course MTH 151
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22:44:33 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 10 136 / 10 = 13 Rem 6 = 6 mod 10
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22:44:47 ** 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, I got this one.
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22:48:54 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 --> I have read the text on this and I think I understand how to do the steps. I'm stuck on the 2/3 part. This does not give you a whole number, so I'm not sure of where to go from there.
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22:52:19 ** 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 --> Oh Ok, I think I see here what to do. I knew that you would probably have to use multiplication, but I wasn't sure how. I think I get this now.
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22:54:30 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 = -24 -24/5 = 4&4/5 = 4.8 mod 5
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22:55:14 ** (3-27) mod 5 = -24 mod 5.
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RESPONSE --> OK, I got this. Was I not supposed to continue my problem as I did?
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22:56:42 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 4. **
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RESPONSE --> Oh, ok. I actually didn't use the 5-hour clock to do this one.
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22:58:24 query 5.4.20 Pos Integer solns (5x-3) = 7 (mod 4)
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RESPONSE --> Ok, I figured out that x=2, but I'm stuck from here. Would this make it 2 mod 4?
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23:00:56 ** 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, I kinda understand this. I should have started with 7 mod 4 (which is 28) is 3 on the clock. I need to read over this section again.
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23:03:40 ** 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, I was a little confused on how to answer this, but I see that I needed to make and addition table, mod 7. I think I understand this. I have made my own table and it correctly compares with this one.
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23:05:42 5.4.33 table for mult mod 4 and properties of operation
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RESPONSE --> This table would be: 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1
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23:07:20 ** 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, I forgot to give the properties. My table is correct. I understand the reasonings you have given here.
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23:10:00 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 --> Ok, after I worked this out, I still don't think that my operations were correct. I will compare with the solution given.
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23:12:39 ** 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 --> I think I made this a little more complicated that it should have been. Seeing it worked out here makes a little more sense. I think my main problem was remembering the order of operations. I'll try to work this out again.
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