In the (88⁷ mod 187) it's easier if you realize that both numbers are divisible by 11, i.e. 88=8*11 and 187=17*11. So one needs to solve a much easier problem which is (8⁷ × 11⁶ mod 17) ≡ (8¹8²8⁴ × 11²11⁴ mod 17) ≡ 8(-4)(-1) × (-2)(4) mod 17 ≡ (32) × (8) mod 17 ≡ (-2) × (8) mod 17 ≡ (-16) mod 17 ≡ -(-1) mod 17 ≡ 1 mod 17. Because N = D*Q + R ⇒ R = N - D*Q. If we divided the RHS by 11, for the equation to remain true the LHS (the remainder) must be divided by 11 too. That means that if the remainder of (8⁷ × 11⁶ ÷ 17) = 1 then the remainder of (88⁷ ÷ 187) = 1 × 11. Therefore (88⁷ mod 187) ≡ 11 mod 187
sir your session is excellent.... no way to say about your session... amazing teaching ever in my life. It is helping my teaching professional. sir one request please update upcoming session soon... we are waiting..... Thank you so mauch.
Thank so much. You are amazing. Ive tried all sort of materials and videos but couldn't get a better explanation like you did in this video. You are an outstanding teacher.
for calculator sake u can change the last step to 175^209 mod 1000 = [(175^128) * (175^64) * (175^16) * (175^1)] mod 1000 = (375*375*375*175) mod 1000 = 375 Ans 625 mod 1000 == 375 mod 1000 😁
11^2 mod 187 will be -66 or 121 11^4 mod 187 will be 55 or -132 11^8 mod 187 will be 33 or -154 11^16 mod 187 will be -33 or 153 So 11^23 will be 88 or -99
11^23 mod 187 => 88; When i solved first i got 108; due to i used -66 instead of 121; minus confused me; at the 2nd example -> i got pretty straightforward answer -> 375; as it's everywhere 625
11^2 mod 187 will be -66 or 121 11^4 mod 187 will be 55 or -132 11^8 mod 187 will be 33 or -154 11^16 mod 187 will be -33 or 153 So 11^23 will be 88 or -99
In video instructor says "Don't use the calculator" but i coded python program for solving example haha. python code. exp = 1 num = 88 ans = 1 while (exp != 8): ans *= num exp = exp + 1 mod = ans % 187 print(mod)
My brain clicked at 4:40 while I was applying this logic to a particularly difficult problem (for myself... as I've never done this before) The problem I was solving was 2116^17 mod(3233) if anyone was wondering... That was my **first** problem I tried to solve. Yes, I do hate myself.
3 mod 10 gives last digit of i.e 3 itself, 12 mod 10 leads 2 which is its last digit, 2457 mod 100 leads to 57 which leads to 57 as its last two digits
In the (88⁷ mod 187) it's easier if you realize that both numbers are divisible by 11, i.e. 88=8*11 and 187=17*11.
So one needs to solve a much easier problem which is (8⁷ × 11⁶ mod 17) ≡ (8¹8²8⁴ × 11²11⁴ mod 17) ≡ 8(-4)(-1) × (-2)(4) mod 17 ≡ (32) × (8) mod 17 ≡ (-2) × (8) mod 17 ≡ (-16) mod 17 ≡ -(-1) mod 17 ≡ 1 mod 17.
Because N = D*Q + R ⇒ R = N - D*Q.
If we divided the RHS by 11, for the equation to remain true the LHS (the remainder) must be divided by 11 too.
That means that if the remainder of (8⁷ × 11⁶ ÷ 17) = 1 then the remainder of (88⁷ ÷ 187) = 1 × 11.
Therefore (88⁷ mod 187) ≡ 11 mod 187
You're awesome
Well but the answer is 88
sir your session is excellent.... no way to say about your session... amazing teaching ever in my life. It is helping my teaching professional. sir one request please update upcoming session soon... we are waiting..... Thank you so mauch.
Thank so much. You are amazing. Ive tried all sort of materials and videos but couldn't get a better explanation like you did in this video. You are an outstanding teacher.
1. 11^23 mod 187 = ?
11^1 mod 187 = 11
11^2 mod 187 = [(11^1) * (11^1)] mod 187 = (11*11) mod 187 = 121
11^4 mod 187 = [(11^2) * (11^2)] mod 187 = (121*121) mod 187 = 55
11^16 mod 187 = [(11^2) * (11^2) * (11^2) * (11^2)] mod 187 = (55*55*55*55) mod 187 = 154
11^23 mod 187 = [(11^16) * (11^4) * (11^2) * (11^1)] mod 187 = (154*55*121*11) mod 187 = 88 Ans
2. 175^209 mod 1000 = ?
175^1 mod 1000 = 175
175^2 mod 1000 = [(175^1) * (175^1)] mod 1000 = (175*175) mod 1000 = 625
175^4 mod 1000 = [(175^2) * (175^2)] mod 1000 = (625*625) mod 1000 = 625
175^8 mod 1000 = [(175^4) * (175^4)] mod 1000 = (625*625) mod 1000 = 625
175^16 mod 1000 = [(175^8) * (175^8)] mod 1000 = (625*625) mod 1000 = 625
175^32 mod 1000 = [(175^16) * (175^16)] mod 1000 = (625*625) mod 1000 = 625
175^64 mod 1000 = [(175^32) * (175^32)] mod 1000 = (625*625) mod 1000 = 625
175^128 mod 1000 = [(175^64) * (175^64)] mod 1000 = (625*625) mod 1000 = 625
175^209 mod 1000 = [(175^128) * (175^64) * (175^16) * (175^1)] mod 1000 = (625*625*625*175) mod 1000 = 375 Ans
how did you calculate (625*625*625*175) my calculator is not processing it🥲🥲
@@freakpunk8178 coming zero .. why
for calculator sake u can change the last step to
175^209 mod 1000 = [(175^128) * (175^64) * (175^16) * (175^1)] mod 1000 = (375*375*375*175) mod 1000 = 375 Ans
625 mod 1000 == 375 mod 1000
😁
please upload videos of the network security daily 🙏🏼 because we are having this subject in our current semester
This video is much helpful for solving modular related problems like return with mod 1000000007
1. 99 mod 187.
2. 375.
Also, I really liked the fact that it was told to compare a and p-a mod p in order to do a fast calculation.
11^2 mod 187 will be -66 or 121
11^4 mod 187 will be 55 or -132
11^8 mod 187 will be 33 or -154
11^16 mod 187 will be -33 or 153
So 11^23 will be 88 or -99
WATCHED THIS 6 TIMES, STILL HAVE NO CLUE
That's sad
Sir, I think you should also provide answers of the homework ques …it will be easier than.
thank you sir, you are a life saver.
Beautiful instruction. Just remember: 26/6 is 6 into 26: and 6x26 is 6 times 26, or 6 by 26.
I finally understand this. Thank you!
11²³ mod 187 = 88
Last 3 digits of 175²⁰⁹ = 375
i think it would be 625 instead of 375
Homework Question Solution:
Ans. 1. (11)^23 mod 187 = 88
Ans.2. last three digit of (175)^209 will be 375
hey i confused at last steps can i get the hw steps
please complete this lectures.
Thank you very much
11^23 mod 187 => 88; When i solved first i got 108; due to i used -66 instead of 121; minus confused me;
at the 2nd example -> i got pretty straightforward answer -> 375; as it's everywhere 625
11^2 mod 187 will be -66 or 121
11^4 mod 187 will be 55 or -132
11^8 mod 187 will be 33 or -154
11^16 mod 187 will be -33 or 153
So 11^23 will be 88 or -99
88 or -99 1st one
11:32 how we got -51 form -551??
@neso
1st answer is 88
No answer is -88
But in answer we can write 99mod187
88 and 375 are the correct answers.
You are correct, but how did you solve for the second one please?
In video instructor says "Don't use the calculator" but i coded python program for solving example haha.
python code.
exp = 1
num = 88
ans = 1
while (exp != 8):
ans *= num
exp = exp + 1
mod = ans % 187
print(mod)
Try writing for large numbers in c++ or java
@@shivasai7707 Segmentation error goes BRRRR
Best. Thanks a lot, sir.
Thank you 💫💫💫
in example 2 how you are calculating negative values ?
answer of 1st question is -> 88
answer of 2nd question is -> 375
1. 88, 2. 375
My brain clicked at 4:40 while I was applying this logic to a particularly difficult problem (for myself... as I've never done this before)
The problem I was solving was 2116^17 mod(3233) if anyone was wondering... That was my **first** problem I tried to solve. Yes, I do hate myself.
He keeps saying don't use calculator while im seeing this to write into a calculator
Anyway very good explanation
I've been asked to solve: 78^859 mod 1829.
Even after watching this video I'm having a hard time as to how to step through this problem. Any advice?
Why do you take mod 100 in the example 29^5 (mod100 )??
I can't get it that how are you taking the base in negative sign?
and why?
how did you change the value?
Sir can you please point me to a website or document where I can find different practice questions?
Will you please explain how to solve 3^5^3^5 mod35
And 2nd 375 😊
Thankyou!
please 🙏🏼🙏🏼🙏🏼complete 🙏🏼😢❤😭😫java programming 🙏🏼😢course
Please teach c++
It is available.
My ans for 11^23 mod 187 is 3. Is it correct? 3 or 88mod17
And my second ans is 3or - 5..
I think I am doing it wrong. Can anyone verify?
@@ruhinpatel7257 i used calculator and i got 11^23 mod 187 = 88
Sir you did not talk about calculation of mod ? Why
thnx
and how exactly are we supposed so think those different strategies ourselves? apparently there is not one-size fits all
Q1. 99 or -88 Q2. 375
what is the answer for 11^23 mod 187??
17
88
how can we find 30^1000 using binary exponentiation
88 and 375
Second answer is 625
216^2011 mod 3127
bro how did u get -59 ?
Sir why u put the last digit sum using mod 100
3 mod 10 gives last digit of i.e 3 itself, 12 mod 10 leads 2 which is its last digit, 2457 mod 100 leads to 57 which leads to 57 as its last two digits
ultenative
hw ques1=88?
3^48 mod 257???
How 841%100 = -59 ?
841 mod 100 = 41 mod 100 = 41 0r -59
idk where I am wrong or there is an error (5929 mod 187=12)
Do you mean that 539 mod 17 = 12
Answer to H.W :
I used Simple Calculator.
1. 88
2. 375
Anyone how 7744 mod 187 = 77
When we divide 7744 with 187 we will get remainder as 77
187*41=7667
7667-7744=77..
印度人 我的超人
Zero and zero sadly i give wrong answer.
Such a shiity explain ATI