2. (👍)
Calculons N seul:
N = 1111111000000000 + 88888888 + 1. (au début 7 fois le 1 et 9 fois le 0)
N = 11111110.10^8 + 8.11111111 + 1 et on introduit X
Or 11111110 = X - 1
N = (X - 1).10^8 + 8X + 1 on développe
N = 10^8.X - 10^8 + 8X + 1 = 10^8.X + 8X - 10^8 + 1
or - 10^8 + 1 = - 99999999 = - 9X
N = 10^8.X + 8X - 9X = 10^8.X - X
N = X(10^8 - 1) = X.9X (voir ligne d'avant) = 9X^2 (cool)
donc N/D = 9X^2 / X^2 = 9 donc le résultat est rac(9) = 3.
Beaucoup plus court, plus clair, plus rapide, donc moins de fautes et beaucoup plus facile à refaire. 😉. Même en maths la rédaction ça compte. L'ABSTRACTION est indispensable et ça aussi ça S'APPREND .
Encore merci pour ton joli calcul.
Took be about 10 seconds to divide mentally both num. and den. by 11111111
Very nice solution !
well, that was a lot more complicated than you had to make it. If you observe number in the square root is (11111111*10^8-11111111)/11111111^2, you can reduce it to 10^8-1/11111111, which is 99999999/11111111, and the square root of 9 is +/- 3
Sqrt(9) = 3.
Not ±3.
You always Take the Principle (positive) root.
You only Go that Route if Your using The inverse Of a Number Squared.
eg x² =9
x =±3
Exvellent work my friend
Thank you very much!!
I participante at the Brazilian math olypmics of public Schools
Много единиц и восьмерок - проверяем делимость на 9. Числитель делится, знаменатель нет. Делим числитель на 9 в столбик - получаем знаменатель.
Получаем, что значение выражения - корень из 9, т.е. 3
Remember shortcuts and make records. Then call them super student 😂
Starting with the assumption that this is only going to be given as a no calculator problem if you're dealing with whole numbers .. . takes 3 seconds to solve. Answer is clearly under 10 but clearly just about 1 under 10. And denominator ends in 1, numerator 9 so that checks. Answer is 3. If you're going to use a "trick" why use a trick that takes that much math . . .
Yeah... Try writing "The answer is clearly 3" in a math copy, see if the corrector likes that.
В следующущий раз попробую поделить в столбик😊
Пример специально подобран так, чтоб не всякий калькулятор осилил)
Красивая задачка)))
придуманная с конца
Wow unbelievable😂
I solved this in a 20 seconds. super easy! 1.11. 10^15 / 1.23.10^14. Must be 9. 11.1 / 1.23 is roughly 9.
If it's not 9, it's simply impossible to calculate on paper or in the head.
Yeah, that is not solving it. that is an approximation and you will get zero points for that in the Math Olympiad. You have an approach that is useless in Mathematics, but you could do very well in engineering.
This should help you to realize the answer is 9, but now you need to prove it. Your trick will be useful for this though, as your proof can now simply be to divide the numerator by 9 and quickly get the denominator as the remaining factor (which would be a difficult approach if you didn’t know 9 was the factor)
Lo resolví en 5 segundos y no por hábil sino por lo predecible. Me dí cuenta que, multiplicando el denominador por 9 daba como resultado el numerador. Raíz de nueve tres. Fin
Wow😊
This is so useful, like when... Wait. 🤔
Mən 5 saniyədə tapdım cavabı. Yuxarıdakının sonu 9, aşağıdakının sonu 1 edir. Məntiqlə 9-a bölünüb. Kök altı 9 da edir 3.
Nice. Anyway it is stupid to use the same symbol (x) for multiplication and as a number.
=sqrt((1111111088888889*9)/(123456787654321*9))=sqrt(9*1111111088888889/1111111088888889)=sqrt(9)=3
Much simpler
Why multiply both by 9: answer must be simple, sqrt(11/1,2) is approximatly 3 so try 3^2=9.
perfect question
Я решил так.
Я знаю, что числа 11, 111, 1111 во 2ой степени имеют вид:
11^2=121
111^2=12321
1111^2=1234321
И так до 11111111^2=123456787654321
В числителе сумма цифр кратна 3, значит делим в столбик 1111111088888889, получаем 37037032962963, опять сумма цифр кратна 3, опять делим на 3, получаем число 123456787654321.
Итак теперь выражение имеет следующий вид: sqrt(11111111^2 × 3 × 3/11111111^2) = sqrt(9) = 3
Потратил всего 1.5 минуты на данное выражение.
корень из 9 не может быть отрицательным?
33^2=1089 , 333^2=110889, 3333^2=111088889, 33333^2=11110888889,....., 33333333^2=11111110888888889
Numerator being divisible by 9 is crazy
beautiful :)
By count ing the digits, 16 digits / 15 digits number, assuming the answer is integer, i guess it is sqrt(9) = 3
divide numerator by 9 manually, and you will see the magic
Very good, but be careful when you also use the 'x' symbol for multiplication, because you might get confused!!
√...8+1/...2-1→ √...9/1 → √9 = 3.
Oh my.... wonderful math
Thank you very much!!
Amazing
Thank you very much!!
Great. Thanks!
Thank you very much!!