I did via the 1st approach only but wouldn't take more than 2 mins because very comfortable with the calcs and series and its kinda very similar to nested loops problems.
I did with second and first method both, second method slightly different not adding straight in beginning but with seeing both series got that add them and get the answer without writing the full series which saved time
All you have to do is calculate the sum of pairs above the diagonal. Which is S = 2*(1*9 +2*8 + 3*7 +4*6) + 5*5 = 165 A becomes 2*S + 55. 2 times the pairs + the diagonal sum. B becomes 10 * Σ(1:10) + S. (Subtract 1 - 10 sum from every row and you'll see S is left. ) Hence 3*165 + 11*55 Max 2 min lage ye solve karne ko.
direct dikh ra tha ki puri matrix ke saare no. ka sum pucha hai islye 20 x 55 direct bol skte hai as ek A set aur B SET ek dusre ke conjugate hi puche hai
Looking back after 2 years, it's hilarious to think how challenging I once found this. Now I can solve it in under a minute with a quick script! 😂😂😂 int main() { int sum=0; for(int i=1;i
suppose i take a value that is equal to 10*(1+2+...10). in A I am subtracting something from it and in B I am adding the same thing in it so when you add it they get cancelled and answer is 2*10*(1+2...10). no need to calculate anything if you can see this
jab i aur j me lerahe h toh if min(i,j) = i then max(i,j) = j, i=j ho ya na ho. So adding both min and max meaning taking both i and j. So very easy to solve.. Ye mere mind me 2 min ke baad strike hua.
bhai mera 1 min ho gya agar meri approach se chle toh assume i>j fir min=j and max=i ke hiasab se solve ho jayega ,bcz sykmetry hai to j>i ke liye bhi same ans ayega
Two equations for finding value of 'x' and 'y' 1) ax + by = C 2) px + qy =C (If "C" is the Real Number as- 0,1,2,3......). Can we write or say. "ax+by = px+qy = C" ("ax+by = px+qy") ( Because both are equal to "C"). Is it true or not. please help with example 🙏
M a neet aspirant and i used a super lengthy method to solve this Min wala to jaldi aa gaya, max wale me lamba chala gaya But for me, a = 440, b = 660, a+b = 1100 First lets solve for a, We get a series 1*10+ 1+2*9 + 1+2+3*8.... Notice, we have 2 series in here, writing them as x and y, we have A= x+y, Where x = 10*1+2*9+3*8... And y= 1+1+2+1+2+3.... But if u would notice, x = y So, a= 2x, or a = 2(1*10+2*9....) Here the series x repeat itself after 5*6... So we dont have to calculate every value, but rather, X = 2(10+18+24+28+30), Thats 110, so x = 220 and a = 440 A = 440 Now solving for b, For j=1, we have the summation to be 55 For j = 2 we summation to be 55-1+2 = 56 For j = 3 we have 55-3+6, thats 58 Or more precisely, if j = x, (x cannot be greater than 10) The summation= 55 +x(x-1) - x(x-1)/2 Or 55+x(x-1)/2 Now we have the series, 10*55 + 1(1-1)/2 + 2(1)/2+ 3(2)/2 + 4(3)/2 + 5(4)/2 + 6(5)/2 + 7(6)/2 + 8(7)/2 + 9(8)/2 + 10(9)/2 550 + 1+3+6+10+15+21+28+36+45 Theres a pattern in the series, 1+(1+2)+(1+2+3)+(1+2+3+4)... 10*1 + 2*9 + 3*8 same as x ! we solved it in a and the answer was 110, This means, b = 550+110 B= 660 Now a+b = 1100 Ik its very lengthy but i enjoyed it (took me 10 mins, ik m noob)
Isse ghatiya sets ka question mene solve nhi kia phle lga tha double sigma h BT ka fir sets dekha to smjh aya fir jb row aur column wise khol k question smjh rha tha to pta chla isme time barbaad hone wala h alg hi ghumaya isne ksm se 1x19+2x17+.... Aese kr k gya tha ye😭
Iska seedha approach tha- woh table banane se better yeh hai ki I and u interchangeable hai toh ek pehle kisi ek ko fix karke dusre ke values nikaal do phir double it- par u have to add the elements jaha pe they're same- kafi aasan hai I did it mentally
Sone jaa rha tha fir notification aaya aapke question ka. Neend udd gyi khud se solve krne mein but ban gya ;-; Mai method 1 se kr rha tha, so thanks for alternate method :)
use: max{i,j} = (1/2) ( | i + j | + | i - j | ) and min{i,j} = (1/2)(| i + j | - | i - j| ) Then A + B will be sum \Sigma | i + j | = \Sigma i + \Sigma j = 2 * \Sigma i = 1100
🔵Theorem: Let n be a positive integer. Then, ( Sum_{i,j = 1 to n} min { i , j } ) + ( Sum_{i,j = 1 to n} max { i , j } ) = n^2 (n+1). Proof. Sum_{i,j = 1 to n} min { i , j } + Sum_{i,j = 1 to n} max { i , j } = Sum_{i,j = 1 to n} ( min { i , j } + max { i , j } ) = Sum_{i,j = 1 to n} ( i + j ) > = Sum_{i,j = 1 to n} i + Sum_{i,j = 1 to n} j = 2 Sum_{i,j = 1 to n} i > = 2 Sum_{i = 1 to n} Sum_{j = 1 to n} i = 2 ( Sum_{i = 1 to n} i ) ( Sum_{j = 1 to n} 1) = 2 ( n(n+1)/2 ) ( n ) = n^2 (n+1). ■ _________________________________________________________________________________________ 🔴Corollary: For n=10, the above theorem implies that the given sum = 100 (11) = 1100. ■
@@urlikskull1071 Brah, it was just for fun. 😆 But if you see beyond the sophisticated appearance of these symbols, you will see that I have proved a general result with simple arguments that will allow one to solve the question in the video with ease.
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
Revision Course is gonna be fun 🤩 forms.gle/NcywqBrqSE45PhCUA
intuitive if youve done nested loops in computer science
true
EXACTLYY
Khi toh kaam aaya padhna
good one 😂
Me who took Sangeet 🥶
Bhaiya did it with mentos method and solved within few seconds.
Literally ek dum se click kara and in one go ans bhi aa gya. 😊
good going brother
Thank you bhaiya btw I m girl😃
@@siyapunia-eq9mx sysstummmmmm
What is mentos method?
min(i,j) + max(i,j) = i + j only so we can replace [sum(i=1->10 )sum(j=1->10)(i+j) ] after that we get 1100 answer
I did via the 1st approach only but wouldn't take more than 2 mins because very comfortable with the calcs and series and its kinda very similar to nested loops problems.
I did with second and first method both, second method slightly different not adding straight in beginning but with seeing both series got that add them and get the answer without writing the full series which saved time
Method 2 simple as
Ya to A ayega ya B aayega so end me it sums up
XACTLY!
All you have to do is calculate the sum of pairs above the diagonal.
Which is S = 2*(1*9 +2*8 + 3*7 +4*6) + 5*5 = 165
A becomes 2*S + 55. 2 times the pairs + the diagonal sum.
B becomes 10 * Σ(1:10) + S. (Subtract 1 - 10 sum from every row and you'll see S is left. )
Hence 3*165 + 11*55
Max 2 min lage ye solve karne ko.
Bhai maine under 1 min solve kr liya jaise btaya apne waise hi❤🎉
direct dikh ra tha ki puri matrix ke saare no. ka sum pucha hai islye 20 x 55 direct bol skte hai as ek A set aur B SET ek dusre ke conjugate hi puche hai
Looking back after 2 years, it's hilarious to think how challenging I once found this. Now I can solve it in under a minute with a quick script! 😂😂😂
int main() {
int sum=0;
for(int i=1;i
arre but ye toh computer calculate kar raha hai na fir toh lol ... ye easy kaise hua fir
@@sandy69402 Spend 1 year in engineering you'll get it
Tumhari time complexity sahi nhi hai agar ye question me 10 ko hatake n krdia jaye toh tumhara solution o(n^2) hai jabki actual solution o(1) hai
8th java did help
suppose i take a value that is equal to 10*(1+2+...10).
in A I am subtracting something from it and in B I am adding the same thing in it so when you add it they get cancelled and answer is 2*10*(1+2...10).
no need to calculate anything if you can see this
Before vid:
If you do A+B it will be sigma of sigma i+j which i gives 1100
Same, also my intuition. Btw are you also a programmer, just asking
@@_elusivex_ yep
I read it as A-B and was surprised cuz i was getting answer as -330
ahahaha
Mujhe pehle mentos wala click kiya, but fir sure nhi tha toh 1st method se kiya -_-
Took me 10 minutes , but it was very intresting !
jab i aur j me lerahe h toh if min(i,j) = i then max(i,j) = j, i=j ho ya na ho. So adding both min and max meaning taking both i and j. So very easy to solve.. Ye mere mind me 2 min ke baad strike hua.
mentos method se toh kiya but idea aane mein 10 min lag gaye... total 12 min kuch lag gaye.. but maza toh aa gaya
good work bro 😁
bhai mera 1 min ho gya agar meri approach se chle toh
assume i>j fir min=j and max=i ke hiasab se solve ho jayega ,bcz sykmetry hai to j>i ke liye bhi same ans ayega
did easily in 3 minutes before watchng video by matrix method same as binomial question
Mate Can you explain how please
Two equations for finding value of 'x' and 'y'
1) ax + by = C
2) px + qy =C
(If "C" is the Real Number as- 0,1,2,3......).
Can we write or say.
"ax+by = px+qy = C"
("ax+by = px+qy") ( Because both are equal to "C").
Is it true or not. please help with example 🙏
bruh
what are you?
9th grade dumbass?
Wait how is this hard...?
max + min = i+j, now it's simple enough
maine general term form krke kiya for both a and b
❤Maza aa gaya
Ashish Aggarwal sir effect ❤
fantastic!
Nice question but done in 2 minutes using i + j method i think 75% logo ko click kar jata😊😊
M a neet aspirant and i used a super lengthy method to solve this
Min wala to jaldi aa gaya, max wale me lamba chala gaya
But for me, a = 440, b = 660, a+b = 1100
First lets solve for a,
We get a series 1*10+ 1+2*9 + 1+2+3*8....
Notice, we have 2 series in here, writing them as x and y, we have
A= x+y,
Where x = 10*1+2*9+3*8...
And y= 1+1+2+1+2+3....
But if u would notice, x = y
So, a= 2x, or a = 2(1*10+2*9....)
Here the series x repeat itself after 5*6... So we dont have to calculate every value, but rather,
X = 2(10+18+24+28+30),
Thats 110, so x = 220 and a = 440
A = 440
Now solving for b,
For j=1, we have the summation to be 55
For j = 2 we summation to be 55-1+2 = 56
For j = 3 we have 55-3+6, thats 58
Or more precisely, if j = x, (x cannot be greater than 10)
The summation= 55 +x(x-1) - x(x-1)/2
Or 55+x(x-1)/2
Now we have the series,
10*55 + 1(1-1)/2 + 2(1)/2+ 3(2)/2 + 4(3)/2 + 5(4)/2 + 6(5)/2 + 7(6)/2 + 8(7)/2 + 9(8)/2 + 10(9)/2
550 + 1+3+6+10+15+21+28+36+45
Theres a pattern in the series,
1+(1+2)+(1+2+3)+(1+2+3+4)...
10*1 + 2*9 + 3*8 same as x ! we solved it in a and the answer was 110,
This means, b = 550+110
B= 660
Now a+b = 1100
Ik its very lengthy but i enjoyed it (took me 10 mins, ik m noob)
You're smart. Your pattern-recognition is amazing (especially the 1*10 + 2*9 +3*8 + ... = 1 + (1+2) + (1+2+3) + ...)
Brilliant.
@@jaisabraham6610 huh smart?
I always thought I was as dumb as a rock
@@priyank5161 fr
(i,j) me se ya to ek min me choose karenge ya to max me so sidha hi 55 ×20
Jee mains 2022 26 June shift 1 🎉
Easy question
Pretty ezy stuff for someone who is targeting advanced.. solved it in about 25-30 sec
Isse ghatiya sets ka question mene solve nhi kia phle lga tha double sigma h BT ka fir sets dekha to smjh aya fir jb row aur column wise khol k question smjh rha tha to pta chla isme time barbaad hone wala h alg hi ghumaya isne ksm se 1x19+2x17+.... Aese kr k gya tha ye😭
Hm bhai right
Double Sigma Bhi Last mein Easily Solve Hota Hai By Using Property of Sigma on Both I and j it becomes sum of 22 of 100 terms by 2
@@MiteshJethawa wo to bt me na lekin yha to (1,1) (1,2) (1, 3).... Kr k sets open krna tha
Iska seedha approach tha- woh table banane se better yeh hai ki I and u interchangeable hai toh ek pehle kisi ek ko fix karke dusre ke values nikaal do phir double it- par u have to add the elements jaha pe they're same- kafi aasan hai I did it mentally
Ghatia set nhi ghatia tu hai 😂😂
1100 solved in 1 min....
Bro which chapter is this question from??
Actually I understood 60% and that limit part smjh nhi aya mai abhi 11 mei hu
Concept crunch ka kya hua
bro ma kasam orally solve krliya💀
ab IIT Bombay CSE guaranteed🤡
ab IIT Bombay CSE guaranteed🤡
nice
Crazy
Where is terror ?
Pj sir bhi normal jindagi aur mentos jindagi se solve karwaya karte the
lengthy a bit not hard
Baki kya haal?
Sone jaa rha tha fir notification aaya aapke question ka. Neend udd gyi khud se solve krne mein but ban gya ;-;
Mai method 1 se kr rha tha, so thanks for alternate method :)
Bhaiya me rho dunga
me to neet de raha hu, maine ye kyu dekha
Python se kiya
use: max{i,j} = (1/2) ( | i + j | + | i - j | ) and min{i,j} = (1/2)(| i + j | - | i - j| )
Then A + B will be sum \Sigma | i + j | = \Sigma i + \Sigma j = 2 * \Sigma i = 1100
please use less words
kam shabdo me bataya kariye...
🔵Theorem: Let n be a positive integer. Then, ( Sum_{i,j = 1 to n} min { i , j } ) + ( Sum_{i,j = 1 to n} max { i , j } ) = n^2 (n+1).
Proof.
Sum_{i,j = 1 to n} min { i , j } + Sum_{i,j = 1 to n} max { i , j }
= Sum_{i,j = 1 to n} ( min { i , j } + max { i , j } )
= Sum_{i,j = 1 to n} ( i + j )
>
= Sum_{i,j = 1 to n} i + Sum_{i,j = 1 to n} j
= 2 Sum_{i,j = 1 to n} i
>
= 2 Sum_{i = 1 to n} Sum_{j = 1 to n} i
= 2 ( Sum_{i = 1 to n} i ) ( Sum_{j = 1 to n} 1)
= 2 ( n(n+1)/2 ) ( n )
= n^2 (n+1). ■
_________________________________________________________________________________________
🔴Corollary: For n=10, the above theorem implies that the given sum = 100 (11) = 1100. ■
Brah 😭🙏 it is not that serious
@@urlikskull1071 Brah, it was just for fun. 😆
But if you see beyond the sophisticated appearance of these symbols, you will see that I have proved a general result with simple arguments that will allow one to solve the question in the video with ease.
Discuss some 2024 adv questions
Again waste of time in JEE students should start practicing coding from 11 12 which is more important for professional careers
I'm first - pin me plz ❤❤😅