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571p01 - x a k(mod m k A system of covering congruences...

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MATH 571 ANALYTIC NUMBER THEORY I, SPRING 2011, PROBLEMS 1 Due 18th January 1. Solve where possible (i) 77 x 84 (mod 143); (ii) 77 x 84 (mod 147); 2. Find the last digit of 7 2001 and of 13 143 . 3. Suppose that f : N Z is a totally multiplicative function with f ( n ) = 0 or ± 1. Prove that m | n f ( m ) 0 and m | n 2 f ( m ) 1 . 4. Suppose that 0 a < m and A denotes the non-negative integers in the residue class a (mod m ). Show that if z C and | z | < 1, then n A z n = z a 1 z m . 5. (L. Mirsky and D. J. Newman) Suppose that K 2, 0 a k < m k for 1 k K and that m 1 < m 2 < . . . < m K . This is called a family of covering congruences when ever integer x satisfies at least one of the congruences
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Unformatted text preview: x ≡ a k (mod m k ). A system of covering congruences is called exact when for every value of x there is exactly one value of k such that x ≡ a k (mod m k ). Show that if the system is exact, then K ° k =1 z a k 1 − z m k = 1 1 − z . When z = re (1 /m K ) (where e ( z ) denotes e 2 πiz ) with r ∈ R > and r → 1 − , show that the left hand side above is ∼ e ( a K /m K ) m K (1 − r ) whereas the right hand side is bounded for z in a neighbourhood of e (1 /m K )....
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