#$&* course Mth 151 7/30 10 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** x mod 10 is the remainder when x is divided by 10. So [ (10+7) * (5 + 3) ] mod 10 = ( 17 * 8) mod 10 = 136 mod 10 = 6, since 136 / 10 leaves remainder 6. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qquery 5.4.20 2 / 3 on 5-hour clock What is 2 divided by 3 on a 5-hour clock, and how did you obtain this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For this problem, one would rotate the numbers around to get 3 * x = 2 When using the 5 hour clock, there needs to be a number that would that would result in a remainder of 2 because that is the beginning number in the problem. 3*4=12/5=remainder 2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You have to turn this one into a multiplication problem to get the correct answer. In decimal numbers, for example, 60 / 20 = 3 because 3 * 20 = 60. Whatever you get when you divide 2 by 3, when you multiply it by 3 you get 2. That is, if 2 / 3 = x, then 3 x = 2. So what would you multiply by 3 to get 2 on a 5-hour clock? It turns out that 3 * 4 = 2. So it follows that 2 / 3 = 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qquery 4.4.42 (3 - 27) mod 5 What is (3 - 27) mod 5, and how did you reason out your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 27 + x = 3 - 27 + 1= 28/5 = remainder of 3- answer is 1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** (3-27) mod 5 = -24 mod 5. You would go all the way around around backwards 5 complete times to get -25, then move forward 1 to get to -24. That would put you at 1 on the actual clock. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qquery 5.4.20 Pos Integer solns (5x-3) = 7 (mod 4) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7 (mod 4)=3 We will use 0,1,2,3 for the x to find the solution 5(0)-3=-3 5(1)-3= 2 5 (2)-3=7 5(3)-3=12 I can’t seem to find a solution confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The solutions have to be integers, and the mod makes a difference in the algebra. 7 (mod 4) is 3. Since (5x - 3) mod 4 = 7 mod 4, (5x - 3) mod 4 must be 3. For x = 1, 2, 3, 4, ..., the expression 5x - 3 takes values 2, 7, 12, 17, 22, 27, 32, 37, ... . These numbers, when divided by 4, give remainders 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, ... . Thus every fourth number, mod 4, is equal to 3. This starts with the second number, which occurs when x = 2. Every fourth number, starting with 2, gives us the sequence 2, 6, 10, 14, ... 2 is the first solution, 4 is the difference between solutions. Thus x can be any element in the set {2, 6, 10, 14, . . . }. The general term of this sequence is 2 + 4 n. So we can also say that x = 2 + 4 n, where n = 0, 1, 2, 3, . . . Checking these results: If n = 0 then x = 2 + 4 * 0 = 2. If n = 1 then x = 2 + 4 * 1 = 6. If n = 2 then x = 2 + 4 * 2 = 10. If n = 3 then x = 2 + 4 * 3 = 14. etc. These are the solutions obtained above to the equation. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I know understand the 2 + 4n for x, I was looking for a sing number but the problem is more complex ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `qquery 5.4.30 table for addition mod 7 and properties of operation YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5 This is operation has closure because numbers in the set are present. Commutative and associative properties are shown. The identity of the group is 0 because no matter what you pair 0 zero with, it will always be that number. It also has inverse, because the numbers seem to start from the beginning when they reach a certain point. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Your table should read 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5 The operation is closed, since all numbers mod 7 are between 0 and 6. The only numbers on the table are 0, 1, 2, 3, 4, 5, 6. The operation has an identity, which is 0, because when added to any number 0 doesn't change that number. We can see this in the table because the row corresponding to 0 just repeats the numbers 0123456, as does the column beneath 0. The operation is commutative--order doesn't matter because the table is symmetric about the main diagonal.. The operation has the inverse property because every number can be added to another number to get the identity 0: 0+7 = 0, 1+6=0, 2+5=0, 3+4=0. These numbers form pairs of inverses. This property can be seen from the table because the identity 0 appears exactly once in every row. The operation is associative, since for any a, b, c we know that (a + b ) + c = a + ( b + c), and it follows that [ (a + b ) + c ] mod 7 must equal [ a + ( b + c) ] mod 7. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q5.4.33 table for mult mod 4 and properties of operation YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 This is a closed set, with associative and commutative properties. The identity is 1 and there is no inverse except possibly 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The correct table is 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 For example the row across from 2 is obtained as follows: 2 * 0 = 0 and 2 * 1 = 2, as always. Then 2 * 2 mod 4 = 4 mod 4, which is 0 and 2 * 3 mod 4 = 6 mod 4, which is 2. the operation is closed because the results all come from the set {0, 1, 2, 3} being operated on 1 is the identity because the row and column for 1 both have 0,1,2,3 in that order, so 1 doesn't change a number when multiplied by that number. 0 and 2 lack inverses--they can't be combined with anything else to get 1--so the operation lacks the inverse property. symmetry about the main diagonal implies commutativity associativity follows from associativity of multiplication of real numbers** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qquery 5.4.70 y + [[ (y-1)/4 ]] - [[ (y-1) / 100 ]] + [[ (y-1) / 400 ]]; day of jan 1, 2002; smallest b with a = b (mod 7); b=0 Sunday etc. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If y = year then 2002 + [[ (2002-1)/4 ]] - [[ (2002-1) / 100 ]] + [[ (2002-1)/ 400 ]] = 2002 + [[500.25 ]] - [[ 20.01 ]] + [[ 5.0025 ]]= 2487 2487/ mod 7 =remainder 2 Finding the smallest with 2 and under Tuesday=2 Monday= 1 Sunday= 0 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The calculation is 2002 + [[ 2002-1/4 ]] - [[ 2002-1/100 ]]+ [[ 2002-1/400 ]]. [[ Q ]] means the greatest integer contained in Q. So we get 2002+ [[500.25]] - [[20.01]] + [[5.0025]] = 2002 + 500 - 20 + 5 = 2487. Now 2487 mod 7 is 2. Sunday is 0, Monday is 1 so Tuesday is 2.** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qquery 5.4.70 y + [[ (y-1)/4 ]] - [[ (y-1) / 100 ]] + [[ (y-1) / 400 ]]; day of jan 1, 2002; smallest b with a = b (mod 7); b=0 Sunday etc. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If y = year then 2002 + [[ (2002-1)/4 ]] - [[ (2002-1) / 100 ]] + [[ (2002-1)/ 400 ]] = 2002 + [[500.25 ]] - [[ 20.01 ]] + [[ 5.0025 ]]= 2487 2487/ mod 7 =remainder 2 Finding the smallest with 2 and under Tuesday=2 Monday= 1 Sunday= 0 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The calculation is 2002 + [[ 2002-1/4 ]] - [[ 2002-1/100 ]]+ [[ 2002-1/400 ]]. [[ Q ]] means the greatest integer contained in Q. So we get 2002+ [[500.25]] - [[20.01]] + [[5.0025]] = 2002 + 500 - 20 + 5 = 2487. Now 2487 mod 7 is 2. Sunday is 0, Monday is 1 so Tuesday is 2.** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!