course Mth 151 ˩~yxEҵ[assignment #026
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11:03:01 5.4.12 What is [ (10+7) * (5+3) ] mod 10 and how did you get your result?
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RESPONSE --> [ (10+7) * (5 + 3) ] mod 10 = ( 17 * 8) mod 10 = 136 mod 10 = 6, and 136 / 10 leaves r= 6. This is because mod 10 is the reminder after x is divided by 0 confidence assessment: 3
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11:03:24 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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11:05:35 query 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?
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RESPONSE --> 60 / 20 = 3 so 3 * 20 = 60 3 * 4 = 2 and 2 / 3 = 4 because you have to multiply by 3 to get 2 on the clock, it had to be turned into a multiplication situation before a division. confidence assessment: 3
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11:05:48 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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11:06:52 query 4.4.42 (3 - 27) mod 5 What is (3 - 27) mod 5, and how did you reason out your result?
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RESPONSE --> (3-27) mod 5 = -24 mod 5 You have to go back on the clock 5 times to reach negative 25, then forward again to get -24. Which would be one in that case. confidence assessment: 3
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11:07:07 ** (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. **
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RESPONSE --> ok self critique assessment: 3
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11:07:58 query 5.4.20 Pos Integer solns (5x-3) = 7 (mod 4)
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RESPONSE --> 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. confidence assessment: 3
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11:08:11 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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11:10:51 query 5.4.30 table for addition mod 7 and properties of operation
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RESPONSE --> 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 It's closed because the numbers are from 0-6 0 is the identity because it doesn't change other numbers It's commutative It's inverse because other numbers can be added to get the identity and associative, because for a, b, c: (a + b ) + c = a + ( b + c), and [ (a + b ) + c ] mod 7 must equal [ a + ( b + c) ] mod 7. confidence assessment: 3
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11:11:15 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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11:13:41 5.4.33 table for mult mod 4 and properties of operation
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RESPONSE --> 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 The operation is closed because it come from (0, 1, 2, 3)3 1 is the identity. There is no inverse because 0 and 2 don't have it I was unsure if it was commutative And the associativty comes from the multiplication of the real numbers. confidence assessment: 3
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11:14:02 ** 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**
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RESPONSE --> ok self critique assessment: 3
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11:15:15 query 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.
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RESPONSE --> I was hoping this problem wasn't on here, I honestly didn't reach a conclusion. I originally began to add all numbers by the divided numbers, but the day's threw me off. confidence assessment: 3
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11:15:23 ** 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.**
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RESPONSE --> ok self critique assessment: 3
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