Reminder- Only jee advanced level prepared students can understand this method clearly 😊. method------ 2023 ^(2023) hai- 2023 ×2023= 4092529 means agr ham 2023 ko 4092529 times add krnege to vhi ans aayga jo 2023^(2023). 2023/35 reminder will be 28 and we have to add 2023/35, 4092529 times. Then reminder is 28× 4092529. Now when u will devide 28×4092529 by 35 we will get reminder 7. Now this is the ans.
Congruent modulo agr tmko aata h to bn jayega warna sikhna prega tmko youtube se But better h JEE mai mt use kro yeh method headache hai yeh.... Yek hint de dete hai 0 mod 7 2 mod 5 Yeh dono= (2023)^2023 aayega dono equation ko binomial se solve krke aajyaega ans
bro IOQM is olympiad specially of maths that's why their questions and syllabus is vast as a engineering exam they do not required such level of questions to be solved by students using different concepts , as we have to do other subjects also not only maths .
This is very easy question I just solve this question within 1 minute, just use cyclyity, cyclyity of 3 is 4 , 4 divided by 2023 Remainder is 3 Put the power of 3:- 3^3 = 27
Simple calculation is firstly 2023/7×5 the remainder is 0 when divided by 7 and when divided by 5 it is 3 and then 3²⁰²³ is very difficult to find so apply cyclicity the unit at the power is 3 so the ans is 3³ ka unit that is 7
Answer is 7. Solved in 10 secs by class X student using simple trick. See 2023 ²⁰²³ can be written as 2023 ^ (4n - 1) where n is a natural number. 2023 ends in 3 and when a number which ends in 3 is raised to a number of the format (4n - 1) where n is a natural number, the number formed always ends in 7. And 2023 is also a multiple of 7. A multiple of 35 always ends in 5 or 0. Since the given number 2023 ²⁰²³ is divisible by 7 and odd, the remainder is 7.
Sol:3^n cycle : 3,9,7,1 so it repeates after interval of 4 so divide 2023 by 4 we get remainder 3 so ans is 3rd repeater of cycle i.e 7 (Rmo students may relate )
@minato7kushina would not need to divide it by 35 as you don't know with which multiple of 35 the no. Would be closest to( like the unit digits might be 77 so 35*2=70 would be the closest so we get remainder 7)
@@demongaming1219 yes definitely it is a topic of bsc . It makes a lot lot lot easier. Just referv any olympiadp course (better if for prmo) on number theory or related and you will learn
For those who want to know the answer, the remainder will be 7. I solved it using Remainder Theorem and Euler Totient Function, but sir solved this using some crazy complex binomial theorem method which I don't even understand(I'm still in grade 10)
Ye kuch tricky nahi he 2023 ko 2030-7 likhe phir dekhe qn redduce hojayega 7^2023/35 matlab 7^2022/5 fir toh simple 49^1011/5 hogaya aapko aage aata hi he
question isnt hard 35 is 5x7 one 7 cancels out 2023 and write that number at top and find the last digit of 2023^2022 and multiply with the other number so ull get the last diggit of the numerator as some x divite that by 5
2023 ko 2023 bar multiply kardo kr 35 se devide 4-5 din mai aiga
abey 😂
That is correct 😂😁
"Your answer will be delivered in 4-5 business days"
But if you know vedic maths then it is very easy
I'm on my way for answer but 3 sal to ho gaye hai and sare page bhi khtm ho gye
Students in comments: Arey ye to Falana theorem se ek minute me aaega😂 kaddu question....
Students in exam:🤡
It was literally me last time🤡
Most underrated comment
@@harsh38kaisa gya
😢
Me to maths all time: 🤡
Jee advanced mathematics left the chat
This is toughest by perspective of maine
Adv. Is a different thing. Doing this question with 2.5 or 3 min isn't possible
I think it entered the chat🗿
@@dmop5071bhai binomial applications ka bht easy concept hai
@@dmop5071lol ye sabse saste sawal me se ek h
maturity is when u realise this is the most easiest question ever sir just wanted to boost our confidence
Nv sir : I don't even exist🗿
NV sir can do in simplest way bro???
@@Boards-gv1lhYes
Yah ...search remainder theorem by NV sir
Pehli baat toh peche ye gaana kis logic se dala hai 😅
😂😂😂no logic bro
Sahi baat hai lag raha hai ki sir IIT ki tayari kar rahe ❤❤
@@raamgautampadhai ka content hain toh padhai tak rakho.... Background music can interrupt the audibility.
NV sir laughing in a corner 🗿
Nv sir kaya
Nv style
Vo physics ke teacher h na?
@@JaiBajrangBali1087Nishant Vora sir
@@drdoom.x9exactly harsh sir alag type ke he ❤ and sabhi teachers best he
7 . Just find the patterns and apply cyclicity
Reminder- Only jee advanced level prepared students can understand this method clearly 😊.
method------ 2023 ^(2023) hai- 2023 ×2023= 4092529 means agr ham 2023 ko 4092529 times add krnege to vhi ans aayga jo 2023^(2023). 2023/35 reminder will be 28 and we have to add 2023/35, 4092529 times. Then reminder is 28× 4092529. Now when u will devide 28×4092529 by 35 we will get reminder 7. Now this is the ans.
IOQM students : Chinese remainder theorem yesss!
Bro can u pls explain this theorem😢 how to do
@@deepikaenuganti3431 lectures are there... Congruence modulo... First learn it...
Yes bro, it is one liner by chinese
true
Wohi Chinese , Euler totient , fermat 😂
These question can be solved in 10 to 15 sec by Euler totient function or congruence modulo easily
Answer is 7
Writing 2023 as 35k -7 it only takes 3 lines to solve using basic binomial
IOQM LEFT THE CHAT💀❤️🔥
Take it common then eradicate, try fermats theoram for remainder and then multiply it by the common which was taken
Better to use CRT( Chinese remainder theorem) method 30 sec mai ans aajayega
Ans is 7
Kaisee?
Congruent modulo agr tmko aata h to bn jayega warna sikhna prega tmko youtube se
But better h JEE mai mt use kro yeh method headache hai yeh....
Yek hint de dete hai
0 mod 7
2 mod 5
Yeh dono= (2023)^2023 aayega dono equation ko binomial se solve krke aajyaega ans
Kis chapter se related hai ye question
@@Bonjour24-rksp&c
@@Bonjour24-rksimo prep mein hota hai
7 is the answer ioqm students laughing in the corner, a bit of euler and chinese remanider theorem meanwhile congruence modulo left the chat
bro IOQM is olympiad specially of maths that's why their questions and syllabus is vast as a engineering exam they do not required such level of questions to be solved by students using different concepts , as we have to do other subjects also not only maths .
@minato7kushina remainder is 7
@minato7kushina ok bro . But i use binomial theorem
IOQM prep is not a fulltime thing and can be done before 11th
Bro chinese se hojaya euler ki jarurat hi nahi
25 jan 2nd shift mera hi tha
Chinese remainder theorem left the chat
This is very easy question I just solve this question within 1 minute, just use cyclyity, cyclyity of 3 is 4 ,
4 divided by 2023
Remainder is 3
Put the power of 3:- 3^3 = 27
you have to divide it by 35 not 10
Ans is 7
25 Jan 2nd shift my shift question
U did?
Most easiest question answer is 7
Break it into (2030-7) and 2030 is divisible by 35
Explain it
It doesn't work like this it will be still (2030-7)^2023 we use binomial to solve
🤡
I used (2020+3)²⁰²³, but it takes me more time,so I think (2030-7)²⁰²³is perfect
The answer will be 28 since it is -7
Simple calculation is firstly 2023/7×5 the remainder is 0 when divided by 7 and when divided by 5 it is 3 and then 3²⁰²³ is very difficult to find so apply cyclicity the unit at the power is 3 so the ans is 3³ ka unit that is 7
Sir maine abhi Kara ye ...thoda lengthy gaya but ho gaya ..7 is the remainder
@@themathematician138 7 is correct
@@cochelper1736 no bro it is 3
@@shalvagang951 bro sir ne kudh solve kiya 7 hoga
Use binomial theorem here
*Nv Sir :- Hold up!*
Negative remainder concept of nv sir 🎉😊
@@abhishekpawar2526yes
Can Anyone Share The Link Please?
Nv sir ka pura name kya he
Nishant vora @@AbcdEfgh-pu6nq
when you know euler totient fuction this is a question of a minute only . aur modular maths se bhi ho jayega 2 min mai
this is a walk in the part for olympiad people
Answer is 7. Solved in 10 secs by class X student using simple trick. See 2023 ²⁰²³ can be written as 2023 ^ (4n - 1) where n is a natural number. 2023 ends in 3 and when a number which ends in 3 is raised to a number of the format (4n - 1) where n is a natural number, the number formed always ends in 7. And 2023 is also a multiple of 7. A multiple of 35 always ends in 5 or 0. Since the given number 2023 ²⁰²³ is divisible by 7 and odd, the remainder is 7.
(4n-1) hi kyu?
Kaha se padh raha hai vro
Mai vi 10 me hu. Mujhe 13 sec lage 🙁
@@ButterKhakeFly mujhe subject bhi batado kyoki Mera answer apple aa raha hai 😰
True
@agni404 You can also write (4n+3) because after every 4 consecutive powers, the unit digits repeats(3,9,7,1,3,9,7,1)like this.
Topic?
Euler totient function se hi jaega ye question
Class 10th ans is 7 not difficult chp 1 real no concept hai
Bro kaise ?
Sol:3^n cycle : 3,9,7,1 so it repeates after interval of 4 so divide 2023 by 4 we get remainder 3 so ans is 3rd repeater of cycle i.e 7 (Rmo students may relate )
@minato7kushina would not need to divide it by 35 as you don't know with which multiple of 35 the no. Would be closest to( like the unit digits might be 77 so 35*2=70 would be the closest so we get remainder 7)
Congurence modulo ❤
guys i think any 11th or 12th class student can also solve this easy question this is not that hard
use congruent modulo
usse nhi ho ga remainder 1 ya -1 nhi mil rha
@@Aaravsrivastava117 -1 aa rha hai phir usme 7 se multiply bhi karenge kyuki suru mai usse divide kiye thae toh -7 aaya means 35-7=28
Ans is 7
Done using congruent modulo
@@niveditabhaskar8193 please explain me the process
ISI ASPIRANTS BE LIKE: YEH TOO CLASS 8 MEI KRTE THE
Bro number theory ka basic saval hai. Congruent modulo lagao
bro kia h ye ??
Mains ke question karne k liye ye sab padhna parta hai kya aajkal😭
@@user-iv2dh8vi4znumber theory ka topic... irrelevant for jee but very good tool in general
It is very is problem sir, if we use number Congruencies answer could be 7
Because
2030=7(mod 35)
Can use the fermats little theorem
Number theory se congruency use karna chahiye
Meanwhile JEE ADV:💀💀
If a similar question comes for 2024 how should we solve that?
Oops NV sir tricks are going too viral...💀💀💀
Us IOQM students laughing in the corner:😂😂
Meanwhile nvians🗿💪
youtube நல்லா சொல்லி தரானுங்க நேர்ல வேஸ்ட் ஆப் டைம்
I solved it in 1 min by using Sachin sir's method
konsa chapter se
Modular congruence left the chat!
I literally thought it
@@adityade897 what's that ? Can you tell about it and does that make the question easier?
@@demongaming1219 yes definitely it is a topic of bsc . It makes a lot lot lot easier. Just referv any olympiadp course (better if for prmo) on number theory or related and you will learn
😂
Its the basis of number theory
Note: All in this comment section are IITians , isiliye gyan pela jaa rha h khud usne!
Easy( in 9th class )
Sol- divide 2023(power not base) by 4 then remainder is 1 then 2023¹=2023 then 2023 divided by 35 remainder will be 7
Yes
Nishant vora sir se jisne pada he wo ise aaram se kar dega within 2 minutes
Bro plz help me ye nishant vrora se kaise pdhun mai plz btaao
@@discordff2126vora classes se
@@discordff2126 mobile ya laptop se
Tq sir 😊
NV sir be like jalwa hai hamara 🔥🔥
Why..
29?
IPMAT indore quant aspirants laughing in corner.
Solve it by tricks in a minute.😅
Congruent mosulus and bionomial theorem....
@@Arnav1883 ioqm
Ultralegends will skip this type of questions 😂😂😂
Are ye easy h
7 is the answer
This can be easily done using Binomial Theoram
Even calculator unable to answer 😂
Meanwhile Euler's Totient Modulo : The Clock is so slow huhhh.
Not applicable as a and n are not coprime 😂
2023 and 35 share 7 as a common divisor
@@futurenewton5303 I mean Solve for 2022 & then solve for 2023 ( short trick that can be a better alternative to Chinese remainder theorem )
Ans 4
Ans is 7 Just solved in 2 minutes without looking at solution.
# RMO students
Link toh open hii nhi ho rha sir...please help sirjii
My shift question😢😢
Ye questions bhut zyda standard h aese swal toh olympiad ke papers mein milhi jate h
(27)^2023 answer
Congruence modulo and CRT se agaya 7 hai answer
Sir i did this question in 9 th class
7 remainder
Link to do
Description mai koi link nahi hai😂😂😂
It was pretty simple
Ans:-7.......
(35x58 - 7)^2023
35k - 7^2023
By dividing 7 and 5 with 7^2023
We get 7 as remainder
Ans by its_me_the_kmk
👌
Ye to mai 10th mai ntse ke liye padh rha tha 😂😂
ICSE me 9th me tha😂😂
Sequence and series me
@@newboygram8654bro ntse mai dono 9th and 10th ka syllabus aata hai maine 9th mai padha nhi tha
Answer is 7
Chinese remainder theorem
25 jan 2 shift
Very easy using number theory trick
My shift questions 😢
Background music hata do
For those who want to know the answer, the remainder will be 7. I solved it using Remainder Theorem and Euler Totient Function, but sir solved this using some crazy complex binomial theorem method which I don't even understand(I'm still in grade 10)
IKR I solved it using Fermet's little theorem and then a simple observation
@@AniketKumar-lw6su Are you appearing for IOQM-23 by any chance?
@@mathskafunda4383 yes
Icse board?
Bro totient function apply kaise kara cause 2023 and 7 are not coprime
7 answer ayga 🙏
Hmm
Use Chinese remainder theorem
ioqm students
7 is the answer, tooo easy using modular arithmetic
Ye kuch tricky nahi he 2023 ko 2030-7 likhe phir dekhe qn redduce hojayega 7^2023/35 matlab 7^2022/5 fir toh simple 49^1011/5 hogaya aapko aage aata hi he
NV sir : hold my 🥤 coffee
Le Olympiad students doing it in 5 second
Neet aspirant but still solved it under 1min 👍
Why you are alive bro 😊
Recently nda also asked this type of question
Congruence modulo
Meri shift me aaya tha ye
25 jan shift 2
question isnt hard
35 is 5x7
one 7 cancels out 2023 and write that number at top and find the last digit of 2023^2022 and multiply with the other number so ull get the last diggit of the numerator as some x divite that by 5
Can somebody sent me a video lonk
NV sir 💜😌
he is physics teacher....
@@shriyasachdeva10242 nv sir one phy other maths
Calculator to infinity bta rha h😂😂
Sir ioqm classes in telugu
Maine calculator se try kiya but usne bhi mana kar diya 😅😂
Trickery but soo easy
ioqm students : Remainder theorem
launching in corner
ye toh bacho wala sawal tha
after 1yr when in 11th : aise hi ques kyu nahi puchte ye log ...