MATH 406 Midterm: MATH406 HERB-R FALL2006 0101 MID EXAM 1
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15 Feb 2019
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Math 406 - exam 2 - nov. 3, 2006. You may use a 4 function or scienti c calculator. All variables are assumed to be integers: (20) find a number x such that x 1 (mod 8) and x 5 (mod 23). Solution: x = 1 + 8y, where 1 + 8y 5 (mod 23). 8 3 + 23 ( 1) = 1, we have y = 3 4 12. Thus x = 1 + 8 12 = 97 is a solution. (20) use the method of successive squaring to reduce 314 (mod 25). 0 b < 24 such that 314 b (mod 25). Solution: 32 9 (mod 25), 34 81 6 (mod 25), 38 36 11 (mod 25). 14 = 8 + 4 + 2, we have 314 383432 11 6 9 19 (mod 25): (20) find an x with x23 9 (mod 200). You do not need to reduce your answer.
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