Not a good solution. This is a simple addition of arthimetic progression = n(n+1)/2. The actual formula for any arthimetic progression for a, a+d, a+2d,.....,a+(n-1)d. Sum = n/2[2a + (n − 1) × d]
If I am not wrong, this is the simple formula for the sum of an AP, (S = n/2(a + l)), I don't think it would be right to call this a trick since most students giving olympiad normally know these formulas
Что за прикол с вариантами ответов, почему D не показали сразу. Решил, а ответ ни один не подходит... Подумал, что туплю, пришлось целиком это длинное видео смотреть!))
I tackled it a slightly different way...I ignored the 1st 3 terms (1+2+3 = 6; we'll add this in at the end) and paired off 4+36=40. 5+35=40... to 19+21=40...(the 20th term = 20 and is left unpaired and so, again, we'll add it in at the end. SO... You have 36 terms less the 1st three and the 20th term which that leaves 32 terms that are sorted into 16 pairs that equal 40. 16x40 +20 +6=666 ... The ADVANTAGE is it's easier to multiply round #s like 16X20 rather than 18X37 (in my humble opinion).
1 + 2 + 3 + 4 + 5 + 6 +…+36 36 + 1 = 37 35 + 2 = 37 34 + 3 = 37 33 + 4 = 37 and so on… There are 18 pairs of even and odd numbers between 36 numbers. Therefore, 18 x 37 = 666 So the answer is 666.
The solution Given In video makes No sense and is totally irrelevant. A student may learn the wrong way to attempt such questions and totally screw his learning procedure. Correct way to attempt this question is to Use Sum of n terms of A.P which is n/2 ( 2a + (n-1)d).
The answer is sqrt(666) Here is my trick... I used the sum of arithmetic series and the "Last Digit Trick" Sn= n/2(a1+an) Sn= 36/2(1+36) Sn= 18(37) Sn= 666 So sqrt(666) I solved it like a beast (pun not intended)😮
Ah, that was novel, a multiple choice question in which all the answers are wrong! Until, the correct answer is finally revealed, off screen.
Dave❤️
Thinking outside the box
It can be more easier like you know the trick of finding the sum of natural numbers which is given by “(n(n+1))/2” easy 😙
Dude came in clutch👉
yes
1:46, you'll never see its comiiiiiiing
Not a good solution. This is a simple addition of arthimetic progression = n(n+1)/2. The actual formula for any arthimetic progression for a, a+d, a+2d,.....,a+(n-1)d. Sum = n/2[2a + (n − 1) × d]
You are right. A normal student also knows that the solution is this, let alone an Olympiad student.
n(n+1) /2=36/2*37=18*37=666
if 'a' is taken as the final number and 'b' is taken as the initial number then it can be written as -
[a^2 - b^2 + a + b] / 2
It's just root under 36×37/2
I remember these kinds of problems, thank you very much 🥳🤗
You’re welcome 😊
If I am not wrong, this is the simple formula for the sum of an AP, (S = n/2(a + l)), I don't think it would be right to call this a trick since most students giving olympiad normally know these formulas
Thank you. You are very good and this instance very beautiful ❤
❤️
awesome.
This is no olympia this is a 3rd grader school question for turkish education
Wow amazing
Thank you🥰
Mind blown
❤️
Factors 💯
👍🏻🥰
@@LKLogic 🤸🤸🤸♥️💫
Very nice
Why couldnt i just calculate all of them one by one
Sum of first N natural numbers method is more easy.
I was wondering why all the answers I saw were wrong. 😆 FWIW, I'd simply it to 3 sqrt(74)
Что за прикол с вариантами ответов, почему D не показали сразу. Решил, а ответ ни один не подходит... Подумал, что туплю, пришлось целиком это длинное видео смотреть!))
3√74 is the answer and gaueses law are very effective to solve this problem .
Simple:-
Sn ka formula lagao
Thank me later
3 solutions fausses apparaissent pendant la démonstration ! La 4ème (d), qui est la bonne, n'apparaît que quand tout est terminé !
I tackled it a slightly different way...I ignored the 1st 3 terms (1+2+3 = 6; we'll add this in at the end) and paired off 4+36=40. 5+35=40... to 19+21=40...(the 20th term = 20 and is left unpaired and so, again, we'll add it in at the end. SO... You have 36 terms less the 1st three and the 20th term which that leaves 32 terms that are sorted into 16 pairs that equal 40. 16x40 +20 +6=666 ... The ADVANTAGE is it's easier to multiply round #s like 16X20 rather than 18X37 (in my humble opinion).
A simple ap question...
1 + 2 + 3 + 4 + 5 + 6 +…+36
36 + 1 = 37
35 + 2 = 37
34 + 3 = 37
33 + 4 = 37 and so on…
There are 18 pairs of even and odd numbers between 36 numbers.
Therefore, 18 x 37 = 666
So the answer is 666.
Don’t forget the square root
The solution Given In video makes No sense and is totally irrelevant. A student may learn the wrong way to attempt such questions and totally screw his learning procedure.
Correct way to attempt this question is to Use Sum of n terms of A.P which is n/2 ( 2a + (n-1)d).
The answer is sqrt(666)
Here is my trick...
I used the sum of arithmetic series and the "Last Digit Trick"
Sn= n/2(a1+an)
Sn= 36/2(1+36)
Sn= 18(37)
Sn= 666
So sqrt(666)
I solved it like a beast (pun not intended)😮
Me in .... With u
🙌🏼🥰❤️
We can’t see answer, d
Application
C)=666
I though all answer are wrong🤣
Your options doen't have the Answer
n(n+1)/2
36*37/2 = 666
too easy!! don't need your tricks😂
This is not the right answer (√666) the right answer is 666 😅
I don't believe in anxious writing I think this isn't true 28.80 ≃ √666 ≠ 100%
Or as the memes say
25.80 is the (square) root of all evil