This question could be solved in a very simple way - First we know that the last digit at the end of 2019**4 is going to be 1 Second we know that last digit of 4**2019 is going to be 4, Therefore the end digit of the sum will be 5 and since it will be greater than 10 , it means the number is divisible by 5 and is NOT A PRIMENUMBER.
well i also thought that it could be solved by Cyclicity of Powers:- last digit of 2019 is 9, and cyclicity is 2, therefore for even number power the resulting number's last digit ends with 1 last digit of 4 is 4 (obv.) and cyclicity is 2, therefore, for even number power, the resulting number's last digit ends with 4 Adding both last digits, we get: 4+1 = 5, and the resulting number of above EXPRESSION is a factor of 5 Therefore it is a COMPOSITE Number and NOT PRIME number
@sahil tambe Yes, But the method taught by sir will also not work in every case, In fact there is no foolproof method to solve equations above the power of 3.
We can use just 2 rule 1. Unit digit 2. Divisibility rule of 5 to solve this problem. But the catch in this problem is that how calculate the unit digit of 4^2019. For this, there is rule of Cyclicity of power 4. First we divide the last two digit of 2019 and the remainder would be the power of base. 2019 => 19➗4=> remainder 3 4^3= 4
Sir I have an alternate solution. If we apply the powers to only the unit's place digits ,that is 4^9 & 9^4 and add their unit's place digits ,which gives the unit's place digit of the whole sum,which is '5' and hence is divisible by 5
And when I saw his first video I fell in love..... Now I am just adddicted ......I am ready to watch his videos even at midnight Your and bprp's videos can force anyone to fall in love for maths.....
This question can be solve very easily by using concept of unit digit,which is 5 means this no. is divisible by 5 (which is other than 1 and given no).Hence given no. is not a prime no
Sir,the 2nd factor that you have solved, containing minus in the middle.so there would be a chance to become that value 1.in this case it is not cleared. It would be very easy when we solving it through unit digit addition. Its a composite number.
Your solution is purely algebric but the solution can be find by finding unit digit concept by cyclic method.which is 5.it means the number is not prime.
Sir, There is another way... break them as below: (2020-1)^4 + (5-1)^2019 if we expand it the we get 5 as common so it also divisible by 5 So we can say that it is not a prime no...
Sir it is also necessary to prove that both the factors are other than the 1 and number itself like for example 11 can be written as 11×1 therefore it can be written as (a) ×( b ) But writing it in a×b doesn't prove that it is an composite number because every prime number has 2 factors which is 1 and number itself Either you need to prove that any one of the factor different from 1 and number itself or you can prove it by writing as product of more than 2 factor in which none of them is 1 or number itself
@@sumankumarshah6176 yes you are right too but that is not the thing For instances in geometry you need to prove that for some questions like triangles are similar We can clearly see in that cases that both triangle are exactly same but still we need to prove that .similarly here we also need to provide valid proof that values are more than 1
@@battleongaming4223 no matter whatever prime number you take, their prime factors will be 1 and that number. So if 2019^4 + 4^2019 is prime, one of the two factors has to be 1, no matter what the other number is. and clearly you can see that none of the two number is 1. i mean what prove do you need? even a kid knows that both the numbers are much more than 1. it's like you are asking prove 2+2=4. geometry is a different thing, bro. those are figures. just looking at figures doesn't give anything, what you need is numerical values.
Sir i found the answer in like 10 secs. The last digit of 2019^4 is 1 as 9*9=81, 1*9=9 and 9*9=81. 4^1=4 4^2=16 4^3=64 4^4=256 So 4^2019 always ends with 4 as 2019 is odd. So after after taking the sum of both the numbers, we get the last digit 5. Hence the number is divisible by 5.
Sir, I'm in 8th Standard and I can understand a lot of maths of 11th and 12th, My parents and My elder brother got Surprised seeing me solving Questions from Logarithm. Thank you sir, ♥️♥️
Wow! Now a days Online Teachers Teach us far better than Offline Teachers. I was weak in Maths Before Lockdown, then I started watching Videos on TH-cam and Now my classmates wanna learn maths from me, And I make them all Concepts Clear..
Unit place of 2019⁴ will 1 and Unit place of 4²⁰¹⁹ will 4 Therefore unit place of there sum will 5 and the number will be a multiple of 5 proving not to be a prime
Sir meine kya logic lagaya (2019)^even no = ____1 (even no. Se power raise karoge to 1 hi ayega units place me, similarly agar odd se raise karoge toh units place me hamesha 9 hi ayega) 4^odd no = ____4 (4 ko agar hum odd no. se raise karenge toh even place me 4 aana hi aana hai similarly even no se raise karoge toh apan ko units place me 6 ayega) (2019)^even no + (4)^odd no = ____5 Answer Jo bhi ayega, 5 se divisible ho jayega, hence diye hue expression ka no prime nahi hai.
My method 2019^4 + 4× (16)^1009 16^1009 main last number 6 hi ayega and uske sath 4 multiple ke bad last word 4 ayega And 2019^4 main last number is 1 Now , ( ...........1) + (...............4) = ...................5 i.e. divisible by 5 So the number is not prime number ❤️❤️
1. Odd ^ ever or odd = is always odd. 2. Even ^ even or odd = is always even. 3. Even + odd = is always odd With these three steps we get 2019^4 is odd . And 4^2019 is even . Final answer even + odd is odd .
Sir we can also solve using cyclicity..... The last digit of 2019^4 is 1 and that of 4^2019 is 4. N So the last digit of this number is 5 and hence it is composites.
best method to solve this is to take mod 5. 2019= -1(mod 5), so 2019^4 = 1(mod 5). similarly, 4==-1(mod 5) so 4^2019= 1 (mod 5) hence, their sum== 0 (mod 5). the number is (obviously) not =5, but it is divisible by 5. hence, it is composite
If it is a prime so it should be of form 6K ± 1 so we have to prove (2019)^4 + (4)^2019 ± 1 is divisible by 6 Proof:- Expressions is even so we just have to prove that it is divisible by 3 2019 is divisible by 3 and using binomial on (4)^2019 we can cancle one and expression is divisible by 3 even so by 6 also Hence proved as Prime
Ist part:(2019)⁴ will give a number with unit digit (i.e.,right se last wala digit)=1. (as 2019×2019×2019×2019=Some number with last digit 1,because last digit will be given by this multiple 9×9×9×9=6561) 2nd part:4²⁰¹⁹ will have the unit digit as 4( as 4^odd number, will have its unit digit always 4, e.g.,4³=64 and 4⁵=1024 and this is true for 4²⁰¹⁹ as well, as 2019 is an odd number.) Now,by divisibility rule for divisibility by 5,we know that a number is divisible by 5 if the its unit digit is 5. Since,the number 2019⁴+4²⁰¹⁹ have unit digit as (1+4) ,i.e., 5,it is divisible by 5. Therefore,the given number is not a prime number,as it has a factor 5 other than 1 and itself. P.S: Wrote this long and elaborated,so that any class student can understand.
Sir Issey acccha solution mere paas hai Ye number 5 se divisible hai 2019⁴ + 4²⁰¹⁹ =( 2020 - 1 )⁴ + ( 5 - 1 )²⁰¹⁹ = 2020k + (-1)⁴ + 5t + (-1)²⁰¹⁹ = 2020k + 5t -1+1 = 2020k + 5t Since rhs is divisible by 5 thus lhs will also be divisible by 5
An easier way to solve this: 2019^4 will end with 1(9*9*9*9) and 4^2019 will end with 4 so the sum's unit digit will have 5 which means it's divisible by 5.
technically - there is a lacunae in your arguments. You have to show that both of your factors are greater than 1 (just to avoid p = 1xp is a factorization of a prime).
Done 2019^1= 2019 2019^2= ....1 2019^4= ....1 Last digit 1 is not a prime number. So 2019^4 is not prime number Similary, 4^1= 4 4^2= 16 4^3=...4 4^5=...4 So definitely 4^2019=....4 Now a number ended with 4 is definitely divisble by 2 (sometime 4&7 also). So 4^2019 is not a prime number.
2019^4 has 1 as it’s last digit. This is because the last digit of 2019^4 is the same as the last digit of 9^4. Now, 4^2019 has it’s last digit as 4. This is because every odd power of 4 has the last digit 4, and 2019 is odd. So, the last digit of 2019^4 + 4^2019 is 5. Hence, the number is divisible by 5 and it isn’t prime.
Guys first thing he is teaching an approach…to just get the answer y’all go and see shorts and don’t come here…come here to learn a concept or approach which may be used in other qs
sir i can do this in a more simpler way it consists of 3 steps if done by shortcut but still it's shorter and a lot simplere than the method you gave . where can i send you my solution?
@@Khushi-vz3yg yea see there are three actually two methods to solve this one which he said reduce the powers such that you have a unit digit and a cylicity order(if you know the concept well and good) or my modulo arithmetic simple just find remainders then if only if there is any integer that is divisible then the expression is not prime simple. And yes sofie germain identity the third method
4 ki power odd gives 4 in unit digit … 9 ki power even gives 1 in unit digit If we add both unit digits -> It will be 4+1 = 5 Since unit digit of the sum is 5 .. therefore it is not a prime number
There is a much easier solution here. 2019 power 4 = ends in last digit with 1. Why? 9 x 9 x 9 x 9 ends with 1. 4 power 2019 = ends with 4. Why? 4 power odd number ends with 4 and even power ends in 6. So last digit of the sum would 1 + 4 = 5 which is divisible by 5. Not a prime.
Sir yeah toh direct ho raha tha 4 ki power 3 se divisible hai matlab unit digit 4 hogi Aur 2019 ki power 4 hai toh unit digit is 1 When we add unit digit 4+1 we get 5 Which is divisible by 5..
Sir , Considering Vedic Mathematics, Consider only the Last Number 2019 Multiplied Even Times always ends with Number 1 (Ex 9 * 9 = 81) and Odd Times with number 9 ( Ex 9*9*9 = 729) Hence the First Number Multplied 4 Times should end with Number 1 Consider the Second Number ie 4 - Multiplied even times always ends with 6 and Odd Times ends with 4 ,Hence irrespective of all other Numbers The Total of this will end with Number 1 + 4 = 5 and Hence it is not a Prime Number I hope i am right, Please correct if i am wrong
Checking The Last number for certain problems itself is the concept of Vedic Mathematics Examples When Finding the Square root of any number say, 576, (The Square root number must end with Either 4 or 6) Observing the Last Digit in Table of Nine Etc.9,8,7,6,5,4,3,2,1,0 Etc.,
just write 2019^4 as 3^4 *673^4 and write 4^2019 as (3+1)^2019 ,,,,, clearly we can taken 3^4 common after expanding the binomial... thus it is not prime with 81 as one of its factors..... i think this is simpler method.
Sir wo factors 1 aur 2019^4+4^2019 ho Sakta hai lekin Hume kaise pata chalega Jo factors hum jo factors nikalne wo 1aur it self nahi hai. If we don't have calculator.
its clear that this expression is not a prime number because :- 1) we can see that both the number 2019 and 4 are composite numbers so the answer of any of their power will also be an composite number. 2) if we'll add the unit digit of both the expression then it will come out be equal to 5 which means that the number is divisible by 5.
Sir ek request hai aesi hi choti choti videos in ques ki bna dijiye jo jee main aur advance jyadatar puchta hai plzz sir I hope ki mein limit se jyada nhi bola Agr bola toh sorry sir
By Cyclicity concept Unit digit of 4 when n is odd 4ⁿ= 6 when n is even Here it is odd so Unit digit of 4²⁰¹⁹ is 4 --------(1) Similarly 9ⁿ has unit digit 9 when n is odd and 1 when n is even Here it is even so Unit digit of 2019⁴ is 1---‐-(2) Net Sum has Unit digit 5 so It is divisible by 5 Therefore it may be concluded that it is not a prime no.
This question could be solved in a very simple way -
First we know that the last digit at the end of 2019**4 is going to be 1
Second we know that last digit of 4**2019 is going to be 4,
Therefore the end digit of the sum will be 5 and since it will be greater than 10 , it means the number is divisible by 5 and is NOT A PRIMENUMBER.
well i also thought that it could be solved by Cyclicity of Powers:-
last digit of 2019 is 9, and cyclicity is 2, therefore for even number power the resulting number's last digit ends with 1
last digit of 4 is 4 (obv.) and cyclicity is 2, therefore, for even number power, the resulting number's last digit ends with 4
Adding both last digits, we get: 4+1 = 5, and the resulting number of above EXPRESSION is a factor of 5
Therefore it is a COMPOSITE Number and NOT PRIME number
I did it the same way
I used the same method !!!
@sahil tambe Yes, But the method taught by sir will also not work in every case, In fact there is no foolproof method to solve equations above the power of 3.
I thought in the me wy
We can use just 2 rule
1. Unit digit
2. Divisibility rule of 5
to solve this problem.
But the catch in this problem is that how calculate the unit digit of 4^2019.
For this, there is rule of Cyclicity of power 4.
First we divide the last two digit of 2019 and the remainder would be the power of base.
2019 => 19➗4=> remainder 3
4^3= 4
An easier way of putting it is:
Unit's digit of 4's power will go like 4,6,4,6,4,6..... If power is odd, unit's digit would be 4 else 6.
a^4 + 4b^4 is also known as Sophie-Germain's identity, and it is factorizable for all integers a and b
Best teacher👩🏫 ever
bruh he copied it from Blackpenredpen
th-cam.com/video/zG0azhBHyE0/w-d-xo.html
Sir I have an alternate solution.
If we apply the powers to only the unit's place digits ,that is 4^9 & 9^4 and add their unit's place digits ,which gives the unit's place digit of the whole sum,which is '5' and hence is divisible by 5
Awesome question
Wonderful answer
By a great teacher
A great worker already solved it
Best series. Random questions are best for preperation. Pls continue this for too many years
And when I saw his first video I fell in love..... Now I am just adddicted ......I am ready to watch his videos even at midnight
Your and bprp's videos can force anyone to fall in love for maths.....
so yk bprp huh
cool
@@aniiico1266 Yup bprp fan
Solve it by many ways 1-switch system theory,2-binomial theoram or even by 3-fermat theoram
sir ur method is universal approach. But units digit is bit simpler for this particular question . question=0(mod 5)
This question can be solve very easily by using concept of unit digit,which is 5 means this no. is divisible by 5 (which is other than 1 and given no).Hence given no. is not a prime no
yes you are right sir
SIR MAINE BHI WAISE HI SOLVE KAR LIYA
Yes absolutely !
Sir,the 2nd factor that you have solved, containing minus in the middle.so there would be a chance to become that value 1.in this case it is not cleared.
It would be very easy when we solving it through unit digit addition.
Its a composite number.
Agreed
Aisa hi mene socha, 1 bhi bn skta h
Not possible since 2²⁰¹⁸>>>>>>>>>>>>>>>>>2¹⁰⁰⁹
Your solution is purely algebric but the solution can be find by finding unit digit concept by cyclic method.which is 5.it means the number is not prime.
I didn't even realise until he revealed answer 😂😂😂
Sir,
There is another way...
break them as below:
(2020-1)^4 + (5-1)^2019 if we expand it the we get 5 as common so it also divisible by 5
So we can say that it is not a prime no...
Sir it is also necessary to prove that both the factors are other than the 1 and number itself like for example
11 can be written as 11×1 therefore it can be written as (a) ×( b )
But writing it in a×b doesn't prove that it is an composite number because every prime number has 2 factors which is 1 and number itself
Either you need to prove that any one of the factor different from 1 and number itself or you can prove it by writing as product of more than 2 factor in which none of them is 1 or number itself
Yes you are right but we can easily see the values are way more than 1 so i think he meant that
@@sumankumarshah6176 yes you are right too but that is not the thing
For instances in geometry you need to prove that for some questions like triangles are similar
We can clearly see in that cases that both triangle are exactly same but still we need to prove that .similarly here we also need to provide valid proof that values are more than 1
@@battleongaming4223 no matter whatever prime number you take, their prime factors will be 1 and that number. So if 2019^4 + 4^2019 is prime, one of the two factors has to be 1, no matter what the other number is. and clearly you can see that none of the two number is 1. i mean what prove do you need? even a kid knows that both the numbers are much more than 1. it's like you are asking prove 2+2=4. geometry is a different thing, bro. those are figures. just looking at figures doesn't give anything, what you need is numerical values.
(A⁴ + 4B⁴) ke jaise or kitni unique identities hai, unsab par video banaye sir please
Sir i found the answer in like 10 secs.
The last digit of 2019^4 is 1 as 9*9=81, 1*9=9 and 9*9=81.
4^1=4
4^2=16
4^3=64
4^4=256
So 4^2019 always ends with 4 as 2019 is odd.
So after after taking the sum of both the numbers, we get the last digit 5.
Hence the number is divisible by 5.
Sir, I'm in 8th Standard and I can understand a lot of maths of 11th and 12th, My parents and My elder brother got Surprised seeing me solving Questions from Logarithm. Thank you sir, ♥️♥️
I am also from 8th standard and my friends got stunned seeing me solving cubic equations and quartic equations
Wow! Now a days Online Teachers Teach us far better than Offline Teachers. I was weak in Maths Before Lockdown, then I started watching Videos on TH-cam and Now my classmates wanna learn maths from me, And I make them all Concepts Clear..
If u were so intelligent then u were not here watching in TH-cam 😂
Mee too i have solved ful 10 and and now i am solving diffentiation of class 11 btw i am in class 8
@@shibam4182 how you solve cubic equation
√a+b=7 and a+√b=11 prove sir.
Unit place of 2019⁴ will 1 and
Unit place of 4²⁰¹⁹ will 4
Therefore unit place of there sum will 5 and the number will be a multiple of 5 proving not to be a prime
easy sophie-germain.
could've also be done by showing it's divisible by 5 but the first one is better for obvious reasons.
We should make sure that one of the factor is not equal to 1 , for final conclusion
Yes this one is necessary part
Sir meine kya logic lagaya
(2019)^even no = ____1
(even no. Se power raise karoge to 1 hi ayega units place me, similarly agar odd se raise karoge toh units place me hamesha 9 hi ayega)
4^odd no = ____4
(4 ko agar hum odd no. se raise karenge toh even place me 4 aana hi aana hai similarly even no se raise karoge toh apan ko units place me 6 ayega)
(2019)^even no + (4)^odd no = ____5
Answer Jo bhi ayega, 5 se divisible ho jayega, hence diye hue expression ka no prime nahi hai.
Sir please bring a number theory and geometry series..
After19-20year you revived my math, i think u r the best in math,
Is very simple problem for whom what had ever been prepared for mathematics olympiad
My method
2019^4 + 4× (16)^1009
16^1009 main last number 6 hi ayega and uske sath 4 multiple ke bad last word 4 ayega
And 2019^4 main last number is 1
Now ,
( ...........1) + (...............4)
= ...................5
i.e. divisible by 5
So the number is not prime number
❤️❤️
You're super cool, Sir. 🙏
1. Odd ^ ever or odd = is always odd.
2. Even ^ even or odd = is always even.
3. Even + odd = is always odd
With these three steps we get 2019^4 is odd . And 4^2019 is even .
Final answer even + odd is odd .
Sir we can also solve using cyclicity.....
The last digit of 2019^4 is 1 and that of 4^2019 is 4. N
So the last digit of this number is 5 and hence it is composites.
if cyclicity rule wasnt a thing , very brilliant solution sir hats off aapka channel dekhkar goosebumps aajate he
Taking modulo 5, 2019=-1 and 4=-1.
(-1)^4 + (-1)^2019 = 1-1 = 0
If you know some modular arithmetic, it's easy.
best method to solve this is to take mod 5. 2019= -1(mod 5), so 2019^4 = 1(mod 5). similarly, 4==-1(mod 5) so 4^2019= 1 (mod 5) hence, their sum== 0 (mod 5). the number is (obviously) not =5, but it is divisible by 5. hence, it is composite
If it is a prime so it should be of form 6K ± 1 so we have to prove
(2019)^4 + (4)^2019 ± 1 is divisible by 6
Proof:-
Expressions is even so we just have to prove that it is divisible by 3
2019 is divisible by 3 and using binomial on (4)^2019 we can cancle one and expression is divisible by 3 even so by 6 also
Hence proved as Prime
Ist part:(2019)⁴ will give a number with unit digit (i.e.,right se last wala digit)=1. (as 2019×2019×2019×2019=Some number with last digit 1,because last digit will be given by this multiple 9×9×9×9=6561)
2nd part:4²⁰¹⁹ will have the unit digit as 4( as 4^odd number, will have its unit digit always 4, e.g.,4³=64 and 4⁵=1024 and this is true for 4²⁰¹⁹ as well, as 2019 is an odd number.)
Now,by divisibility rule for divisibility by 5,we know that a number is divisible by 5 if the its unit digit is 5.
Since,the number 2019⁴+4²⁰¹⁹ have unit digit as (1+4) ,i.e., 5,it is divisible by 5.
Therefore,the given number is not a prime number,as it has a factor 5 other than 1 and itself.
P.S: Wrote this long and elaborated,so that any class student can understand.
Already watched it on BPRP channel.
Sir please one video on difference between coprime and relatively prime
Bhai best teacher nhi the best 🔥🔥🔥
Sir Issey acccha solution mere paas hai
Ye number 5 se divisible hai
2019⁴ + 4²⁰¹⁹ =( 2020 - 1 )⁴ + ( 5 - 1 )²⁰¹⁹
= 2020k + (-1)⁴ + 5t + (-1)²⁰¹⁹
= 2020k + 5t -1+1
= 2020k + 5t
Since rhs is divisible by 5 thus lhs will also be divisible by 5
Can u tell me , thus question is of which class ?
@@Nikola_tesla-52 idk but nowadays even 7th std students can also solve these kinds of questions
For expanding your equation u have to use binomial
This isn't the correct way.
@@parthamondal4650 look at my solution carefully, I've used binomial expansion but in short
You are the legend sir ji...........
Bhai gazab ka banda hai yaar mujhe PRMO ,RMO,IMO waale din yaad aa gye
❤Your method of teaching is extraordinary .May your interest in Maths be accelerated ,dear Bhannat .❤❤❤
Thanks sir I am in 1st grade but I am able to solve MIT research paper question
the McDonald's worker he is speaking of is the youtuber blackpenredpen 😄
An easier way to solve this: 2019^4 will end with 1(9*9*9*9) and 4^2019 will end with 4 so the sum's unit digit will have 5 which means it's divisible by 5.
One can use unit digit method to easily solve this question..last digit will be 5 so divisible by 5.Hence not a prime..👍👍🙏🙏
15,20000005,,,12578955,...........
Kya ye bhi prime h , Inka bhi unit digits 5 h
technically - there is a lacunae in your arguments. You have to show that both of your factors are greater than 1 (just to avoid p = 1xp is a factorization of a prime).
9^4 results in unit digit 1. And 4^2019 results in unit digit 4
4+1=5. Any number with unit digit 5 is non-prime. Simple !
Done
2019^1= 2019
2019^2= ....1
2019^4= ....1
Last digit 1 is not a prime number. So 2019^4 is not prime number
Similary, 4^1= 4
4^2= 16
4^3=...4
4^5=...4
So definitely 4^2019=....4
Now a number ended with 4 is definitely divisble by 2 (sometime 4&7 also). So 4^2019 is not a prime number.
Bohut bohut achchha hai apka teaching. I love math and I respect. you very much sir.
Best teacher
Best maths tutor
2019^4 has 1 as it’s last digit. This is because the last digit of 2019^4 is the same as the last digit of 9^4. Now, 4^2019 has it’s last digit as 4. This is because every odd power of 4 has the last digit 4, and 2019 is odd. So, the last digit of 2019^4 + 4^2019 is 5. Hence, the number is divisible by 5 and it isn’t prime.
OmG of Mathematics teacher 🙏
Guys first thing he is teaching an approach…to just get the answer y’all go and see shorts and don’t come here…come here to learn a concept or approach which may be used in other qs
sir i can do this in a more simpler way it consists of 3 steps if done by shortcut but still it's shorter and a lot simplere than the method you gave .
where can i send you my solution?
I used by finding unit digit . But this method is also good.
Just find unit place 5 and
Mini factor
1 , 5 and itself
Amazing
Sir you are realy genious and we don't knowmathes but your patience of solving problem and language is effectively to penitrate to watch videoes .
Sir I did it in much more simpler way.
I computed the unit digit of the number.
It was 5.
So the number is divisible by 5.
Can you explain me little bit?
@@Khushi-vz3yg yea see there are three actually two methods to solve this one which he said reduce the powers such that you have a unit digit and a cylicity order(if you know the concept well and good) or my modulo arithmetic simple just find remainders then if only if there is any integer that is divisible then the expression is not prime simple. And yes sofie germain identity the third method
@@angelinageorge2278 Thank you so much, You cleared all my doubts 😊😊
4 ki power odd gives 4 in unit digit …
9 ki power even gives 1 in unit digit
If we add both unit digits ->
It will be 4+1 = 5
Since unit digit of the sum is 5 .. therefore it is not a prime number
Sir According to your definition of prime number 1 is also a prime number because it is divisible by 1 and itself
First comment sir , you are my best teacher
2019^4 ka uniy digit 1 , 4^2019=2^4038, Unit digit is 4, 1+4=5 , it will divisible by 5 , hence not prime
May be प्राइम If A^2+ 2B^2- 2AB=1
Or (A-B)^2 = 1-B^2 which is
Only possible if
B=1 or 0 ie A = B or A=B +/- (1)
Which is not the case.
There is a much easier solution here.
2019 power 4 = ends in last digit with 1. Why? 9 x 9 x 9 x 9 ends with 1.
4 power 2019 = ends with 4. Why? 4 power odd number ends with 4 and even power ends in 6.
So last digit of the sum would 1 + 4 = 5 which is divisible by 5.
Not a prime.
Sir yeah toh direct ho raha tha
4 ki power 3 se divisible hai matlab unit digit 4 hogi
Aur 2019 ki power 4 hai toh unit digit is 1
When we add unit digit 4+1 we get 5
Which is divisible by 5..
Sir ye kaise pta chalega ki jo factors Bane Hain unki value 1 or 2019⁴ + 4²⁰¹⁹ nhi hai?
Agar Aisa hua th to vo prime number ho jayega......
how can i improve my imagination/observation of question
how i come to the first step of solution
Very simple but interesting
Sir ,another approach is one should determine last two digits of this number
Sir ,
Considering Vedic Mathematics, Consider only the Last Number
2019 Multiplied Even Times always ends with Number 1 (Ex 9 * 9 = 81) and Odd Times with number 9 ( Ex 9*9*9 = 729) Hence the First Number Multplied 4 Times should end with Number 1
Consider the Second Number ie 4 - Multiplied even times always ends with 6 and Odd Times ends with 4 ,Hence irrespective of all other Numbers The Total of this will end with Number 1 + 4 = 5 and Hence it is not a Prime Number
I hope i am right, Please correct if i am wrong
Isme Vedic Maths kahan se aayi? Basic Cyclicity ka concept hai. Har cheez mein Vedic nhi lgadete, dost.
Checking The Last number for certain problems itself is the concept of Vedic Mathematics
Examples
When Finding the Square root of any number say, 576, (The Square root number must end with Either 4 or 6)
Observing the Last Digit in Table of Nine Etc.9,8,7,6,5,4,3,2,1,0
Etc.,
Excellent. I could not understand the language (is it Hindi??) but i can follow it mathematically. How i wish you can make these video in English too.
yeah it is hindi,where are you from
just write 2019^4 as 3^4 *673^4 and write 4^2019 as (3+1)^2019 ,,,,, clearly we can taken 3^4 common after expanding the binomial... thus it is not prime with 81 as one of its factors..... i think this is simpler method.
Very very interesting videos you are making hats off
It can be found easily by modular congurency
Modular congurency+euler's formula
Other way :
unit digit
9 's power even : 8
4's power odd : 4
Unit digit of given no. : 8+4= 12 i.e. 2
A no. With unit digit two is not prime
Sir it is so simple
we can solve it by congurence modulo concept of divisibility
Aman sir is nostalgia pace series
just write (2019)^4 + (4)^2019 as (2020-1)^4 + (5-1)^2019
use binomial expansion
u'll have 5 multiplied by another number, so it is not prime
My instant instinct was Sophie Germaine but that unit digit trick is slick.
Sir aapki hair style bahut genius lagti hai 🥰
Matlab agar dusara minus wala factor 1 ke barabar ho gaya to kya hoga .
Prime ko is tarah se bhi to likh sakte
H like 13=13*1
11=11*1
Sir isse bhi easy way me ye solve krliya hai hmne... aap is comment pe reply kariya to mai aapke saath share karu!
Nice sir you do the things in a unique way
Gajab👌👌👌
Maza aagaya sir
Sir wo factors 1 aur 2019^4+4^2019 ho Sakta hai lekin Hume kaise pata chalega Jo factors hum jo factors nikalne wo 1aur it self nahi hai. If we don't have calculator.
Sir taking log base 10 both side se bhi ho jaayega
its clear that this expression is not a prime number because :-
1) we can see that both the number 2019 and 4 are composite numbers so the answer of any of their power will also be an composite number.
2) if we'll add the unit digit of both the expression then it will come out be equal to 5 which means that the number is divisible by 5.
I was thinking we could have solved it using logarithms. I don't care about it being prime or not. I just wanted to know what 4^2019+2019^4 would be
Sir, ye blackpenredpen channel ne 4 saal pehele karwaay tha. Is video k baad recommendations me aaya.
Wow sir ji
Math song kb ayega
Very good
Sir ek request hai aesi hi choti choti videos in ques ki bna dijiye jo jee main aur advance jyadatar puchta hai plzz sir
I hope ki mein limit se jyada nhi bola
Agr bola toh sorry sir
By Cyclicity concept
Unit digit of
4 when n is odd
4ⁿ=
6 when n is even
Here it is odd so Unit digit of 4²⁰¹⁹ is 4 --------(1)
Similarly
9ⁿ has unit digit 9 when n is odd and 1 when n is even
Here it is even so Unit digit of 2019⁴ is 1---‐-(2)
Net Sum has Unit digit 5 so
It is divisible by 5
Therefore it may be concluded that it is not a prime no.
Sir aaapppa ioqm ke liye rmo ke liye bhi course teyar kijia