Question

The last two digits of 7 to the power 81 are​

Answer 1

Step-by-step explanation:

72=49=50−172=49=50−1

⟹74=(72)2=(50−1)2=502−2⋅50⋅1+1≡1(mod100)⟹74=(72)2=(50−1)2=502−2⋅50⋅1+1≡1(mod100)

⟹781=7⋅(74)20≡7⋅120(mod100)≡7

⟹781=7⋅(74)20≡7⋅120(mod100)≡7

In general, Euler's totient theorem or Carmichael's theorem can be used in such cases.

Here

ϕ(100)=10⋅ϕ(10)=10⋅ϕ(2)⋅ϕ(5)=40ϕ(100)=10⋅ϕ(10)=10⋅ϕ(2)⋅ϕ(5)=40

⟹740≡1(mod100)⟹740≡1(mod100) as (7,100)=1(7,100)=1

So, 781=(740)2⋅7≡12⋅7(mod100)≡7781=(740)2⋅7≡12⋅7(mod100)≡7

and λ(100)=λ(100)=lcm(λ(25),λ(4))=(λ(25),λ(4))=lcm(20,2)=20(20,2)=20

So, 781=(720)4⋅7≡14⋅7(mod100)≡7