everything is fine, but once the factorization has been obtained, Vieta's theorem can be applied. You can also simply factorize the free term so that they differ by one 1111122222=11111*2*3*16667=33333*33334. And also apply Vieta’s theorem.
If you add 3*3+3,you get 12,in the same way if you multiply 33 with 33 and add 33 you get 1122,in the same way if you need to get 1111122222(that is,five 1's and five 2's)your answer must be 33333(five 3's)
I solved by realzing it is X(X+1) which is two consecute numbers. The square root of 1111122222 will be between them andthe answer is the lower of the two numbers.
everything is fine, but once the factorization has been obtained, Vieta's theorem can be applied. You can also simply factorize the free term so that they differ by one 1111122222=11111*2*3*16667=33333*33334. And also apply Vieta’s theorem.
Simple pq formula: x1,2 = - (p/2) +/- SQR((p/2)²-q)
x1 = 33333, x2 = - 33334
Excelente "saída" para a equação. Está aí a beleza da matemática. Parabens professor.
Thank you very much!!!
Решение хорошее,но очень много лишних преобразований,к тому же сложных.Две скобки (7:02) можно было получить значительно проще.
If you add 3*3+3,you get 12,in the same way if you multiply 33 with 33 and add 33 you get 1122,in the same way if you need to get 1111122222(that is,five 1's and five 2's)your answer must be 33333(five 3's)
What about the other value -33334?
The sound of the roster in the background makes it even more exotic :)
Thanks 👍👍
Excelente explicación en el idioma de los números y sus reglas. COMO SE DICE EN MI PAÍS: MÁS CLARO ECHALE AGUA👌👌👍👍👍
Thank you very much!!!
Beautiful prob.
Thank you very much!!!
I solve it💪
Great 👍👍
Same idea but I had assigned y = 3*11111 along the way. Then x(x+1) = y(y+1), (2x+1)^2=(2y+1)^2, so x=y or x=-(y+1). Nice problem.
Cool: 33333*33334=1111122222 just as 33*34=1122. Would you please explain how you arrived from x(x+1)=y(y+1) to (2x+1)^2=(2y+1)^2 ?
I solved by realzing it is X(X+1) which is two consecute numbers. The square root of 1111122222 will be between them andthe answer is the lower of the two numbers.
Log2 + x = 1,111,122,222
Sem finalidade pratica
Танзанийские мотематеги оказались круче индийских мотематегов!
х1=33333; х2=-33334.
Why you substitute or used the y nothing to do with the equation
brother just solve the quadratic the usual way at this point
استاد عزیز سرعت شما زیاد است وپرش داری
11111
이해안가, 인수분해가 어떻개돠지????
3y + 1을 쓰기 위해서 x의 인수를 (3y+1-3y)로 만든거에요
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X=33333.
who cares?? not pratical