course MTH 151 ???b?????H???x?assignment #025025. GCF, LCM
......!!!!!!!!...................................
07:46:11 `q001. There are three questions in this assignment. 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105. What do the prime factorizations of 60 and 105 having common? What is the prime factorization of the smallest number which contains within its prime factorization the prime factorizations of both 60 and 105?
......!!!!!!!!...................................
RESPONSE --> 60,105 have 3,5 in common the GCF= 15 the LCM= 3 confidence assessment: 0
.................................................
......!!!!!!!!...................................
07:48:19 The prime factorizations 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105 have in common the product 3 * 5 = 15. This is the largest number that will divide evenly into both 60 and 105, and is called the greatest common divisor of 60 and 105. In order to contain to both of the prime factorizations 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 105 a number must contain in its prime factorizations the entire prime factorization 2 * 2 * 3 * 5, and in addition the 7 still necessary in order to contain 3 * 5 * 7. Thus the number must be 2 * 2 * 3 * 5 * 7 = 420. This number is a multiple of both 2 * 2 * 3 * 5 = 60 and 3 * 5 * 7 = 120, and is the smallest number which is a multiple of both. We therefore call 420 the Least Common Multiple of 60 and 105.
......!!!!!!!!...................................
RESPONSE --> ok, i was confused on the wording for LCM. self critique assessment: 0
.................................................
......!!!!!!!!...................................
07:57:53 `q002. What are the prime factorizations of 84 and 126, and how can they be used to find the greatest common divisor and the least common multiple of these two numbers?
......!!!!!!!!...................................
RESPONSE --> 84= 2*42 2*2*3*7 2^2*3*7 126= 2*63 2*3*3*7 2*3^2*7 GCF= 2*3*7= 42 LCM=4*9*7= 252 confidence assessment: 1
.................................................
......!!!!!!!!...................................
07:58:02 The prime factorization of 84 is 2 * 2 * 3 * 7, and the prime factorization of 126 is 2 * 3 * 3 * 7. The greatest common divisor of these numbers is the number we build up from all the primes that are common to both of these prime factorizations. The two prime factorizations having common 2, 3 and 7, which give us the greatest common divisor 2 * 3 * 7 = 42. The least common multiple is made up of just those primes which are absolutely necessary to contain the two given numbers. This number would have to contain the first number 2 * 2 * 3 * 7, and would in addition need another 3 in order to contain 2 * 3 * 3 * 7. The least common multiple is therefore 2 * 2 * 3 * 3 * 7 = 252.
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:05:53 `q003. Find the greatest common divisor and least common multiple of 504 and 378.
......!!!!!!!!...................................
RESPONSE --> 378= 2*189 2*3*3*3*7 2*3^3*7 504= 2*252 2*2*126 2*2*2*3*3*7 2^3*3^3*7 GCF= 126 LCM=1,512 confidence assessment: 1
.................................................
......!!!!!!!!...................................
08:06:02 We find that 504 = 2 * 2 * 2 * 3 * 3 * 7 and 378 = 2 * 3 * 3 * 3 * 7. The greatest common divisor can contain a single 2 since 378 has only a single 2 in its factorization, two 3's since both numbers contain at least two 3's, and a single 7. The greatest common divisor is therefore 2 * 3 * 3 * 7 = 126. The least common multiple must contain the first number, 2 * 2 * 2 * 3 * 3 * 7, and another 3 because of the third 3 in 378. The least common multiple is therefore 2 * 2 * 2 * 3 * 3 * 3 * 7 = 1512.
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment: 2
.................................................
?????v{D?W???? assignment #026 026. Modular systems Liberal Arts Mathematics I 08-02-2008
......!!!!!!!!...................................
14:45:49 `q001. There are six questions in this assignment. We defined an operation as follows: x * y (mod 4) = remainder when x * y is divided by 4. Find 3 * 9 (mod 4); 7 * 12 (mod 4) and 11 * 13 (mod 4).
......!!!!!!!!...................................
RESPONSE --> 3*9[mod4] = 27/4=7 7*12[mod 4]= 84/4=21 11*13[mod 4]= 143/4=36 confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:48:16 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.
......!!!!!!!!...................................
RESPONSE --> ok, i divided wron. used a calculator with settings that rounds up with no remainder self critique assessment: 1
.................................................
......!!!!!!!!...................................
14:59:37 `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.
......!!!!!!!!...................................
RESPONSE --> mod 4 * 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 6 8 3 0 3 6 9 12 4 0 4 8 12 16 all are properties confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:00:06 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}.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:09:02 `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.
......!!!!!!!!...................................
RESPONSE --> mod 5 * 1 2 3 4 5 1 0 0 0 0 0 2 0 0 1 3 0 3 0 1 4 7 0 4 0 1 2 1 0 5 1 0 0 0 0 all properties met confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:12:10 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.
......!!!!!!!!...................................
RESPONSE --> i missed multiplied, carelessness self critique assessment: 0
.................................................
......!!!!!!!!...................................
15:27:11 `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?
......!!!!!!!!...................................
RESPONSE --> 1 confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:31:19 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).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 0
.................................................
......!!!!!!!!...................................
15:38:23 `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?
......!!!!!!!!...................................
RESPONSE --> 6 is the solution 3x+7=9[mod 5] 3(6)+7= 18+7= 25/5= 5 so that is 0 [mod 5] confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:39:57 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 0
.................................................
......!!!!!!!!...................................
15:41:29 `q006. What are the first five positive values of x which solve the equation 3x + 7 = 9 (mod 5) of the preceding problem?
......!!!!!!!!...................................
RESPONSE --> 2,4,6,8,10 confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:42:07 We just saw that x = 4 and x = 9 are solutions, and we saw that because we are solving an equation mod 5 solutions have to occur at intervals of 5. Thus the first five solutions are x = 4, 9, 14, 19 and 24.
......!!!!!!!!...................................
RESPONSE --> ok but 9, 19 are not positive self critique assessment: 0
.................................................
???????y???g??o? assignment #027 027. Liberal Arts Mathematics I 08-02-2008