Hogya bhaiya Critical step toh bs a,b,c wali identity identity krni thi baaki toh easy tha Kyoki pehli equation me hi lhs me cube horha aur rhs me digits ka sum (wo sn³ bhi ghnta farq nhi daal paaega jb 3 digit number aaega) Aur 1,9 toh pehle hi discard krliya tha mann me , kyoki again uss case me n³=sn³ wagera hojata aur no natural sol
Maine 2nd part of the question pehle hi equation me lagaya, calculation laga kafi but prove hogya ki 3 ya more digits satisfy nahi karega and for 1 digit s(n)=n so 2 digit me fir divide kiya like 1-50 and 51-100 so answer aa gaya
for a two digit number (N), if 1 is in tens place, (Sn)= N-9 if 2 is in tens place,(Sn)=N-18 and so on now,N=2Sn +1 -> N=2(N-9K)+1 ->18K=N+1 here we can deduce that we may only have 1 s tens digit ->18=N+1 ->N=17
It took me 1 hr 40 mins to solve this question At first I was trying things with parity and proved n is odd I thought about primes being expressed as 6q +1 And I saw that n must be odd But when I realized that the expression is written similar to( y - 2S )^3 then I wrote its expansion and put it into the original expression then through some algebra manipulation I saw ( n - 2s - 1 )( k)=0 where k was an expression easily proven greater than 0 after that the question was a cake walk I took n as an expression of a1 a 2 a3 multiplied to their respective 10^x powers and proved that only 2 digits numbers will work after that I just 10a + b = n and S= a+b and the question was easily solved
I solved it within 5 minutes. Honesty, not a really good problem, bcoz both the steps are so intuitive , 1st step was really easy as we often deal with conditional identity, second step was also thinkable as after taking cases of 1,2,3 digit number and finding maximum and minimum like analysing functions and making is the basic for jee advanced questions. So not very happy about the solution.
Being an olympiad aspirant a general method to solve these questions based on n and S(n) is to place limits on the digits of n and the use some modular arithmetic to place special conditions. This reduces the sample space to some 15-30 numbers and then you can use hit and trial. Since these questions are common in rmo and you have 30 minutes to solve each question given that you attempt all the questions which only 5-6 people do you have plenty of time to do hit and trial on 15 to 30 numbers.
i solved it as follows: i observed the terms following a^3+b^3+c^3=3abc hence s(n)=(n-1)/2 now for 2 digits max value of S(n) is 18 and n for that would be 37 but for numbers till 37 max S(n)= 11 ( for 29) ( very easy to observe pattern ) since s(n) is linear function for s(n)=11 we need n=23 till 23 max value is 10 at 19 so for for 19 s(n) from n-1 /2 =9 ( hits that we r close since 9 is close to 10 and since we obviously know the answer is going to be a odd number thats it 17 is the answer hasta la vista!
I'm pretty sure a similar question was asked in last year's Fall Berkeley Tournament held in U.C Berkeley, I do remember my approach but this question is easier as it reduces to a linear My solution: We have n=2S(n)+1 and if we could solve it using a recursion kind of formula, lets say n is a 3 digit or higher number, then S(n) will max to max be 27( because of 999) and therefore n is max to max 2*27+1=55 , now if n cannot be greater than 55 then S(n) cannot be greater than 13 ( because of 49) , now if S(n) cannot be greater than 13 then n cannot be greater than 27(2*13+1) , so now S(n) is again max 10 ( cuz 19) , so n is max to max 21 and S(n) is still maximum at 19 so we have finally reached the saturation point and we can see that 17 is now the answer Edit: I know we can create a formula of this recursion to directly get the saturation point but doing it manually is more convenient and easier to explain
are u in college because ive never heard of this approach in my life (i did solve it using mod theory and basic reductions as ive done these kind of problems during my rmo prep)
I am in love with your voice #nohomo the way you simplify things and explain is just like a friend is teaching who is most close to heart you are so underrated and intelligent. KEEP IT UP BHAIYA!!! and keep bringing all those brain boosters for us.
Assume n is a two digit number with digits a(not equal to zero and can takr values from 1 to 9) and b (takes values from 1 to 9). So, n = 10a + b and S(n) = a + b. Put these in n = 2S(n) + 1. You will find that 17 is the only possibility. (8a = b +1, put a=1, you will get b=7. a=2 implies b=15, which is not possible, so only 17 is the solution. Similarly, no 3 digit or beyond number exists)
Sach me maza aa gaya , identity nahi dikhi isliye maine 1 digit , 2 digit .. no ke cases rakhe lekin voh mess mein badal gaya toh last me soln dekh liya
Nice question my approach was n³-1 = 2Sn ( 4(sn)²+3) (n-1)(n²+n+1) = 2sn ( 4(sn)²+3) Niw on comparing Kyuki rhs even no hain toh lhs bhi ek even no hoga so either n-1 or n²+n+1 or both must be even but n²+n+1 can never be a even no becoz for n = even (even)² + even +1 = odd And similar for n= odd Now n-1 must be even than This implies n must be a odd number Now again coming to eqn we can similarly prove that 4(sn)² +3n is always odd (since n is odd ) Than this shows either n-1 = 2 This implies n=3 not satisfying ( comparing even part) Or 2sn = n-1 and than rest your method
Bhai maths se pyaar karne waale aise chapter ko jo 15 din baithke ache se padhna chahiye aur uska Mazza Lena chahiye woh usse 7-8 ghante me nahi padhayega
bhai prepration and dedication se koi bhi uss stage tak pauh sakta hai mai khud daily ke 1-2 imo aur usamo ke ques krta geo ke aur alegbra and nt rmo ke aur mai 9th se prep kr rha hu . im not mathematicallt gifted but training se kuch bhi ho sakta hai
I was solving parabola, some pyqs mainly and was about to sleep, your notification came, I was like, छोड़ो नींद और मज़े लो सवाल का😂😂😂. Well now I am going to sleep 😴
Bhaiya, Maine 8S³ ko shift kiya aur factorize kar diya Toh (n-2S)(n+S)² = 6nS + 1 Ab 6nS +1 is either a prime or prime ka square Aur uske factors is (n-2S)(n+S)² Ab agar ye possible hona hai toh factors mai se koi ek 1 hona chahiye Toh n-2s = 1 Toh n = 2s+1 Aur thoda analysis karke pata chala ki S is even Fir jaise aapne discard waise Maine bhi kardiya Aur n= 17 milgaya
One of the best questions of these types in which logical thinking is required rather than more simplification or solving more and more (I wasn't able to solve).... Just pure Mathematics... Loved it ❤
Just took a hint by seeing the rearrangement of terms and then solved it in about 5 mins, randomly did calculation for three digit instead of observation
I am a Olympiad student, ye problem me pehle toh mujhe n³ = 2s(n)³+6n.s(n)+1 dikha toh maine socha ki a³+b³ bna leta hu but then I realised ki 6n.s(n) ke sath a+b nhi hai phir aur thodi analysis ke baad I cracked it and got the ans 17
Bhai ek identity hoti hai a³+b³+c³-3abc=1/2(a+b+c)((a-b)²+(b-c)²+(c-a)²) Toh yaha pe agar a³+b³+c³-3abc ko zero hona hai toh do hi possibility hai ya toh a+b+c=0 Ya phir a=b=c ho taaki woh perfect squares wala factor zero hojaye
Choti moti errors ko ignore karna
like, solution mai vo -2S(n) hoga
👍
Hogya bhaiya
Critical step toh bs a,b,c wali identity identity krni thi baaki toh easy tha
Kyoki pehli equation me hi lhs me cube horha aur rhs me digits ka sum (wo sn³ bhi ghnta farq nhi daal paaega jb 3 digit number aaega)
Aur 1,9 toh pehle hi discard krliya tha mann me , kyoki again uss case me n³=sn³ wagera hojata aur no natural sol
Maine 2nd part of the question pehle hi equation me lagaya, calculation laga kafi but prove hogya ki 3 ya more digits satisfy nahi karega and for 1 digit s(n)=n so 2 digit me fir divide kiya like 1-50 and 51-100 so answer aa gaya
Bhaiya bahut khubsurat question tha love it ❤❤
Bhai 'Pyaar' blikul shai word hai
MAza aa gaya
Choti moti erros ko ignore karne,
like vo 2S(n) negative ka hoga soln mai
Congruent to mod 3 lelo overr
Easiest question of straight lines ever
Solved it in 3 and half minutes.
n=2S(n)+1 ke aage ka nhi hua tha
Aur yha tk ke liye jyada time nhi lga tha
lovely
for a two digit number (N),
if 1 is in tens place, (Sn)= N-9
if 2 is in tens place,(Sn)=N-18 and so on
now,N=2Sn +1
-> N=2(N-9K)+1
->18K=N+1
here we can deduce that we may only have 1 s tens digit
->18=N+1
->N=17
This channel is the best gold mine I have ever found.
Obv
It took me 1 hr 40 mins to solve this question
At first I was trying things with parity and proved n is odd
I thought about primes being expressed as 6q +1
And I saw that n must be odd
But when I realized that the expression is written similar to( y - 2S )^3 then I wrote its expansion and put it into the original expression then through some algebra manipulation I saw ( n - 2s - 1 )( k)=0 where k was an expression easily proven greater than 0 after that the question was a cake walk I took n as an expression of a1 a 2 a3 multiplied to their respective 10^x powers and proved that only 2 digits numbers will work after that I just 10a + b = n and S= a+b and the question was easily solved
Clickbait karne ka tareeka thoda kezual hai
Sexual tha
😂
Muje tho pehle se hi pta tha
Love this series please extend this series of advanced problems
S(n)= summation =k=0 to (n-1) (-1)^k cos((k pi)/n).Anything on this? (General term)
I solved it within 5 minutes. Honesty, not a really good problem, bcoz both the steps are so intuitive , 1st step was really easy as we often deal with conditional identity, second step was also thinkable as after taking cases of 1,2,3 digit number and finding maximum and minimum like analysing functions and making is the basic for jee advanced questions. So not very happy about the solution.
Bro konsa institute sa ho??
Late night with maths
This is what called love ❤
Beautiful ❤️
Couldnt solve it as I'm not used to these kinda probs,
But soln dekh ke maza aa gaya ❤✨
Being an olympiad aspirant a general method to solve these questions based on n and S(n) is to place limits on the digits of n and the use some modular arithmetic to place special conditions. This reduces the sample space to some 15-30 numbers and then you can use hit and trial. Since these questions are common in rmo and you have 30 minutes to solve each question given that you attempt all the questions which only 5-6 people do you have plenty of time to do hit and trial on 15 to 30 numbers.
Ya bro one could also use this s(n) is congruent to n mod 9
Which yields only one solution n=17
(Sinmilar to 2018 RMO
i solved it as follows: i observed the terms following a^3+b^3+c^3=3abc hence s(n)=(n-1)/2 now for 2 digits max value of S(n) is 18 and n for that would be 37 but for numbers till 37 max S(n)= 11 ( for 29) ( very easy to observe pattern ) since s(n) is linear function for s(n)=11 we need n=23 till 23 max value is 10 at 19 so for for 19 s(n) from n-1 /2 =9 ( hits that we r close since 9 is close to 10 and since we obviously know the answer is going to be a odd number thats it 17 is the answer
hasta la vista!
@@shalvagang951 btw i gave inmo in 2019 so ex olmpiyadicion current IITian (rusty tho) guess kitne time me socha hoga hai
@@RocketsNRoversMaine bhi abhi ioqm Diya tha 2023 mein 10 th grade mein (bohot hi late pata. Chala ) usme bhi 13 hi aaye kyki nervousness mein bohot silly mistake kar di
@@RocketsNRovers btw I have one question for you
|1/1/2022+1/2023+...+1/2064|
Ye kya hoga
N=17, sn =8
Maza aya
Bana kuch nhi isme....BUT
Solution dekh ke dil ko khushi si hui ekdm...maja sa agya😅
This is what I call simply complex.
N=17
10min
solved it correctly in approx 40 mins
If you enjoyed this prob.. definitely try out RMO P3 2023!
😮❤
yes i solved that problem in test
day same type of concept
this is a rmo pyq(rmo 2017 and hence 17 as an answer)
No
I am Olympiad student to ye kaafi routine problem sa laga.
Problem like his came in RMO 2023
was able to solve under 5 mins , interesting regardless . This is the type of questions i like do bring more
This is a RMO pyq..thats wt i call the beauty of olys
IT WAS BEAUTIfUL.
You are a genius brother
Bhai 10th mein hu tabhi solution samaj aa gaya
9th mein padhate hai ye wale identity 😂:
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ac)
Elon musk wants to know your location !
Us bro us 9th ka proff bhi yaad aagya
Bahut easy tha
1 min me ho gaya
I couldn't solve but it was amazing question
Literally bro said literally so many times.
I'm pretty sure a similar question was asked in last year's Fall Berkeley Tournament held in U.C Berkeley, I do remember my approach but this question is easier as it reduces to a linear
My solution: We have n=2S(n)+1 and if we could solve it using a recursion kind of formula, lets say n is a 3 digit or higher number, then S(n) will max to max be 27( because of 999) and therefore n is max to max 2*27+1=55 , now if n cannot be greater than 55 then S(n) cannot be greater than 13 ( because of 49) , now if S(n) cannot be greater than 13 then n cannot be greater than 27(2*13+1) , so now S(n) is again max 10 ( cuz 19) , so n is max to max 21 and S(n) is still maximum at 19 so we have finally reached the saturation point and we can see that 17 is now the answer
Edit: I know we can create a formula of this recursion to directly get the saturation point but doing it manually is more convenient and easier to explain
Which uni are you in?
Nice bro mene pda toh nhi pr meko pta h kitni mushkil hoti h ye type krne me 🤝
BlackPenRedPen ??
are u in college because ive never heard of this approach in my life (i did solve it using mod theory and basic reductions as ive done these kind of problems during my rmo prep)
Matlab khud se toh ghanta nahi hota leken har step samaj aya. Alag maza tha.
17❤❤❤❤
Maja aagya bhai
I am in love with your voice #nohomo the way you simplify things and explain is just like a friend is teaching who is most close to heart you are so underrated and intelligent. KEEP IT UP BHAIYA!!! and keep bringing all those brain boosters for us.
concept ke scroop kholdiya
i qualified ioqm and i was in some akaash south india rmo badge and they did this question, its easy for the olympiad people i would say
Assume n is a two digit number with digits a(not equal to zero and can takr values from 1 to 9) and b (takes values from 1 to 9). So, n = 10a + b and S(n) = a + b. Put these in n = 2S(n) + 1. You will find that 17 is the only possibility. (8a = b +1, put a=1, you will get b=7. a=2 implies b=15, which is not possible, so only 17 is the solution. Similarly, no 3 digit or beyond number exists)
I have solved thia question. Once our teacher posed this to us.... and few of us were able to solve it 😂
Sach me maza aa gaya , identity nahi dikhi isliye maine 1 digit , 2 digit .. no ke cases rakhe lekin voh mess mein badal gaya toh last me soln dekh liya
Nice question my approach was
n³-1 = 2Sn ( 4(sn)²+3)
(n-1)(n²+n+1) = 2sn ( 4(sn)²+3)
Niw on comparing
Kyuki rhs even no hain toh lhs bhi ek even no hoga so either n-1 or n²+n+1 or both must be even but n²+n+1 can never be a even no becoz for n = even
(even)² + even +1 = odd
And similar for n= odd
Now n-1 must be even than
This implies n must be a odd number
Now again coming to eqn we can similarly prove that 4(sn)² +3n is always odd (since n is odd )
Than this shows either n-1 = 2
This implies n=3 not satisfying ( comparing even part)
Or 2sn = n-1 and than rest your method
Yep same approach
No it's an easy for beginner non rutione maths
Took some time but hogya.. mast swal tha..😇🖤
pta nhi pen utha nhi paa rha
I think you have to make one short videos for jee maths chapters it really help students 🤫
Bhai maths se pyaar karne waale aise chapter ko jo 15 din baithke ache se padhna chahiye aur uska Mazza Lena chahiye woh usse 7-8 ghante me nahi padhayega
@@shivanshnigam4015 bro can do anything 🤫
A classic number theory problem, took some time as i get stuck in s(n) most times but cracked it😅
Same question ig like this ask in rmo and this year nmtc
Try problem from different Olympiad....you may try the non geometry part if you like...
Ho gya mujhse but because i preped for math Olympiad only numner theory for inmo this year 😂. Nice question
bhaiya hogya in 5 minutes mujhe bhi a^3+b^3+c^3=3abc dikh gaya tha aur fir toh kuch bacha hii nhi maine aise questions kaafi kare hai isliye hogya
isko to programming se bana dungaa
Have studied congruence modulo so I used that n is congruent to s(n) mod (9) but it was time taking😊 urs was faster thanks
You caught me off-guard 🤕
Please make an detailed video on Po Shen Lo method to solve Jee advance quadratic problem.
Looks beutiful and hard but is easy if you crack the first stage
From which college are you from bhaiya 😊
Wowwww😮😮😮😮
Ye konse chapter ka sawal tha 💀
Ekdum Chumashewari question
Abhi Paata chaala bhai kyu nahi so paye raat bhaar
Ha Bhai jara rkmo inmo ka pyq lago to tumhara satyatha sa parichay ho jayga ig not sSying about attitude but this is truth not take in that way bhayia
bhai prepration and dedication se koi bhi uss stage tak pauh sakta hai mai khud daily ke 1-2 imo aur usamo ke ques krta geo ke aur alegbra and nt rmo ke aur mai 9th se prep kr rha hu . im not mathematicallt gifted but training se kuch bhi ho sakta hai
"Proof merko nahin pata hai" ARE YOU KIDDING ME? chhih
I was solving parabola, some pyqs mainly and was about to sleep, your notification came, I was like, छोड़ो नींद और मज़े लो सवाल का😂😂😂. Well now I am going to sleep 😴
Bhaiya,
Maine 8S³ ko shift kiya aur factorize kar diya
Toh (n-2S)(n+S)² = 6nS + 1
Ab 6nS +1 is either a prime or prime ka square
Aur uske factors is
(n-2S)(n+S)²
Ab agar ye possible hona hai toh factors mai se koi ek 1 hona chahiye
Toh n-2s = 1
Toh n = 2s+1
Aur thoda analysis karke pata chala ki
S is even
Fir jaise aapne discard waise Maine bhi kardiya
Aur n= 17 milgaya
Factors galt nhi he?
bhaiya iden tity k baare me batao
Maza aaya Bhai
amazing qn
Main toh yaha thumbnail dekhkar aaya tha 🙂,socha ki kisi ke break up 💔 ki dukhat kahani sunne milegi 🥲🥲....
One of the best questions of these types in which logical thinking is required rather than more simplification or solving more and more (I wasn't able to solve).... Just pure Mathematics... Loved it ❤
Done 8 min calculation krne mein rechecking kr rha tha
sir qn mein pakka gadbad h
please recheck it
isme identity nahi ban rahi
Agreed
thanks bhaiya majeee agai
Yeh toh rmo ka questions hai, kiya hua hai halwa sawal
Solved
Bhaiya maths m 11th m kuch nhi padha 12th m achi krni h kuch btado?
Amazing question bro
Understood well 😃😄
Just took a hint by seeing the rearrangement of terms and then solved it in about 5 mins, randomly did calculation for three digit instead of observation
Bhaiya clickbait mat kiya karo long term me problem hogi baki questions badiya tha 😂
Niw thats something insane 🔥
RMO 2016 problem 3.
Thumbnail 😅
I am a Olympiad student, ye problem me pehle toh mujhe n³ = 2s(n)³+6n.s(n)+1 dikha toh maine socha ki a³+b³ bna leta hu but then I realised ki 6n.s(n) ke sath a+b nhi hai phir aur thodi analysis ke baad I cracked it and got the ans 17
Par vai tujhe har question try karne ka time kyon nahin milta
Amazed😮
sir ye maths olympiad kequestion laake jee adv kyun likh rahe ho exactly same question rmo batch mai karwaya tha sir ne
Ye to rmo ka sawal h
1
❤❤❤
Kya bhaiya IPL match dekh rhe the 😶😶
Ye aap roj olympiad ka que dete hain😂😂😂
So jao bhaiya 💓💓
Keep secret for some time
👀
Only one doubt why a+b+c = 0😢
Bhai ek identity hoti hai
a³+b³+c³-3abc=1/2(a+b+c)((a-b)²+(b-c)²+(c-a)²)
Toh yaha pe agar a³+b³+c³-3abc ko zero hona hai toh do hi possibility hai ya toh a+b+c=0
Ya phir a=b=c ho taaki woh perfect squares wala factor zero hojaye
@@shivanshnigam4015 thank you
Present❤
1 st comment brother😅😅😅
date aa gayi date
Bhaiya idhar meri halat kharab hai 4 april ko hai mera
Or aap yha 1 baje question kra rhe ho
sabko 4 april de diya hai kya.. idhar sab vhi likh rhe hai .. mera bhi same hai
@@thefireyphoenix history will repeat
Remember first day first show 27 Januray 🤯🤯