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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.
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!
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!
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.
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.
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.
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.
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.
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😂
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
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.
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.
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
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
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.
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.
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.
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
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.
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
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.
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
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.
The sum of the digits in the numerator is 1*7+0+8*8+1=72. 72 is divisible by 9 without a remainder. Divide the numerator by 9 and get a number equal to the number in the denominator. As a result, we get the square root of 9*1, and we get 3. Magic? No - mathematics.
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.
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
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.
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.
@@justaboutanything9378 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
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.
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
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 . . .
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
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
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.
