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<Title><![CDATA[Lesson 4: Solving Modular Problems]]></Title>
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<Title><![CDATA[Solving Modular Problems]]></Title>
<Notes><![CDATA[Chapter 10   Section 3]]></Notes>
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<SlideText><![CDATA[Solving Modular Problems Chapter 10   Section 3  ]]></SlideText>
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<Title><![CDATA[Simple Modular Equations]]></Title>
<Notes><![CDATA[No in-class assignment problem]]></Notes>
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<SlideText><![CDATA[Simple Modular Equations The solution to a modular equation is a set of all numbers in one or more equivalence classes.   It is possible that the solution is the empty set. A simple modular equation is of the nature  x a" a (mod m) No in-class assignment problem ]]></SlideText>
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<Title><![CDATA[Solving a Simple Equation]]></Title>
<Notes><![CDATA[In-class Assignment 25 - 1]]></Notes>
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<SlideText><![CDATA[Solving a Simple Equation x a" 5 (mod 6) 5 is the name of an equivalence class in mod 6. The solution is the equivalence class 5. x = {5, 11, 17, 23, & } Remember each to find the next number in an equivalenc class add the mod. x a" 25 (mod 4) 25 is not less than 4. Change 25 to the name of the equivalence class 25 is in.  25 a" 1 (mod 4) x = {1, 5, 9, 13, &  } In-class Assignment 25 - 1 ]]></SlideText>
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<Title><![CDATA[Solving More Complicated Modular Equations]]></Title>
<Notes><![CDATA[In-class Assignment 25 - 2]]></Notes>
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<SlideText><![CDATA[Solving More Complicated Modular Equations 5x a" 17 (mod 4) Change 17 to the name of the class it belongs to.  17 a" 1 (mod4) Write 4 statements because the mod is 4 replacing x with the name of an equivalence class. 5 · 0 a" 1 (mod 4)  False 5 · 1 a" 1 (mod 4)  True 5 · 2 a" 1 (mod 4)  False 5 · 3 a" 1 (mod 4)  False The solution is equivalence class 1. x = {1, 5, 9, 13, & }    In-class Assignment 25 - 2 ]]></SlideText>
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<Title><![CDATA[Another Equation to Solve]]></Title>
<Notes><![CDATA[In-class Assignment 25 - 2]]></Notes>
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<SlideText><![CDATA[Another Equation to Solve 2x a" 4 (mod 8) Equivalence classes 2 and 6 are solutions. x ={2,10,18, & }           {6, 14, 22, & } Notice the sets can be joined as one set. x = {2, 6, 10, 14, & } 2 · 0 a" 4 (mod 8) False 2 · 1 a" 4 (mod 8) False 2 · 2 a" 4 (mod 8) True 2 · 3 a" 4 (mod 8) False 2 · 4 a" 4 (mod 8) False 2 · 5 a" 4 (mod 8) False 2 · 6 a" 4 (mod 8) True 2 · 7 a" 4 (mod 8) False In-class Assignment 25 - 2 ]]></SlideText>
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<Title><![CDATA[A Word Problem]]></Title>
<Notes><![CDATA[In-class Assignment 25 - 3]]></Notes>
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<SlideText><![CDATA[A Word Problem Ann wants to see a play.  The date of the play is 65 days from today.  Ann is a nurse and her schedule is 5   10 hour days on and 3 days off.  Will she be off on the day of the play if she is in the second day of her 5 days on? On = 1 and Off = 2 8 day schedule  1 1 1 1 1 2 2 2 Remainder =1 1 day past     2nd day of     schedule.  Ann will be on.    In-class Assignment 25 - 3 ]]></SlideText>
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<Title><![CDATA[Symbols of Inequalities]]></Title>
<Notes><![CDATA[No in-class assignment problem]]></Notes>
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<SlideText><![CDATA[Symbols of Inequalities    < means  less than.     d" means is  less than or equal to  or         no more than.     > means  greater than.     e" means  greater than or equal to  or         no less than.    No in-class assignment problem ]]></SlideText>
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<Title><![CDATA[Solving Restricted Modular Equations]]></Title>
<Notes><![CDATA[In-class Assignment 25 - 4]]></Notes>
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<SlideText><![CDATA[Solving Restricted Modular Equations Find the set of whole number solutions given  that      x < 40          x a" 1 (mod 7)          x a" 4 (mod 5)          x a" 5 (mod 8)   Find all the solution sets for the 3 equations. x = {1, 8, 15, 22, 29, 36} x = {4, 9, 13, 17, 21, 25, 29, 33, 37}  x = {5, 13, 21, 29, 37} Note   only numbers < 40. Find the intersection of the solution sets. x = 29 In-class Assignment 25 - 4 ]]></SlideText>
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<Title><![CDATA[Solving a Verbal Problem in Mod Systems]]></Title>
<Notes><![CDATA[In-class Assignment 25   5, 6]]></Notes>
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<SlideText><![CDATA[Solving a Verbal Problem in Mod Systems The football team has 50 members. When the coach put them in drill teams of 7 there were 4 players left.  When the coach had them in drill teams of 3 all players were used. When the coach put them in teams of 6 there were 3 players left.  How many were at practice on this day? Mods are the groupings Numbers are whats left.    x d" 50   x a" 4 (mod 7)   x a" 0 (mod 3)   x a" 3 (mod 6) Solve as before. 39 players were at practice on this day. In-class Assignment 25   5, 6 ]]></SlideText>
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<Title><![CDATA[A Short Cut]]></Title>
<Notes><![CDATA[In-class Assignment 25   4 (short cut)]]></Notes>
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<SlideText><![CDATA[A Short Cut x < 40     x a" 1 (mod 7)     x a" 4 (mod 5)     x a" 5 (mod 8) Find the solution set for the largest mod.            x = {5, 13, 21, 29, 37} Cross out any number that does not leave a remainder of 1 when divided by 7  (5, 13, 21, 37 crossed out) or a remainder of 4 when divided by  5 (the same numbers crossed out). x = {29) In-class Assignment 25   4 (short cut) ]]></SlideText>
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