Analyze (6, 9, 20)-accessibility. 1. We have broken the positive integers into t

Question

Analyze (6, 9, 20)-accessibility. 1. We have broken the positive integers into two families, based on remainders when divided by 2, and three families, based on remainders when divided by 3. Observe that the positive integers can also be broken into six families, depending on what their remainders are when divided by 6. 2. Write down the first few positive integers in each of the six families. For instance, one family begins with 1, 7, 13, 19, 25, 31, 37, 43, 49. 3. For each of the six lists, find the first number in the list that is (6, 9, 20)- accessible. 4. Explain why every number after that number in the list is also (6, 9, 20)- accessible. 5. Produce a complete list of the (6, 9, 20)-inaccessible positive integers. 6. Using the list, deduce that there is a largest integer that is (6, 9, 20)- inaccessible and find the positive integers that are (6, 9, 20)-accessible. 7. What ideas went into this solution? How did these ideas allow you to reduce the calculations necessary to a rather small amount?

Explanation / Answer

2.

The first few positive integers of the 6 families are as under:

        i.     1,7,13,19,25,31,37,43,49,55,61,….

      ii.    2,8,14,20,26,32,38,44,51,58,64,…

        iii.   3,9,15,21,27,33,39,45,51,57,63,69,….

        iv.   4, 10,16,22,28,34,40,46,52,58,64,70,…

        v.    5, 11,17,23,29,35,51,57,63,69,75,82,…

        vi.    6,12,18,24,30,36,42,48,54,60,66,72,78,…

3.

i.     49( 9,20,20)

        ii.   20 (20)

        iii.   9(9)

        iv.   40(20,20)

       v.    29(9,20)

       vi.   6(6)

4.  Every succeeding number in each family after first number in the list that is (6, 9, 20)- accessible is 6 + this number. Hence every such number is (6,9,20) – accessible.

5. 1,2,3,4,5,7,8,10,11,13,14,16,17,19,22,23,25,28,31,34,37,43.

6. The largest number that is 6,9,20- inaccessible is 43. The positive integers that are 6,9,20-accessible are 6, 9,12,15,18,20,21,24,26,27,29,30,32,33,35,36,38,39,40,41,42,44 and every positive integer greater than 44.

7. We can start with remainders 1,2,3,4,5 and 6.Once we know the smallest number in each of the six families, the next number in a familiy is the 1st number + 6.Once the 6,9,20-inaccesible list is ready, the remaining numbers are in the 6,9,20 accesible list.

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