Math Olympiad Problem | 90% Failed to solve | You should know this Trick ! 

Let a=11111111. The denominator is trivially a². The numerator can be split into trunks: 1111110 × 10⁸ + 88888888 + 1 = (a-1) 10⁸ + 8a + 1. But 10⁸ = 99999999 + 1 = 9a +1. So the numerator becomes (a-1)(9a+1) + 8a + 1 = 9a² after multiplying out the parentheses.

I'm in the lower 50% and solved it by reasoning. 11111111^2=123456787654321, so 123456787654322-1=11111111^2. The numerator is a bit more challenging. The sum of its digits is 72, so it is divisible by 9. Guess it is 33333333^2. Use induction to verify. 33^2=1089; 333^2=110889; 3333^2=11108889; ...; 33333333^2=1111111088888889. Thus, √[(111111108888889)/(123456787654321)]=√(33333333^2/11111111^2)=√(3^2/1^2)=3

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excellent sir

One of the few cases where i went the exact same way (first raw estimating, then realizing the pattern), except first taking the numerator to be 33'333'333 ^ 2. But as always: "All roads lead to Rome!" :D

First thing that came to my mind was “this numerator looks like a multiple of the denominator, let me try 9”. Bingo, solved it in a minute.

Please speak English, so that you have global audience. The video and the topic were good, but the verbal explanations were incomprehensible.

😮

Great presentation! Seems that the biggest difficulty in this challenge was trying to keep track of the number of digits.

I think that 99.5% failed to solve, thanks sir.

👏👏👏

√(33333333/11111111 = √3

Multiple 123456787654321 with 9 and get the answer

For general people, it is unlikely to rationalize the sequence of getting 11111111x 11111111, not to mention 99999999x 11111111, without using a calculator------------- Yet once you find out the magic number of 11111111, you can derive that 1111111088888888+1= 11111111x ( 100000000-1) to get the answer, instead of 'guessing the 99999999------

Сколько лишнего бормотания

Intresting problem 👍

It is wrong answer! True is (+/-)3.

I'm about 10sec 😂

Nice explanation 👍

Super👍👍👍

(x+2x-3) (x+2x-3)

brilliant !

Thinks sir