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a1_2008sols

# a1_2008sols - 2 The University of Sydney MATH2068 Number...

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The University of Sydney MATH2068 Number Theory and Cryptography (http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/) Semester 2, 2008 Lecturer: R. Howlett Assignment 1 1. Determine the number of positive integers that are divisors of 15!, and their sum. Solution. 15! = (3 · 5) · (2 · 7) · 13 · (2 2 · 3) · 11 · (2 · 5) · 3 2 · 2 3 · 7 · (2 · 3) · 5 · 2 2 · 3 · 2 = 2 11 · 3 6 · 5 3 · 7 2 · 11 · 13. As proved in lectures, if the prime factorization of the number n is p k 1 1 p k 2 2 · · · p k r r then the values of the number of divisors function τ and the sum of divisors function σ are given by τ ( n ) = r i =1 ( k i + 1) and σ ( n ) = r i =1 p k i +1 i - 1 p i - 1 . Thus we see that τ (15!) = 12 · 7 · 4 · 3 · 2 · 2 = 4032 , and σ (15!) = 2 12 - 1 1 · 3 7 - 1 2 · 5 4 - 1 4 · 7 3 - 1 6 · 11 2 - 1 10 · 13 2 - 1 12 = 4095 · 1093 · 156 · 57 · 12 · 14 = 6686252969760 . 2. ( i ) Use Fermat’s Little Theorem to show that 2 2004 1 (mod 13), and hence determine the residue of 2 2008 modulo 13. ( ii ) Find the residue of 2 2008 modulo 11 and the residue of 2 2008 modulo 7. ( iii ) Noting that 1001 = 7 × 11 × 13, use your answers to the first two parts to find the residue of 2 2008 modulo 1001. Solution. Fermat’s Little Theorem says that a p - 1 1 (mod p ) if p is prime and p a . So 2 12 1 (mod 13). So (2 12 ) 167 1 167 1 (mod 13); that is, 2 2004 1 (mod 13). Hence 2 2008 = 2 2004 × 2 4 1 × 2 4 = 16 3 (mod 13). Hence 3 is the residue of 2 2008 modulo 13. By Fermat, 2 10 1 (mod 11), and so 2 2000 = (2 10 ) 200 1 200 1 (mod 11). So 2 2008 1 × 2 8 = 256 3 (mod 11). Similarly, 2 6 1 (mod 7); so 2 2004 1 (mod 7) and 2 2008 2 4 2 (mod 7). So the residues of 2 2008 modulo 11 and 7 are 3 and 2 respectively. 2 Write 2 2008 = N for short. We have shown that N 3 mod 13 , (1) N 3 mod 11 , (2) N 2 mod 7 . (3) Now (1) gives N = 13 k + 3 for some integer k , and then (2) gives 13 k + 3 3 (mod 11). So 13 k 0 13 × 0 (mod 11), and by coprime cancellation it

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