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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 iz ) 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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- Spring '08