Obviously, doesn't work for more complex cases, but i just kept multiplying by 3, only saving the last two digits, until it repeated the pattern of the last two digits, which happened at 3^21. So, 2024 mod 20 is 4 and 3^4 is 81. In this case, seemed a lot faster.
My method of solving this: 3^2024 = 9^1012 = 81^506 = 6561^253 = 6561*6561^252. Since only the last two digits of the numbers multiplied together affect the last two digits of the product, this has the same last two digits as 61*61^252 = 61*3721^126. This in turn has the same last two digits as 61*21^126. We continue this reasoning while reducing the size of the exponent until we have 61*41^3 = 2501*41^2 which has the same two digits as 1*1681 = 1681. So, the tens digit of 3^2024 is 8.
@@maxhagenauer24 Because for example the third digit can only effect the third or higher digits. You'd have to multiply 500 by 0.1 to have 50 for example, but all multiplicants here are integers so that doesn't happen. So, 500 * 1 = 500, 500 * 2 = 1000 etc. No effect on first 2 digits.
Not sure what you mean ... 81*81=6561 => 61; 61*81=4941 => 41; 41*81=3321 => 21; 21*81=1701 => 01 So the system repeats every 81^5. 506=1 mod 5 so ans =1*81 = 81.
For phi function , there is equation when is phi(n ) = n× (p1-1)/p1 ×(p2-1)/p2 ..(pk-1)/pk Where p1,p2,...,pk are the prime factor of n For n=100, the prime factors are 2 and 5 Phi(100) = 100×1/2× 4/5 = 40 When we do mod 100 the exponent will be mod phi(100) =40 3^2024 (mod 100 ) = 3^(2024 mod 40) mod 100 = 3^24 mod 100 =(3^6)^4 mod 100 = 29^4 mod 100 = (29^2)^2 mod 100 = 41^2 mod 100 =81 The tenth place digit is 8
I think you mean the last two members - the third from last would have 10^2 multiplier so makes no contribution. So you get 1012*(-1)^1011*10 + (-1)^1012 = -10120+1 = -10119 = -19 mod 100 =81 mod 100.
0:12 ok there have to be a way to find it without calculate the whole number i allready have an idear what could be an option lets see 0:41 yes sth. like this 1:00 sounds great 1:14 that was my first question! 1:35 coprime ok 1:53 yes please i got never tought sth. about number teorie 2:00 ok i should research this video but the problem is you make to much videos in to less time it is even difficult for me finding time to watch and comment any actuel videos :D 4:10 ok the halv is understood by me 6:56 that was my verry first idea to split it 8:59 allright! 10:11 i would had choosen the same way if i had to calculate that without a calculator 11:59 nice video again my only problem is that my english is not as good as nessessery so i can not understand anything but luckyly enought for commenting! LG K.Furry
There is a formula for phi based on the prime factorization of the number. phi(n) = n (1-1/p1)(1-1/p2)...(1-1pn) where p1, p2..pn are the prime factors of n. So phi(100) = 100(1/2) (4/5) = 40. Alternatively you can use the principle of inclusion-exclusion to count them: 100 - (100/2 + 100/5 -100/10) =40 . But the formula above is easier when the number has a lot of prime factors. Since phi is multiplicative you can also exploit that to find phi for larger n: phi(ab) = phi(a) phi(b) when gcd(a,b)=1
I like how you always pick questions which are trickier then Olympiad questions
With these questions you need to know the trick
Obviously, doesn't work for more complex cases, but i just kept multiplying by 3, only saving the last two digits, until it repeated the pattern of the last two digits, which happened at 3^21. So, 2024 mod 20 is 4 and 3^4 is 81. In this case, seemed a lot faster.
i did the same, but then i studied for a couple hours how modular arithmetic works and i could easily solve it for 11^2024
unless i did some kind of error, it should be 41
There is also reduced totient function, λ(n).
λ(100) = 20
My method of solving this: 3^2024 = 9^1012 = 81^506 = 6561^253 = 6561*6561^252. Since only the last two digits of the numbers multiplied together affect the last two digits of the product, this has the same last two digits as 61*61^252 = 61*3721^126. This in turn has the same last two digits as 61*21^126. We continue this reasoning while reducing the size of the exponent until we have 61*41^3 = 2501*41^2 which has the same two digits as 1*1681 = 1681. So, the tens digit of 3^2024 is 8.
How do you know that only last 2 digits of 6561 effect the last 2 digits of the product?
@@maxhagenauer24 Because for example the third digit can only effect the third or higher digits. You'd have to multiply 500 by 0.1 to have 50 for example, but all multiplicants here are integers so that doesn't happen. So, 500 * 1 = 500, 500 * 2 = 1000 etc. No effect on first 2 digits.
@pojuantsalo3475 Ok, so why did you even split 6561^253 into 6561*6561^252?
(3^4)^506
(81)^506
Unit digit will be 1 always in this case
And ten digit will be unit digit of 8×6=8
So 81 is answer.
Can you tell us how it works?
Not sure what you mean ... 81*81=6561 => 61; 61*81=4941 => 41; 41*81=3321 => 21; 21*81=1701 => 01
So the system repeats every 81^5. 506=1 mod 5 so ans =1*81 = 81.
For phi function , there is equation when is phi(n ) = n× (p1-1)/p1 ×(p2-1)/p2 ..(pk-1)/pk
Where p1,p2,...,pk are the prime factor of n
For n=100, the prime factors are 2 and 5
Phi(100) = 100×1/2× 4/5 = 40
When we do mod 100 the exponent will be mod phi(100) =40
3^2024 (mod 100 ) = 3^(2024 mod 40) mod 100
= 3^24 mod 100
=(3^6)^4 mod 100
= 29^4 mod 100
= (29^2)^2 mod 100
= 41^2 mod 100
=81
The tenth place digit is 8
Or we can take 3^2024 = 9^1012 = (10-1)^1012 ,expand and take the last three members and compute the last two figures from them.
I think you mean the last two members - the third from last would have 10^2 multiplier so makes no contribution.
So you get 1012*(-1)^1011*10 + (-1)^1012 = -10120+1 = -10119 = -19 mod 100 =81 mod 100.
@@franciscook5819 Right you are, buddy, two members are enough :-)
0:12 ok there have to be a way to find it without calculate the whole number i allready have an idear what could be an option lets see
0:41 yes sth. like this
1:00 sounds great
1:14 that was my first question!
1:35 coprime ok
1:53 yes please i got never tought sth. about number teorie
2:00 ok i should research this video but the problem is you make to much videos in to less time it is even difficult for me finding time to watch and comment any actuel videos :D
4:10 ok the halv is understood by me
6:56 that was my verry first idea to split it
8:59 allright!
10:11 i would had choosen the same way if i had to calculate that without a calculator
11:59 nice video again my only problem is that my english is not as good as nessessery so i can not understand anything but luckyly enought for commenting!
LG K.Furry
There is a formula for phi based on the prime factorization of the number.
phi(n) = n (1-1/p1)(1-1/p2)...(1-1pn)
where p1, p2..pn are the prime factors of n.
So phi(100) = 100(1/2) (4/5) = 40.
Alternatively you can use the principle of inclusion-exclusion to count them: 100 - (100/2 + 100/5 -100/10) =40 . But the formula above is easier when the number has a lot of prime factors. Since phi is multiplicative you can also exploit that to find phi for larger n:
phi(ab) = phi(a) phi(b) when gcd(a,b)=1
Respect from Cameroon 3:55
But I think for all n>0
Is zero not a digit......?
n starts from 1
tens digit of 3^2024 is 8
Assume that positive number d is the least value such that
a^d = 1 (mod n) then d is divisor of Euler's totient function
last 2 digits of 3^2024 is 81.
hi, just wanted to say that belphegor prime vid is pvt, and i was kinda interested...so maybe change the permissions?
gracias
Tens digit or tenth digit?
Ohhhh digit in penultimate position. I'd never have guessed
Obvious to me
Whats a tens digit
Second digit from the right
白いブランコって、blanco自体がスペイン語で白いだから、くどいと父親が言ってたのを思い出しました。
I think this should read "tenth".
❤
phi
First