Math 115 - Fall 1999 - Ribet - Midterm 2

Math 115 - Fall 1999 - Ribet - Midterm 2 - f (-18) is...

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Math 115 Professor K. A. Ribet Second Midterm Exam October 28, 1999 This is a closed-book exam: no notes, books or calculators are allowed. Explain your answers in complete English sentences. No credit will be given for a “correct answer” that is not explained fully. Don’t worry too much about simplifying arithmetical expressions; “3 · 5+1” is the same answer as “16” in most contexts. 1 (5 points) . Suppose that n and m are positive integers, that p is a prime and that α is a non-negative integer. Assume that n is divisible by p α , that m is prime to p and that F = n m is an integer. Show that F is divisible by p α . 2 (6 points) . Let f ( x ) be a polynomial with integer coefficients that satisfies f (1) = f 0 (1) = 3. Calculate the remainder when
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Unformatted text preview: f (-18) is divided by 19 2 . 3 (5 points) . Determine the number of solutions to the congruence x 2 + x +1 mod 7 11 . 4 (6 points) . Find an integer n 1 so that a 3 n a mod 85 for all integers a that are divisible neither by 5 nor by 17. 5 (6 points) . Find the number of solutions mod 120 to the system of congruences x 2 mod 4 3 mod 5 4 mod 6 . 6 (7 points) . If m = 15709, we have 2 ( m-1) / 2 1 mod m and 2 ( m-1) / 4 2048 mod m . With the aid of these congruences, one can nd quite easily a positive divisor of m that is neither 1 nor m . Explain concisely: how to nd such a divisor, and why your method works....
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