Fermat's Little Theorem 

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 !

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

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.

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

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}

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

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Thank you man, really helpful video!

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

Thank you sir, nice explanation

nice explanation sir. continue like in this way

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

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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 ..😊

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

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THANKS FOR THIS VIDEO

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

This is great! Thanks❤

Yes

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

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

Start the series of calculus

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

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

What if p is not prime number?

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

Hold true p=11 and a=5

Who are from class 11th 😅

5^10=1mod11 Rem=1 True

Fatmet theorem hold ? P=3, a=3

super

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

True, tqs

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nice

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

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1st comment 😀😀

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

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