15. Using Fermat's Theorem, compute the remainder when 2^(2^23)+1 were divided by 29 . ( .5 pts)

Name: 15. Using Fermat's Theorem, compute the remainder when 2^(2^23)+1 were divided by 29 . ( .5 pts)
Uploaded: 2019-10-29T00:06:43-08:00
Duration: 2 min 11 s
Channel: James Chok
Description: 15. Using Fermat's Theorem, compute the remainder when 2^(2^23)+1 were divided by 29 . ( .5 pts)

First, let's find the remainder when (2^{23}) is divided by 28. We can use the fact that (2^6

Question

15. Using Fermat's Theorem, compute the remainder when \( 2^{\left(2^{23}\right)}+1 \) were divided by 29 . ( \( \left.5 \mathrm{pts}\right) \)

          15. Using Fermat's Theorem, compute the remainder when \( 2^{\left(2^{23}\right)}+1 \) were divided by 29 . ( \( \left.5 \mathrm{pts}\right) \)

15. Using Fermat's Theorem, compute the remainder when 2^(2^23)+1 were divided by 29 . ( .5 pts)

Added by Abril

Discrete Mathematics and its Applications

Kenneth Rosen 8th Edition

Chapter 4

Instant Answer

Solved by Expert James Chok

10/29/2019

Step 1

We can use the fact that \(2^6 \equiv 1 \pmod{29}\) (which can be easily verified by calculating \(2^6\) and checking that it leaves a remainder of 1 when divided by 29). So, we have \(2^{23} = 2^{6 \cdot 3 + 5} = (2^6)^3 \cdot 2^5 \equiv 1^3 \cdot 2^5 \equiv 2^5 Show more…

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