We hope that you learned how fundamental properties can occur at unexpected times, a mix of such unique illustrations must be practices by you in case aiming for top ranks. for that only we offer {SO60} to those who struggle with problem solving Here's the link for more details courses.jeesimplified.com/courses
Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17.. Is my method correct?
it can be proved using binomial theorem: we write a^n = (a - b + b)^n expanding using binomial thm: a^n = k^n + nC1.k^n-1.b + ...... + b^n (where k=(a-b)) so, a^n - b^n = k {k^n-1 + nC1.k^n-2.b + .....} as combination operator always result in integer the whole bracket on RHS is an integer (say p) so, a^n - b^n = (a - b) . p thus a^n - b^n is always divisible by (a - b)
I have some confusion here hope you could resolve. In last expansion of binomial there is only 2nC2nb²^n then how did you take k as common from it??????????
@@jeesimplified Bhaiya a^(2n) - b^(2n) ko we can rewrite as (a^n)² - (b^n)² fir x²-y²= (x-y)(x+y) property use krke (a^n-b^n)(a^n+b^n) likh skte hai and (a^n-b^n) is divisible by a-b and hence pura a^(2n) - b^(2n) is divisible by a-b
Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17.. Is my method correct?
3:01 proof By remainder theorem Let f(a,b)=a^2n-b^2n Put a=b so the function becomes zero which shows that a-b is a factor of this expression Put a=-b Again the function becomes zero so a+b is also a factor
Everybody is probably just copying from Google and using that modulo thingy/lengthy binomial expression But here is a person who used a basic method to prove this logically
@@dheerendrayadav8335 yes , see first forget about the big 190 power. Start with 1, the answer is 0. Then put 2 as power for 25, 19, 8 and 2. You'll get 204, which is divisible by 17, so it is also divisible by 34, you can check further by 3, but then you'll reach the same conclusion. The above I don't like to get into abstraction cause I'm lazy, anyone can derive the formula or something. But this is just as easy as that.
(a^n)²-(b^n)²=(a^n-b^n)(a^n+b^n) (a^n-b^n) can easily be proved as divisible by (a-b) by long division While dividing we get few terms of quotient as a^(n-1)b⁰ + a^(n-2) b^1 + a^(n-3)b^2 ........a¹b^(n-2)+a⁰b^(n-1) And remainder as zero!
We know that A^n-1 * B^1 + A^n-2 * B^2 ...... + A^1 * B^n-1 = (A^n - B^n) /( A - B) [by gp sum] Hence (A^n - B^n) is divisible by (A - B) [ LHS is just a some of integers i.e. also integer]
We can write b^n as ( a-(a-b))^n and then if we open the binomial , considering a = x and (a-b) = y then this goes as b^n = a^n -nC1.a^n-1.(a-b) + ... Multiply both sides by - and shift a^n on the left side then this becomes a^n - b^n = nC1.a^n-1.(a-b).. Take a-b as common from the binomial then we get a^n - b^n = (a-b)k where k is some integer Therefore hence proved a^n - b^n is divisible by a-b , ( I know there might be some error in the proof but still gave it a fair try though😅)
Here we can take pairs satisfying (x^2n - y^2n) which will be divisible by x-y hence following this method for each possible pair we will get that the number will be divisible by 17,6 and their LCM(17,6)=106 will be the least number for which we will have to check divisibility hence answer A This was the solution I found to be shorter.
Same approach just using congruent modulo or direct writing the remainders is faster. Like for checking with 17 the expression will be congruent to: (-8)^(190) -(-2) ^190 - 8^190 +2^190. U can see Everything cancels out and u get 0 as remainder. Similar approach for 7 gives: 4^190 - (-2)^190 - 1^190 +2 ^190 So we are left with only 4^190 -1 And from there same approach as yours to get the remainder as 3.
1:40 but n ki value Even ya odd dono ho sakti hai, example n = 3 leke bhi a^3-b^3 divisible ho raha hai a-b se also a^n + b^n is divisible by a+b for n belongs to odd number, this can easily be proved by expanding a^n +b^n/a+b as sum of gp i.e a^n-1[1-(-b/a)^n]/1-(-b/a)
Mera (D) aaya Pehle 25 and 8 aur 19 and 2 mein a^2-b^2 = (a+b)(a-b) lagaya isse 17 ka multiple aa gaya then second factor mein a^odd-b^odd is divisible by a-b lagaya
Bhaiya 4:02 pe (-19¹⁹⁰) hai aur fir jab use 21-2 likhenge to bahar ke minus ki vajah se ultimately (+2¹⁹⁰) rahega Aur firr +2¹⁹⁰ cross cancel nahi hoga
when we divide apower2n-bpowe2n by a-b. a^2n-b^2n= (a-b)(a^(2n-1)+ b^(2n-1)) + ab(a^(2n-2) - b^(2n-2)....now i replaced 2n with 2n-2 to obtain a^(2n-2) - b^(2n-2)=(a-b)(a^(2n-3)+b^(2n-3) + ab(a^(2n-4) - b^(2n-4)).substitue this in prev eqn.a series is formed with 2 power dec(hence even power required) till power o.adding one component of a-b each time...
a^2n-b^2n=(a^n+b^n)(a^n-b^n) a^n-b^n can be further simplified into (a^n/2-b^n/2)(a^n/2+b^n/2) so on and so forth so if n is even eventually this will simplify to (a+b)(a-b) and hence u ll get a long list of all the addition terms times (a-b) so and a and b are integers and hence it will be some k(a-b) hence proved hope this makes sense lol hard to explain via comments
3:02 (a²)^n-(b²)^n=(a²-b²)[a^2(n-1)+a^2(n-2)b^2+.......+b^2(n-1)] And since (a²-b²)=(a-b)(a+b) thus (a^2n-b^2n) is divisible by (a-b).... Correct me if I am wrong somewhere.
Ans-d The given expression is divisible by 2,3,6,17 also, it is not divisible by 4 hence, the only option that satisfies is d. It should also be divisible by 34 i suppose.
Lekin bhaiya hum aise bhi kar sakte kya ki kyoki kisis bhi number ka divisibility toh hum check kar hi sakte hai toh isliye woh unsure waale options toh nikal hi Gaye aur phir kyoki b bhi nhi hoga toh directly answer d hi deduce ho jata hai Isliye answer d. Ye within less than 15 sec ho jata.
i am 11th grader jee aspirant what i found was ki a^2n-b^2n ek tarah se aise kuch hoga (a-b)(a+b)(a^2+b^2)(a^4+b^4) by simple algebraic identity of a^2-b^2 and from here we see this expression is divisible by (a-b)
If you are a jee asirant ek important theorem yaad karlo aise kisi bhi question mai kam aayega agar tumhe ek p prime se divide hota hai ya nahi ye nikalna hai to jo power hain usko us number se replace kardo jo uska remaider aayega (p-1) se or base number ko remainder of p se replace kardo jaise is question mai 7 ka check karne ke liye above operation ke baad(190=4:6) 256-625-1+16=-354 which is not divisible by 7 ye ek approach ye sare remainder wale q tackle ho jaenge aur tumhe binmial ki shayad hi jarurat padegi
for shorter method u can use modulo method, it is a property generally used in olympiads but that property is basically derived from binomial expansions and it is just what u said but by property it is considered to be known already basically u can write any number as the no it will leave remainder and leave the power as it is and try converting all numbers into +1,+2,-1,-2 to ease down calculation and u can solve almost every remainder question within 2 min, this one i especially solved before u started solving without pausing vid
for proof you asked : let take common a^2n from expresssion, then divide , and multply term with (1-b/a ) , this create a normal gp(a^2n+.....b^2n) *(1-b/a) , clearly div, by (a-b )
Idk I enjoyed? Why this happened as I See u didn't edit the blundered part. I think u were fully immersed into it That's why u made it soo smooth for us.
It is a part of number theory which is not taught in routine or jee maths it is taught in Olympiad preparation it is a very fast method for finding remainders I am in class 10 and this q in less than 2 mints
Consider the polynomial (P(x) = a^n - b^n). Notice that (P(b) = 0), which means (x - b) is a factor of (P(x)). Thus, (a - b) is a factor of (a^n - b^n). In more straightforward terms, for any integers (a) and (b), when you subtract (b^n) from (a^n), the result can always be expressed as a product of (a - b) and another polynomial. Hence, (a^n - b^n) is always divisible by (a - b). Does this make sense or not 😅 Please reply ❤ . Thank you ❤ And yes I have a shorter method then this maybe 5to6 lines .
25 ≡ 8 (mod 17) 19 ≡ 2 (mod 17) 8 ≡ 8 (mod 17) 2 ≡ 2 (mod 17) Ab isme ((8^(190) -2^(190) -8^(190) +2^(190)) (mod 17) Simply kr lo 0(mod 17) Matlb hai ab 28 mein bhi kr lo nahi hoga I know ki yeh topic OLYMPIAD ka hai but ek true JEE aspirant tabhi hota hai jab woh har approach ko seekhe aur chota sa hai yaar yeh toh tumhaara B. T. easy kr dega seekh lo jaake 😊
Can we prove it this way? We know, a²-b² = (a+b)(a-b) Similarly, a⁴-b⁴ = (a²+b²)(a²-b²) Hence.. since in a^2n - b^2n The 2n part always gives an even number.. which in turn gives an even factor in terms like this.. like say for example In this, a^2n - b^2n .. n=5 then, a¹⁰ - b¹⁰ = (a⁵-b⁵)(a⁵+b⁵) This a⁵-b⁵ can be further simplified into (a-b)(a⁴ + a³b + a²b² + ab³ + b⁴) Thus.. we will always get (a-b) as a factor of a^2n - b^2n
Sir pls pls explain more of revision ... i mean explain more about revision course in how much time will it over and all I wanna buy it but just a little confused...
Well you can use modulu as you know many aspirant give exam of ioqm these are the basic question like remainder so if you use moduluit will take around 2 min but it will get the answer but i love your method
Bhai property nahi bhi aati ho toh we can split 25 and 19 as 17+8 and 17-2, binomial se open karne ke baad the 8^190 and 2^190 automatically get cut so it becomes divisble by 17. Thoda dhyaan se dekhein toh it becomes divisible by 34 also as all terms have a multiple of 2
Bhaiya a^(2n) - b^(2n) ko we can rewrite as (a^n)² - (b^n)² fir x²-y²= (x-y)(x+y) property use krke (a^n-b^n)(a^n+b^n) likh skte hai and (a^n-b^n) is divisible by a-b and hence pura a^(2n) - b^(2n) is divisible by a-b
a^n = (a-b+b)^n Now, applying binomial theorem a^n = b^n + (a-b)k, where k is an integer Subtracting b^n from both sides a^n-b^n = (a-b)k Which implies a^n-b^n will be divisible by a-b Is the proof okay?
Pta nahi sahi proof hai ya nahi but I just tried (a^2n -b^2n)/a-b. Isko a^2-b^2 identify se break kiya toh {(a^n -b^n)(a^n+b^n)}/a-b aise break karte gaye toh iske baad n/2 banega uske baad n/4 till n/n which is equal to 1 toh ekdam last step mein a-b numerator mein banega toh wo cut hojayega that why divisible
Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17.. Is my method correct?
no need of this for to prove it's divisible by 17.. remainder of 1st one is 16, second one is 4, third one is 1 and fourth one is 13.. add all and it's divisible by 17...
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We hope that you learned how fundamental properties can occur at unexpected times, a mix of such unique illustrations must be practices by you in case aiming for top ranks. for that only we offer {SO60} to those who struggle with problem solving
Here's the link for more details
courses.jeesimplified.com/courses
Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17..
Is my method correct?
it can be proved using binomial theorem:
we write a^n = (a - b + b)^n
expanding using binomial thm:
a^n = k^n + nC1.k^n-1.b + ...... + b^n (where k=(a-b))
so, a^n - b^n = k {k^n-1 + nC1.k^n-2.b + .....}
as combination operator always result in integer the whole bracket on RHS is an integer (say p)
so, a^n - b^n = (a - b) . p
thus a^n - b^n is always divisible by (a - b)
I have some confusion here hope you could resolve. In last expansion of binomial there is only 2nC2nb²^n then how did you take k as common from it??????????
It is copied from chatgpt😂
@@crazymathematician88 last term se K common nahi nikla hai i.e
a^n = k {k^n-1 + nC1.k^n-2.b + .....} + b^n
then usne B^n ko left hand side laya hai
@@strangeboysam2594 got it. Matlab left side lake k common (jo ki (a-b) hain) common liya hain. 👍
How was it 👀?
@@jeesimplified Bhaiya a^(2n) - b^(2n) ko we can rewrite as (a^n)² - (b^n)² fir x²-y²= (x-y)(x+y) property use krke (a^n-b^n)(a^n+b^n) likh skte hai and (a^n-b^n) is divisible by a-b and hence pura a^(2n) - b^(2n) is divisible by a-b
Aye, fake account, Original channel pe comment karte hue😂
@@Naitik_Barnwal 😭😭😭🙏🏻
@@jeesimplified bhai, fake account leke, students attract mat karo, hume original channel ka naam pata hai😆
Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17..
Is my method correct?
3:01 proof
By remainder theorem
Let f(a,b)=a^2n-b^2n
Put a=b so the function becomes zero which shows that a-b is a factor of this expression
Put a=-b
Again the function becomes zero so a+b is also a factor
Everybody is probably just copying from Google and using that modulo thingy/lengthy binomial expression
But here is a person who used a basic method to prove this logically
@@dheerendrayadav8335 yes , see first forget about the big 190 power. Start with 1, the answer is 0. Then put 2 as power for 25, 19, 8 and 2. You'll get 204, which is divisible by 17, so it is also divisible by 34, you can check further by 3, but then you'll reach the same conclusion. The above
I don't like to get into abstraction cause I'm lazy, anyone can derive the formula or something. But this is just as easy as that.
Bro but this is verfication not proof 😅 , there is difference between those two@@dheerendrayadav8335
Congruence Modulo joins the chat 👽
(a^n)²-(b^n)²=(a^n-b^n)(a^n+b^n)
(a^n-b^n) can easily be proved as divisible by (a-b) by long division
While dividing we get few terms of quotient as
a^(n-1)b⁰ + a^(n-2) b^1 + a^(n-3)b^2 ........a¹b^(n-2)+a⁰b^(n-1)
And remainder as zero!
We know that
A^n-1 * B^1 + A^n-2 * B^2 ...... + A^1 * B^n-1 = (A^n - B^n) /( A - B) [by gp sum]
Hence (A^n - B^n) is divisible by (A - B) [ LHS is just a some of integers i.e. also integer]
We can write b^n as ( a-(a-b))^n and then if we open the binomial , considering a = x and (a-b) = y then this goes as
b^n = a^n -nC1.a^n-1.(a-b) + ...
Multiply both sides by - and shift a^n on the left side then this becomes
a^n - b^n = nC1.a^n-1.(a-b)..
Take a-b as common from the binomial then we get
a^n - b^n = (a-b)k where k is some integer
Therefore hence proved a^n - b^n is divisible by a-b , ( I know there might be some error in the proof but still gave it a fair try though😅)
Here we can take pairs satisfying (x^2n - y^2n) which will be divisible by x-y hence following this method for each possible pair we will get that the number will be divisible by 17,6 and their LCM(17,6)=106 will be the least number for which we will have to check divisibility hence answer A
This was the solution I found to be shorter.
Congruence modulo
Olympiad😅
Here we ioqm student get ahead. Congurence modulo🔥
tu tg pe famous hogya
@vamshitarun4399 kha par?
Sahi kaha 😅
Wahi, congruence se aasani se ho jayega
Mai bhi yahi soch raha tha
Same approach just using congruent modulo or direct writing the remainders is faster.
Like for checking with 17 the expression will be congruent to:
(-8)^(190) -(-2) ^190 - 8^190 +2^190. U can see Everything cancels out and u get 0 as remainder.
Similar approach for 7 gives:
4^190 - (-2)^190 - 1^190 +2 ^190
So we are left with only 4^190 -1
And from there same approach as yours to get the remainder as 3.
1:40 but n ki value Even ya odd dono ho sakti hai, example n = 3 leke bhi a^3-b^3 divisible ho raha hai a-b se
also a^n + b^n is divisible by a+b for n belongs to odd number, this can easily be proved by expanding a^n +b^n/a+b as sum of gp
i.e a^n-1[1-(-b/a)^n]/1-(-b/a)
Proof:
Consider
f(x)=x^2n-b^2n
It is divisible by x-b
Put x=a and we proved it
Umm brother.. we have to prove that how your f(x) = x^2n - b^2n is divisible by x-b..
I think.
Mera (D) aaya
Pehle 25 and 8 aur 19 and 2 mein a^2-b^2 = (a+b)(a-b) lagaya
isse 17 ka multiple aa gaya
then second factor mein a^odd-b^odd is divisible by a-b lagaya
Bhaiya 4:02 pe (-19¹⁹⁰) hai aur fir jab use 21-2 likhenge to bahar ke minus ki vajah se ultimately (+2¹⁹⁰) rahega Aur firr +2¹⁹⁰ cross cancel nahi hoga
Can be easily prooved by congruent modullo
when we divide apower2n-bpowe2n by a-b. a^2n-b^2n= (a-b)(a^(2n-1)+ b^(2n-1)) + ab(a^(2n-2) - b^(2n-2)....now i replaced 2n with 2n-2 to obtain a^(2n-2) - b^(2n-2)=(a-b)(a^(2n-3)+b^(2n-3) + ab(a^(2n-4) - b^(2n-4)).substitue this in prev eqn.a series is formed with 2 power dec(hence even power required) till power o.adding one component of a-b each time...
a^2n-b^2n=(a^n+b^n)(a^n-b^n)
a^n-b^n can be further simplified into (a^n/2-b^n/2)(a^n/2+b^n/2) so on and so forth
so if n is even eventually this will simplify to (a+b)(a-b) and hence u ll get a long list of all the addition terms times (a-b) so and a and b are integers and hence it will be some k(a-b)
hence proved
hope this makes sense lol hard to explain via comments
Bro your given proof is done with 2 methods and thankx for such a op ques bro your ques literally open brain
6:51 bhaiya ka whatsapp profile description, battery about to die😂
3:02 (a²)^n-(b²)^n=(a²-b²)[a^2(n-1)+a^2(n-2)b^2+.......+b^2(n-1)]
And since (a²-b²)=(a-b)(a+b) thus (a^2n-b^2n) is divisible by (a-b)....
Correct me if I am wrong somewhere.
dimag me ek approach aa raha hai ki sayad summation of gp ke kuch ho isse ache se dekhta hun
yes, absolutely correct 💯
@@jeesimplified-subject Thank you bhaiya
Ans-d
The given expression is divisible by 2,3,6,17 also, it is not divisible by 4 hence, the only option that satisfies is d.
It should also be divisible by 34 i suppose.
a^2n-b^2n is divisible by a-b can be proved by binomial theorem. It's probably one of the core things you're taught in this chapter.
Lekin bhaiya hum aise bhi kar sakte kya ki kyoki kisis bhi number ka divisibility toh hum check kar hi sakte hai toh isliye woh unsure waale options toh nikal hi Gaye aur phir kyoki b bhi nhi hoga toh directly answer d hi deduce ho jata hai
Isliye answer d. Ye within less than 15 sec ho jata.
If n=1 then it will become a^2-b^2 which will be (a-b)(a+b) which is dividible by a-b and clearly all even power will be divisible the same way
i am 11th grader jee aspirant what i found was ki a^2n-b^2n ek tarah se aise kuch hoga (a-b)(a+b)(a^2+b^2)(a^4+b^4) by simple algebraic identity of a^2-b^2 and from here we see this expression is divisible by (a-b)
divisible by 25-8, -19+2,
If you use congruence modulo,then this question is pretty easy
Bhaiya conics concept cruch when ?
If you are a jee asirant ek important theorem yaad karlo aise kisi bhi question mai kam aayega agar tumhe ek p prime se divide hota hai ya nahi ye nikalna hai to jo power hain usko us number se replace kardo jo uska remaider aayega (p-1) se or base number ko remainder of p se replace kardo jaise is question mai 7 ka check karne ke liye above operation ke baad(190=4:6) 256-625-1+16=-354 which is not divisible by 7 ye ek approach ye sare remainder wale q tackle ho jaenge aur tumhe binmial ki shayad hi jarurat padegi
Fermat little theorem
Shortest method use these theorems:
1) Fermat's little theorem
2) Euler's totient theorem
Of course not in jee syllabi
Bro ek baar karke bhej sakte ho, modulo se?
Bro ek baar kar sakte modulo se?
@@foreveradirectioner5985take mod17 entire equation, 17 is prime so from fermat little theorem
a^16=1mod17
Followed by simplification.
@@nilusingh377 Ohk, thanks so much!
for shorter method u can use modulo method, it is a property generally used in olympiads but that property is basically derived from binomial expansions and it is just what u said but by property it is considered to be known already
basically u can write any number as the no it will leave remainder and leave the power as it is and try converting all numbers into +1,+2,-1,-2 to ease down calculation and u can solve almost every remainder question within 2 min, this one i especially solved before u started solving without pausing vid
exactly congurence modulo se easily ho jaega but uske liye number theory padhna padega
Congduence modulo se ho jata hai
for proof you asked : let take common a^2n from expresssion, then divide , and multply term with (1-b/a ) , this create a normal gp(a^2n+.....b^2n) *(1-b/a) , clearly div, by (a-b )
Idk I enjoyed? Why this happened as I See u didn't edit the blundered part. I think u were fully immersed into it That's why u made it soo smooth for us.
Proof
Consider a GP
x^n-1+(x^n-2)y+(x^n-3)y^2+......+y^n-1
Do its sum and simplify it result will be magnificient
ANS IS (D)
3:59 pe + 2^190 hoga na sir?
What the hell is this modulo??????
Please explain it!!!!!!!
It is a part of number theory which is not taught
in routine or jee maths it is taught in Olympiad preparation it is a very fast method for finding remainders I am in class 10 and this q in less than 2 mints
Consider the polynomial (P(x) = a^n - b^n). Notice that (P(b) = 0), which means (x - b) is a factor of (P(x)). Thus, (a - b) is a factor of (a^n - b^n).
In more straightforward terms, for any integers (a) and (b), when you subtract (b^n) from (a^n), the result can always be expressed as a product of (a - b) and another polynomial. Hence, (a^n - b^n) is always divisible by (a - b).
Does this make sense or not 😅
Please reply ❤ .
Thank you ❤
And yes I have a shorter method then this maybe 5to6 lines .
25 ≡ 8 (mod 17)
19 ≡ 2 (mod 17)
8 ≡ 8 (mod 17)
2 ≡ 2 (mod 17)
Ab isme
((8^(190) -2^(190) -8^(190) +2^(190)) (mod 17)
Simply kr lo
0(mod 17)
Matlb hai ab 28 mein bhi kr lo nahi hoga
I know ki yeh topic OLYMPIAD ka hai but ek true JEE aspirant tabhi hota hai jab woh har approach ko seekhe aur chota sa hai yaar yeh toh tumhaara B. T. easy kr dega seekh lo jaake 😊
4:23 kaise bhaiya??? wo to 2^190 hi hoga...bahar ek minus tha...aapne 4:28 me bhi to -21Y bithaya hai
aise kaise chalega😭😭
Woh 17 and 34 wale toh ho gye the pr 28 wala nhi hua tha
Can we prove it this way?
We know, a²-b² = (a+b)(a-b)
Similarly, a⁴-b⁴ = (a²+b²)(a²-b²)
Hence.. since in a^2n - b^2n
The 2n part always gives an even number.. which in turn gives an even factor in terms like this.. like say for example
In this, a^2n - b^2n .. n=5
then, a¹⁰ - b¹⁰ = (a⁵-b⁵)(a⁵+b⁵)
This a⁵-b⁵ can be further simplified into (a-b)(a⁴ + a³b + a²b² + ab³ + b⁴)
Thus.. we will always get (a-b) as a factor of a^2n - b^2n
Solve this using modulo concept
Sir pls pls explain more of revision ... i mean explain more about revision course in how much time will it over and all
I wanna buy it but just a little confused...
Well you can use modulu as you know many aspirant give exam of ioqm these are the basic question like remainder so if you use moduluit will take around 2 min but it will get the answer but i love your method
Bhai property nahi bhi aati ho toh we can split 25 and 19 as 17+8 and 17-2, binomial se open karne ke baad the 8^190 and 2^190 automatically get cut so it becomes divisble by 17. Thoda dhyaan se dekhein toh it becomes divisible by 34 also as all terms have a multiple of 2
a2n -b2n ÷ a-b is actually g.p sum
3:01 by factor theorem we can say
Answer is D
Gp ke sum se bhi proof ho jayega sum of this series
a^n-1+a^n-2b+a^n-3b+.....+b^n-1. Is
can put it in the form x^n-1/x-1 where x is a/b and this is gp sum
mains mei jo sawal tha waha 14 aur 34 mei hi options diye the yaha alag kyu hai?
Bhai maths ke liye kaunsi achi book rhegi??
Bhaiya a^(2n) - b^(2n) ko we can rewrite as (a^n)² - (b^n)² fir x²-y²= (x-y)(x+y) property use krke (a^n-b^n)(a^n+b^n) likh skte hai and (a^n-b^n) is divisible by a-b and hence pura a^(2n) - b^(2n) is divisible by a-b
areee bhai a^n-b*n is divisible by a-b isko kese assume kariya, isko bhi u’ll have to prove.
@@jeesimplified got it , will try it again
a^n = (a-b+b)^n
Now, applying binomial theorem
a^n = b^n + (a-b)k, where k is an integer
Subtracting b^n from both sides
a^n-b^n = (a-b)k
Which implies a^n-b^n will be divisible by a-b
Is the proof okay?
Ye maine pyqs krte time kiya tha
First attempt me to nhi hua pr soln dekhke mja ayega
You are a good question😂
a^2n wala
is it because after factoring we get a^n-b^n which divides a-b??
3:05 gp k sum se kiya jaaskta h
Sir just learn modulo, mod 17 is so easily then look at mod 7 very easy
a^n-b^n=(a-b)(a^n-1+a^n-2b+a^n-3b^2+........+a.b^n-2+b^n-1) .hence, it is divisible by a-b
pls like and pin.
A piece of cake for oly asp
2:34 how to confirm that it is divisible by 34???
Bcz 34 is a multiple of 17
Pta nahi sahi proof hai ya nahi but I just tried
(a^2n -b^2n)/a-b. Isko a^2-b^2 identify se break kiya toh {(a^n -b^n)(a^n+b^n)}/a-b aise break karte gaye toh iske baad n/2 banega uske baad n/4 till n/n which is equal to 1 toh ekdam last step mein a-b numerator mein banega toh wo cut hojayega that why divisible
Which software are you using.
Bhai poori calculation galat hai, 2^190 cancel nahi hoga. - sign bhi hai 19^190 pe. In fact 2^191 ho jaayega
Sir we can solve it by binomial method by breaking 25 as 17+8, 19 as 17+2, 2^some as (17-1)^some. At the end we only get multiple of 17..
Is my method correct?
Sir you can solve using euler & fermat theorem
Congruence modulo and ye without pen paper solved
This theorem was in cengage book but we ignored like our crush used to ignore us 😅 and jee gave question on this concept
indeed 💯
did it, as a IOQM student
Bhaiya pls continue concept crunch series
Bro use cogurence Modulo IT WILL become more easy
Your prove can be proved by mathematical induction
Congruence Modulo
Can we prove it by lmvt ?
Sandal bhaiya 2024 cc op
Easier than my dpp.
fermats little theorem aur khatam karo isko
But I got it correct 😌
no need of this for to prove it's divisible by 17.. remainder of 1st one is 16, second one is 4, third one is 1 and fourth one is 13.. add all and it's divisible by 17...
Ye wali property aati h. 🌚🐮
Euler and fermat laughing in corner
Yes
Solved your question for the first time just by looking at it (it was done in our class)