Numerical analysis homework help

MATH 312 Final Exam Past Due Date Final Exam Will Not Be accepted
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(Show all work.) [1] Find the small positive integer solution of each problem: (38 pts)
(Hint: you may find any integer solution first, then add/subtract a multiple of n)
[1a] 7x โ‰ก 6 (mod 41), 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 40
 
[1b] 35x โ‰ก 15 (mod 64), 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 64
[1c] 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 17 โ€ข 19 = 323
๏ฟฝ x โ‰ก 5 (mod 17) x โ‰ก 11 (mod 19)
 
 
 
 
 
 
 
 
 
 
 
 
MATH 312 Final Exam Past Due Date Final Exam Will Not Be accepted
Due on Sunday, December 06, 2020 Print your name: _________________________________.
Page 2 of 3
[2] Given ๐‘๐‘ = 41 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž ๐‘”๐‘” = 9. Solve each equation: (20 pts) [2a] Find ๐‘Ž๐‘Ž satisfying 1 โ‰ค ๐‘Ž๐‘Ž < 41 and ๐‘Ž๐‘Ž๐‘”๐‘” โ‰ก 1 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 41)
 
[2b] Find ๐‘š๐‘š satisfying 1 โ‰ค ๐‘š๐‘š < 41 and ๐‘š๐‘š๐‘”๐‘” โ‰ก 28 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 41) (Use [2a] is easier.)
[3] Find the set of quadratic residues and the set of nonresidues of p = 17. 12 pts
 
 
MATH 312 Final Exam Past Due Date Final Exam Will Not Be accepted
Due on Sunday, December 06, 2020 Print your name: _________________________________.
Page 3 of 3
[4] Use [4*] to find the smallest positive integer solutions (if any) of each quadratic congruence equation: (30 pts)
[4*] 12 โ‰ก 1 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 22 โ‰ก 4 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 32 โ‰ก 9 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 42 โ‰ก 5 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11),
52 โ‰ก 3 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 62 โ‰ก 3 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 72 โ‰ก 5 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 82 โ‰ก 9 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11),
92 โ‰ก 4 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 102 โ‰ก 1 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11)
[4a] Solve the congruence equation: 9๐‘ฅ๐‘ฅ2 + 6๐‘ฅ๐‘ฅ + 1 โ‰ก 0(๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 10
 
 
 
 
 
 
[4b] Solve the congruence equation: 5๐‘ฅ๐‘ฅ2 + 3๐‘ฅ๐‘ฅ + 2 โ‰ก 0 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 10
 
 
 
 
 
[4c] Solve the congruence equation: 3๐‘ฅ๐‘ฅ2 โˆ’ 4๐‘ฅ๐‘ฅ + 1 โ‰ก 0 (๐‘š๐‘š๐‘š๐‘š๐‘Ž๐‘Ž 11), 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 10