#$&* course mth151 12/7/14 12:20 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: 3 * 9 (mod 4) is the remainder when 3 * 9 is divided by 4. Since 3 * 9 = 27 and 27 / 4 leaves remainder 3, we see that 3 * 9 (mod 4) = 3. 7 * 12 (mod 4) is the remainder when 7 * 12 is divided by 4. Since 7 * 12 = 84 and 84 / 4 leaves remainder 0, we see that 7 * 12 (mod 4) = 0. 11 * 13 (mod 4) is the remainder when 11 * 13 is divided by 4. Since 11 * 13 = 143 and 143 / 4 leaves remainder 3,we see that 11 * 13 (mod 4) = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q002. Make a table for the x * y mod 4 operation, which we will call '* mod 4', operating on the set {0, 1, 2, 3}. Determine which of the properties, including commutativity, associative, identity, inverse and closure properties, are properties of this operation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: * mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 if x is 0 then anything multiplied by x is 0 and therefore when divided by 4 has a remainder of 0 1*1=1 and when divided by 4 leaves a remainder of 1 1*2=2 when divided by 4 leaves remainder of 2 1*3=3 when divided by 4 leaves a remainder of 3 2*2=4 when divided by 4 leaves a remainder of 0 2*3=6 when divided by 4 leaves a remainder of 2 3*3=6 when divided by 4 leaves a remainder of 1 there is symmetry accross the diagonal line so the operation has the commutative property we can also see that it has the closer property since in the set {0,1,2,3} the remainder will always be 0, 1, 2, or 3 since 0 and 2 do no have an additive inverse in the table then the operation does not have the inverse property this also comtains the identity property since the number 1, when provided, with any other element provide that same element. 1*2=2, 1*0=0 multuiplication is associative and therefore the operation has the associative property confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Whatever x is, 0 * x = x * 0 = 0, which when divided by 4 leaves remainder 0. Whatever x is, 1 * x = x * 1 = x, and if x is in the set {0, 1, 2, 3} we have get remainder x when dividing by 4 (e.g., 4 divides into 0, 1, 2 or 3 zero times, leaving that number as the remainder) and x mod 4 = x. From this we can see that 1 is the identity for this operation. Multiplying 0, 1, 2, and 3 by 2 we get 0, 2, 4, and 6, which when divided by 4 leave remainders 0, 2, 0 and 2, respectively. Multiplying 0, 1, 2, and 3 by 2 we get 0, 3, 6, and 9, which when divided by 4 leave remainders 0, 3, 2 and 1, respectively. The table for this operation is therefore * mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 We note that this operation does contain identity 1, but since neither 0 nor 2 can be combined with any of the elements of the set to give us the identity, the operation on this set does not have the inverse property. We do see from the symmetry of the table about the main diagonal that it has the commutative property, which we could in any event have concluded from the fact that multiplication is commutative so that the product we get before calculating the remainder is independent of the order of the two numbers. In a similar matter we can reason that the operation is associative. The operation is also closed, since the remainder upon dividing by 4 must always be 0, 1, 2 or 3 and hence in the set {0, 1, 2, 3}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q003. Repeat the preceding exercise for the operation x * y mod 5, defined to give the remainder when x * y is divided by 5, on the set {1, 2, 3, 4}. Determine which of the properties are exhibited by this operation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: * mod 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 1*1 divided by 5 leaves 1 1*2 divided by 5 leaves 2 1*3 leaves 3 1*4 leaves 4 2*2 leaves 4 2*3 leaves 1 3*3 leaves 4 3*4 leaves 2 4*4 leaves 1 we can see that if we drew a diagonal line across the table there would be symmetry on either side so it has commutative property as before we can say that the operation has associative property because of the associative properties of multiplication the results stay within the set {1, 2, 3, 4} so it has the closure property it has the identity 1 preoperty because 1 can be multiplied with any other of the numbers to reslut in that same number, ex. 1*1=1, 1*4=4,... we can see from the table that each number appears once in each column and each row and therefore has the inverse property confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First we might wish to do a couple of example calculations to get familiar with the operation. For example: 2 * 3 mod 5 = 6, which when divided by 5 gives us remainder 1. 3 * 4 mod 5 = 12 which when divided by 5 gives us remainder 2. 2 * 4 mod 5 = 8 which when divided by 5 gives us remainder 3. The table is * mod 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 We immediately see that all the results are in the set {1, 2, 3, 4}, so that the operation is closed. This operation has identity 1, as we can see from the row and the column across from and beneath 1. We easily see from the table that the identity appears exactly once in each row and in each column, which assures us that the operation has the inverse property. Specifically we see that 1 * 1 mod 5 = 1 so that 1 is its own inverse, that 2 * 3 mod 5 = 1 so that 2 and 3 are inverses, and that 4 * 4 mod 5 = 1, so that 4 is its own inverse. The associativity and commutativity of the operation follow from the associative and commutative properties of multiplication on real numbers, as discussed in the preceding problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q004. The equation 3x + 7 = 9 (mod 5) has an integer solution for x = 0, 1, 2, 3 or 4. Which value of x is a solution to this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3x + 7 =9 (mod5) is 3x + 7 -9 = 0 (mod5) 3x-2=0 (mod5) this means that we must solve 3x-2 and then divide that by 5 and get a remainder of 0 if x=0 then 3(0)-2=-2 x=1 then 3(1)-2=1 x=2 then 3(2)-2=4 x=3 then 3(3)-2=7 x=4 then 3(4)-2=10 the only one of these that is a multiple of 5 and would therefor give us a remainder of 0 would be x=4 3(4)+7=9 (mod5) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 3x + 7 = 9 (mod 4) means that 3x + 7 - 9 = 0 (mod 5) so 3x - 2 = 0 (mod 5). If 3x - 2 = 0 (mod 5) then when we divide 3x - 2 by 5 we should get remainder 0. So we substitute the different possible values for x into the expression 3x - 2 until we get a number of which when divided by 5 gives us remainder 0. If x = 0 then 3x - 2 = -2, and -2 (mod 5) = 3 (if you don't understand why -2 mod 5 = 3, think of the 5-hour clock in the text; but for now it should be obvious that -2 is not a multiple of 5 so that you cannot get remainder 0 when dividing -2 by 5). If x = 1 then 3x - 2= 1, and 1 (mod 5) = 1. If x = 2 then 3x - 2= 4, and 4 (mod 5) = 4. If x = 3 then 3x - 2= 7, and 7 (mod 5) = 2. If x = 4 then 3x - 2= 10, and 10 (mod 5) = 0. Thus x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). STUDENT QUESTION What is that 5-hour clock? INSTRUCTOR RESPONSE A 5-hour clock is depicted below: We label the usual 12:00 position 0. Moving 1 hour at a time in the clockwise direction we encounter hours 1, 2, 3 and 4 before finally returning after 5 hours to the 0 position. To calculate 3 + 4 on this clock, we start at 0 and move 3 hours, which puts us at position 0. Then we move 4 more hours, which takes us through positions 4, 0 and 1 before ending up at 2. Thus, on this clock, 3 + 4 = 2. It's easy to see how this gives us a picture for the calculation (3 + 4) mod 5, and why the result is the remainder when we divide 3 + 4 by 5. To calculate 3 * 4 on this clock, we would start at 0 and move 4 hours to position 4, then 4 more hours to position 3, then 4 more hours to position 2. Thus on this clock 3 * 4 = 2, and this gives us a picture of (3 * 4) mod 5. It's easy to see that we have moved 3 * 4 = 12 hours, and that where we end up is the remainder when 12 divided by 5. To divide 3 by 4 on this clock, we observe that for ordinary base-10 division, (3 / 4) * 4 = 3. So we ask what number multiplied by 4 on this clock will give us 3. If we start at 0 and move 4 hours around the clock we end up at 4; if we move 4 more hours we end up at 3. Thus 4 * 2 = 3, and we can therefore say that 2 = 3/4. That is, 3 divided by 4 is 2. To solve the equation 3x - 2 = 0: Next we try different numbers for x: If x = 1 we get 3 * 1 - 2 = 1. If x = 2 we get 3 * 2 - 2 = 4. If x = 3 we get 3 * 3 - 2 = 7. Since 7 hours on this clock takes us from our starting position at 0 to position 2, we conclude that when x = 3 our result is 3 * 3 - 2 = 2. If x = 4 we get 3 * 4 - 2 = 10. Since 10 hours on this clock takes us from our starting position at 0, twice around the clock and back to position 0, we conclude that when x = 4 our result is 3 * 4 - 2 = 0. So x = 4 is a solution to our equation 3 * x - 2 = 0 (mod 5). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q005. You see that x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). One of the numbers x = 5, 6, 7, 8, 9 is also a solution. Which one is it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if x=5 then 3(5)-2=13 x=6 then 3(6)-2=16 x=7 then 3(7)-2=19 x=8 then 3(8)-2=22 x=9 then 3(9)-2=25 the only one that is a multiple of 5 and would therefore give us a remainder of 0 is x=9 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We recall that 3x + 7 = 9 (mod 5) is equivalent to 3x - 2 = 0 (mod 5). We evaluate 3x - 2 (mod 5) for x = 5, 6, 7, 8 and 9 and we find that the results are 3, 1, 4, 2, and 0. So x = 9 is our next solution. We might also note that the series of results 3, 1, 4, 2, 0 is the same as the series we got for x = 0, 1, 2, 3, 4. Our results therefore seem to indicate a repeating pattern in which the remainder 0 occurs every fifth number starting with 4. This is in fact what happens, and you might wish to think about why this happens. However, you should in a case remember that this is what happens. In general when we have an equation of the form A x + B = C (mod n), integer solutions happen at intervals of n. for some values of A, B and C integer solutions can also occur at shorter intervals, but they always do occur at intervals of n. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. What are the first five positive values of x which solve the equation 3x + 7 = 9 (mod 5) of the preceding problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: in the previous problems we noted a pattern with the remainders being 3,1,4,2,0 in respects to x being 0,1,2,3,4 and again with x being 5,6,7,8,9 from this we can gather that the remainder will be 0 every fifth number starting from 0, so x= 4, 9, 14, 19, 24 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We just saw that x = 4 and x = 9 are solutions, and we saw that because we are solving an equation mod 5, the solutions have to occur at intervals of 5. Thus the first five solutions are x = 4, 9, 14, 19 and 24. ********************************************* Question: `q007. Can you find a solution to the equation 3x + 7 = 9 (mod 6)? Show how you reason out your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: from previous questions we know that 3x+7=9 (mod6) will be 3x-2=0 (mod6) i cannot find a solution to this equation. everythings that going to be a multiple of 6 is also going to be a multiple of 3 and then you subtract 2 and you will always have a remainder for ex... x=4 3(4)-2= 12-2=10 which leaves you with a remainder of 4 x=36 3(36)-2= 108-2=106 which again leaves a remainder of 4 and the list can go on and on confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:ok
#$&* course mth151 12/7 1:35 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 4.4.20 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: 2/3=x is equivilant to 3x=2 on a five hour clock you must turn the hand 4 times to get to 2 3(4)=2 therefore 2/3=4 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You have to turn this one into a multiplication problem to get the correct answer. In decimal numbers, for example, 60 / 20 = 3 because 3 * 20 = 60. Whatever you get when you divide 2 by 3, when you multiply it by 3 you get 2. That is, if 2 / 3 = x, then 3 x = 2. So what would you multiply by 3 to get 2 on a 5-hour clock? It turns out that 3 * 4 = 2. So it follows that 2 / 3 = 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery 4.4.42 (3 - 27) mod 5 What is (3 - 27) mod 5, and how did you reason out your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (3-27) (mod5) = -24(mod5) on the clock go backwards around the clock 5 times and then go forward one to put you at -24 it would put you at 1 on the clcok confidence rating #$&*: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 4.4.56 Find the positive integer solutions of the equation (5x-3) = 7 (mod 4) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5x-3=7 (mod4) we can go ahead and divide 7 by 4 which gives us a remainder of 3 therefore (5x-3) would have to be 3 after being divided by 4 if x=1 5(1)-3=2 rem. 2 x=2 5(2)-3=7 rem. 3 x=3 5(3)-3=12 rem. 0 x=4 5(4)-3=17 rem. 1 x=5 5(5)-3=22 rem. 2 x=6 5(6)-3=27 rem. 3 x=7 5(7)-3=32 rem. 0 x=8 5(8)-3=37 rem. 1 x=9 5(9)-3=42 rem. 2 x=9 5(10)-3=47 rem. 3 as we see here it equals 3 starting at x=2 and every fourth number after that {2,6,10,14,18...} confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery 5.4.47 Give the table for addition mod 7 and list the properties of operation; show how the properties are proven YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: + mod 7 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 is closed since all numbers are between 0 and 6 it has the communative property because its symmetric if you drew a diagonal line it has identity of 0 because you can add 0 to any number and still be the same it has the inverse property because every number is in each column and row its associative because of the associative property of addition 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: `q4.4.50 table for mult mod 4 and properties of operation YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: *mod4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 closed bcause all of the numbers stay within 0-3 associative because of the associative property of multiplication communitative because it has symmetry if we drew a diagonal line it has identity 1 because 1 can be multiplied by any other number to get the same number does not have inverse property because 0 an 2 do not have inverses confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The correct table is 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 For example the row across from 2 is obtained as follows: 2 * 0 = 0 and 2 * 1 = 2, as always. Then 2 * 2 mod 4 = 4 mod 4, which is 0 and 2 * 3 mod 4 = 6 mod 4, which is 2. the operation is closed because the results all come from the set {0, 1, 2, 3} being operated on 1 is the identity because the row and column for 1 both have 0,1,2,3 in that order, so 1 doesn't change a number when multiplied by that number. 0 and 2 lack inverses--they can't be combined with anything else to get 1--so the operation lacks the inverse property. symmetry about the main diagonal implies commutativity associativity follows from associativity of multiplication of real numbers** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `qquery 5.4.66 y + [[ (y-1)/4 ]] - [[ (y-1) / 100 ]] + [[ (y-1) / 400 ]]; day of jan 1, 2002; smallest b with a = b (mod 7); b=0 Sunday etc. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2002 + [[(2002-1)/4]] - [[(2002-1)/100]] + [[(2002-1)/400]] 2002 +[[500.25]] - [[20.01]] + [[5.0025] 2002 + 500 - 20 + 5 2487 2487(mod7) = 2 sunday is 0 ???How do you get that Monday is 1, Tuesday is 2? What do you mean ""smallest b""? Also i understand what you are doing with the [[Q]] thing but where are you getting that? how are we supposed to know what you are talking about without looking at the answer for guidance??? confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:
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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:
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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: