Present each of the members of the set T= (2019,2020,2021)
as a sum of n of its factors (n>1), not necessarily distinct.
Example: 2018=1009+1009 = 2+2+2+...2 (1009 times)
List all possible solutions.
Generalize for number of solutions for different years.
As 2019 = 3*673 and therefore a semiprime:
2019 = 3 + 3 + 3 ... (673 times) = 673 + 673 + 673
Similarly
2021 = 43 * 47,
so just
43 + 43 + ... (47 times)
47 + 47 + ... (43 times)
But 2020 = 2 * 2 * 5 * 101 presents a more difficult case.
If "factor" means any divisor, listing them all would take up much too much space. That leads to the conclusion that what is meant is prime factors. But even then it's quite complicated.
A computer program find even then there are 2,129 solutions, starting out:
1010 * 2
2 * 5 + 1005 * 2
4 * 5 + 1000 * 2
6 * 5 + 995 * 2
8 * 5 + 990 * 2
10 * 5 + 985 * 2
12 * 5 + 980 * 2
14 * 5 + 975 * 2
16 * 5 + 970 * 2
. . .
.
with 1010 * 2 meaning 2 + 2 + ... (1010 times), or 2 * 5 meaning 5 + 5, etc.
through
. . .
1 * 101 + 45 * 5 + 847 * 2
1 * 101 + 47 * 5 + 842 * 2
1 * 101 + 49 * 5 + 837 * 2
1 * 101 + 51 * 5 + 832 * 2
1 * 101 + 53 * 5 + 827 * 2
1 * 101 + 55 * 5 + 822 * 2
1 * 101 + 57 * 5 + 817 * 2
1 * 101 + 59 * 5 + 812 * 2
1 * 101 + 61 * 5 + 807 * 2
1 * 101 + 63 * 5 + 802 * 2
1 * 101 + 65 * 5 + 797 * 2
1 * 101 + 67 * 5 + 792 * 2
1 * 101 + 69 * 5 + 787 * 2
1 * 101 + 71 * 5 + 782 * 2
1 * 101 + 73 * 5 + 777 * 2
1 * 101 + 75 * 5 + 772 * 2
1 * 101 + 77 * 5 + 767 * 2
1 * 101 + 79 * 5 + 762 * 2
1 * 101 + 81 * 5 + 757 * 2
1 * 101 + 83 * 5 + 752 * 2
1 * 101 + 85 * 5 + 747 * 2
1 * 101 + 87 * 5 + 742 * 2
1 * 101 + 89 * 5 + 737 * 2
1 * 101 + 91 * 5 + 732 * 2
1 * 101 + 93 * 5 + 727 * 2
. . .
and ending with
. . .
19 * 101 + 5 * 5 + 38 * 2
19 * 101 + 7 * 5 + 33 * 2
19 * 101 + 9 * 5 + 28 * 2
19 * 101 + 11 * 5 + 23 * 2
19 * 101 + 13 * 5 + 18 * 2
19 * 101 + 15 * 5 + 13 * 2
19 * 101 + 17 * 5 + 8 * 2
19 * 101 + 19 * 5 + 3 * 2
20 * 101
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Posted by Charlie
on 2020-08-26 15:49:59 |
Is this a solution?
