Solutions_MT1sum11_mk

# Solutions_MT1sum11_m - MAT 312/AMS 351 Summer 2011 Solutions to the Midterm I(Make-up exam Maximal grade 100 points 1(15 points Find the prime

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Unformatted text preview: MAT 312/AMS 351, Summer 2011 Solutions to the Midterm I (Make-up exam) Maximal grade: 100 points 1. (15 points) Find the prime decomposition of the number 1! · 2! · 3! · 4! · 5! · 6! Answer: 1! · 2! · 3! · 4! · 5! · 6! = 2 12 · 3 5 · 5 2 Solution: We have 1! = 1 , 2! = 2 , 3! = 2 · 3 , 4! = 2 3 · 3 , 5! = 2 3 · 3 · 5 , 6! = 2 4 · 3 2 · 5 . Therefore 1! · 2! · 3! · 4! · 5! · 6! = 2 · (2 · 3) · (2 3 · 3) · (2 3 · 3 · 5) · (2 4 · 3 2 · 5) = 2 12 · 3 5 · 5 2 . 2. (20 points) Compute 19 37 mod 12. Answer: 19 37 = 7 mod 12. Solutions: a) Since 12 = 3 · 4 , we have ϕ (12) = ϕ (3) ϕ (4) = (3- 1)(4- 2) = 4 . By Euler’s theorem (we can use it since g.c.d (19 , 12) = 1) we have 19 4 = 1 mod 12 . Therefore 19 37 = (19 4 ) 9 · 19 = 1 9 · 7 = 7 mod 12 . b) Remark that 19 = 7 mod 12, so 19 2 = 7 2 = 49 = 1 mod 12 . 1 Therefore 19 37 = (19 2 ) 18 · 19 = 1 18 · 7 = 7 mod 12 ....
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## This note was uploaded on 01/25/2012 for the course MAT 312 taught by Professor Bescher during the Summer '08 term at SUNY Stony Brook.

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Solutions_MT1sum11_m - MAT 312/AMS 351 Summer 2011 Solutions to the Midterm I(Make-up exam Maximal grade 100 points 1(15 points Find the prime

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