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Ivahew [28]
4 years ago
6

Determine n between 0 and 19 such that (2311)(3912) ≡ n mod 20.

Mathematics
1 answer:
sleet_krkn [62]4 years ago
3 0

You can write 2311 and 3912 in the form 20q+r:

2311=115\cdot20+11

3912=125\cdot20+12

Then

2311\cdot3912=(115\cdot20+11)(125\cdot20+12)

2311\cdot3912=115\cdot125\cdot20^2+(11\cdot125+12\cdot115)\cdot20+11\cdot12

Taken modulo 20, the terms containing powers of 20 vanish and you're left with

2311\cdot3912\equiv11\cdot12\equiv132\pmod{20}

We further have

132=6\cdot20+12

so we end up with

2311\cdot3912\equiv12\pmod{20}

and so n=12.

###

If instead you're trying to find 2311^{3912}\pmod{20}, you can apply Euler's theorem. We can show that \mathrm{gcd}(2311,20)=1 using the Euclidean algorithm. Then since \varphi(20)=8, and 8 divides 3912, we have

2311^{3912}\equiv2311^{489\cdot8}\equiv(2311^{489})^8\equiv1\pmod{20}

To show 2311 and 20 are coprime:

2311 = 115*20 + 11

20 = 1*11 + 9

11 = 1*9 + 2

9 = 4*2 + 1   =>  gcd(2311, 20) = 1

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