Fermat's Little Theorem

[@rishith4346]

The simplest way to solve P=11, a=5 *We need to check P should always be a prime number ! * Formula=] a^p-1 congruent (1modp) 5*11-1=(1mod11) 5*10=(1 mod11) The rule for fermat Little problem is The number which is in the left should always be larger than the number which is right of equal to sign So in this case its not satisfying the given condition so we divide it using modular methods The number 5 is subtracted, with p and written as -6 on the right hand side... -6*10=1mod11 Now we split the power of ten As (2*5) (-6^2)(*5)=1mod11 36(*5)=1mod11 Now 36mod11=3 3(*5)=1mod 11 Mod of 3*5 mod 11= 1 Now 1=1mod 11 Its congruent P=11 and a=5 proved !

[@shivamkandhare8897]

Fermat's theorem holds true for p=11, a=5 And thanks for this presentation!

[@sanika6916]

NESO ACADEMY YOU ARE LITERALLYY A SAVIOUR!!!! THANKYOU SO MUCH.

[@rajeshprajapati4863]

5^10 ≡ 1 mod 11 5^(2*5) ≡ 1 mod 11 3^5 ≡ 1 mod 11 243 ≡ 1 mod 11 (Valid) Therefore, FLT holds true for p=11 and a=5.

[@sivasaigunta8924]

3 ala vachindi bro miku

[@ashutoshpatil7330]

@@sivasaigunta8924 how ??

[@janpost8598]

​@@ashutoshpatil7330 5^(2*5) = (5^2)^(5) now 5^2 ≡ 3 mod 11. 5^(2*5) ≡ 1 mod 11 (5^2)^(5) ≡ 1 mod 11 (3)^(5) ≡ 1 mod 11 Hence 3^5 ≡ 1 mod 11

[@lakshitapatnaikuni2143]

12 is congurent to 1 (mod 11) which shows Fermat's theorem holds true for p=11,a=5 Thanks for the explaination:)

[@Stocks_Technical_Analyser]

Your point shows some deep knowledge. Can you please explain it ?

[@dsalgos]

@@Stocks_Technical_Analyser I tried solving step by step and one of them is 12 = 1 (mod 11) 5^10 = 1 (mod 11) => 3^5 = 1 (mod 11) => 81 x 3 = 1 (mod 11) => 12 = 1 (mod 11) Proved.

[@user_unknown04]

Answer for the HW problem in the easiest way possible: 5^10 = 1 (mod 11) will first find 5^1 mod 11 = 5 now 5^2 mod 11 = 3 now 5^4 is simply 5^2 x 5^2, that is 3 x 3 = 9 now 5^8 is simply 5^4 x 5^4, that is 9 x 9 = 81 we need 5^10 now, we can write it as 5^8 x 5^2, which is 81 x 3, gives us 243 now 243 mod 11, will result in 1 !. Hence fermat's little theorem satisfied.

[@Ritesh_kumar773]

Given p=11 ,a=5 a^p-1 = 1(mod11) 5^11-1 =1(mod11) 5^10 =1(mod11) -6^10 =1(mod11) {5-11= -6} -6^5*2 =1(mod11) {6^5=7776 is ÷ by 11 and remainder is 10 10^2 =1(mod11) 100 =1(mod11) {100÷11 and remainder should be 1}

[@Nejiyanesrin-1]

P=11 a=5 1)p is prime 2)a is an integer not divisible by p 11 a^p-1=1modp 5^10=1mod11

[@thanmaijami8962]

Fermat's theorem holds true for p=11, a=5

[@senthilnathan3200]

How to do it?

[@senthilnathan3200]

How to do it?

[@BhojpuriWorld14314]

how

[@neelamyadav7609]

@@senthilnathan3200 5^10=1(mod 11) 5^5*2 = 1(mod 11) (31125)^2 = 1(mod 11) 9765625 =1(mod 11) after 887784 times it gives the reminder 1 so this is value is true.

[@neelamyadav7609]

it anyone have more simple way please suggest.

[@sufiyansyed2846]

Excellent sir ❤❤❤❤

[@markuche1337]

Very great course thank you Neso Academy ❤

[@СанжарАлманов-т3с]

Thank you man, really helpful video!

[@G.VISHNUTEJA24BSD702]

excellent explanation

[@sahilanand30]

3:28 anyone noticed Indian flag? 🇮🇳

[@monicabattacharya6416]

please complete Database management systems fastly. I have in my current semester 😩

[@BhojpuriWorld14314]

🖐🏼

[@dharavathsrikanth8477]

😅😅😅

[@manas-tapas-Art]

Hii..

[@Junedwrites]

Love you sir 🤗 , from🇮🇳🇮🇳🇮🇳

[@alan_johnson_]

He is indian

[@danieldanmola8266]

Fermat little theorem holds for a= 5 and p=11 By the Fermat little theorem a^(p-1)=1(mod p) 5¹⁰=1(mod 11) Let's check using the Euler phi totient function Which states that a^phi(n)=1(mod n) Provided that GCD of (a, n)=1 And in this case a=5 and n=11 Thus ::: 5^phi(11)=1(mod 11) Since the phi(11)=11-1=10 5¹⁰=1(mod 11) Thus Fermat little theorem holds

[@victorymindset-1]

can someone say how -2^(4*3) congruent 1 (mod 13) changes to 3 ^ 3 congruent 1 (mod 13)

[@balajimetla3886]

nice explanation sir. continue like in this way

[@chamin111]

Thank you sir, nice explanation

[@ANMOLKUMAR-ix8di]

p=11 , a=2 , It holds fermet's little theorem (homework question answer)

[@AkashKumar-ff6mx]

THANKS FOR THIS VIDEO

[@IMdrummerTab]

This is great! Thanks❤

[@AkashKumar-ff6mx]

Thanks for this video

[@shivambhardwaj1137]

Fermat theorem true value for p=11 and a=5

[@BookwormUnited]

For home work question I guess p=11 and a=5 so we are using a^p-1=1 mod p 5^11-1=1mod 11 5^10=1 mod 11 then 5 is smaller than 11 so we are reduce 11 from 5 is 6 -6^10=1 mod 11 -6^5*2 = 1 mod 11 here 6^5 = 7776 then divide by 11 so we will get remainder 10 10^2 = 1 mod 11 100 = 1 mod 11 here 100/ mod 11 = 1 so 1= 1 hence proved .. if it's correct like this comment ..😊

[@jesusbosch2720]

Mmm I am not satisfied this video. I see exercises in my classbook that ask me to calculate thibgs like 97 power 200 + 97 power 201 mod 13 (or similar operations with huge exponents). I am explicitly asked to use fermat’s little theory (not the modular exponentiation method used in previous videos). Any advice please?

[@ajay.2461]

P=15 and a=4 it's not holding false it's holding true

[@aathavang5505]

How to solve by using Fermat's theorem, if the question is 27^452 mod 113?

[@Agenkia]

Where is the minus in -2^4*3?

[@gaurav561crazy5]

Start the series of calculus

[@garunkumar8921]

Fermat's Theorem hold true for p=11 and a=5 sir

[@tiamiyuyusuff234]

Fermat's little theorem holds for p=11 and a=5.

[@KuladeepGompa]

why did you take -2 in the place of 11 in the second example

[@love_in_nature8616]

Because 11 mod 13 is 11 or -2, among them min(abs(-2, 11)) is considered... So it is -2.

[@nasserkhamis1690]

is there a way to check my answer using a calculator ?

[@தமிழன்-ர8ள]

Hold true p=11 and a=5

[@angelicaaquino2159]

What if p is not prime number?

[@zanti4132]

If p is not prime, you will usually get a^(p-1) ≠ 1 mod p, but not always. For example, if a = 2 and p = 11 × 31 = 341, you can show 2³⁴⁰ = 1 mod 341 as follows: 2³⁴⁰ = (2¹⁰)³⁴ = 1024³⁴. However, 1024 = 1 mod 341, therefore 2³⁴⁰ = 1³⁴ mod 341 = 1 mod 341, q.e.d. Hence, once in a great while a composite number gives the same result as a prime number for certain values of a, making the composite number what's known as a pseudoprime in base a. There are even composite numbers for which a^(p-1) = 1 mod p regardless of the value chosen for a. These numbers are known as Carmichael numbers, the smallest of which is 561.

[@mihirmathur5855]

Fermat's Theorem holds true for p=11 and a=5

[@kowshik_reddy_iguturi]

5^10=1mod11 Rem=1 True

[@udayakiran8965]

Fatmet theorem hold ? P=3, a=3

[@vrajpatel8302]

Yes

[@KaivalyaGawande]

Who are from class 11th 😅

[@idledogtunes]

Me

[@BRIDGETNGATO]

Me 😂😂❤❤🇨🇲🤴

[@mohankumarmuthukumar2949]

Fermat theorem true for given value

[@Utkarshkushwaha-ld8xh]

Yes bcz remainder is 1 Buddy

[@vaibhavsharma9318]

5^10 = -6^10 = -6^2*5 = 12^5 = 1^5 If you think how 12 becomes 1 ,the reminder of 12÷11 or 11÷12 😅 is 1 Then the reminder of 1÷11 or 11÷1 🤫 is 1 So the fermat theorem is true Hence proved

[@explorer4793]

how 6^2 becomes 12

[@vaibhavsharma9318]

​@@explorer4793 I can't remember how it becomes 12 , sorry...

[@hanishvenkat9496]

super

[@chandukmit]

True, tqs

[@karunkarna9397]

nice

[@raghavsingh4257]

1

[@BhojpuriWorld14314]

fermat's theorem does not hold ture for p=11 and a=5

[@neelamyadav7609]

how?

[@SmirthikaShri]

Y88

[@ayeshasiddika8552]

1