{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW10-solutions

# HW10-solutions - EECS 203 DISCRETE MATHEMATICS Homework 10...

This preview shows pages 1–2. Sign up to view the full content.

EECS 203: DISCRETE MATHEMATICS Homework 10 Solutions 1. (4 points) Section 3.5, Problem 8. Consider the n integers starting with ( n +1)!+2 , ( n +1)!+3 , . . . , ( n +1)!+( n +1) where n Z + . Notice that k | ( n +1)! for 1 k n +1. Thus, 2 | [( n +1)!+2], 3 | [( n +1)!+3] , . . . , ( n +1) | [( n +1)!+( n +1)] and each of these numbers is composite. 2. (4 points) Calculate 3 530 mod 53. Indicate which theorems or properties of modular arithmetic you are using to arrive at an answer. Since 53 is prime we may apply Fermat’s little theorem. 3 530 (3 52 ) 10 · 3 10 (mod 53) [(1) 10 mod 53] · [3 10 mod 53] (mod 53) [(3 2 ) 5 mod 53] (mod 53) [(9) 2 mod 53][(9) 3 mod 53] (mod 53) [28][40] (mod 53) [56][20] (mod 53) [3][20] (mod 53) 7 (mod 53) 3. (4 points) Prove that 2 a + 1 and 4 a 2 + 1 are relatively prime. Let’s assume a N so that 2 a + 1 and 4 a 2 + 1 are non-negative. We use the Euclidean algorithm to ﬁnd gcd(2 a + 1 , 4 a 2 + 1). Notice that (2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

HW10-solutions - EECS 203 DISCRETE MATHEMATICS Homework 10...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online