Tricky Question of JEE Main 2023 | 98% Failed to solve this #shorts #jee #jee2023 #jeemains

[ashishchalcalanikivines]

maturity is when u realise this is the most easiest question ever sir just wanted to boost our confidence

[markzcodm]

2023 ko 2023 bar multiply kardo kr 35 se devide 4-5 din mai aiga

[jaishreebalaji]

Reminder- Only jee advanced level prepared students can understand this method clearly 😊.
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.

[aparnakan]

7 . Just find the patterns and apply cyclicity

[shobhit]

Jee advanced mathematics left the chat

[RRtBoss]

IOQM students : Chinese remainder theorem yesss!

[Ayush]

Pehli baat toh peche ye gaana kis logic se dala hai 😅

[arpitraj]

Better to use CRT( Chinese remainder theorem) method 30 sec mai ans aajayega

Ans is 7

[notyourtypefanpage]

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

[crckterblue]

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.

[rachnashrivastava]

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

[Filmyclips]

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 )

[SounitBose]

Writing 2023 as 35k -7 it only takes 3 lines to solve using basic binomial

[mehulagrawal]

Sir maine abhi Kara ye ...thoda lengthy gaya but ho gaya ..7 is the remainder

[ayankumarsharma]

Most easiest question answer is 7
Break it into (2030-7) and 2030 is divisible by 35

[udayshankar]

7 is the answer ioqm students laughing in the corner, a bit of euler and chinese remanider theorem meanwhile congruence modulo left the chat

[Flamyflake]

Take it common then eradicate, try fermats theoram for remainder and then multiply it by the common which was taken

[muralikaply]

(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

[sahilraj]

These question can be solved in 10 to 15 sec by Euler totient function or congruence modulo easily

[kabirsethi]

Bro number theory ka basic saval hai. Congruent modulo lagao