 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Now, before and later (Posted on 2020-08-26) 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.

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Is this a solution? | Comment 1 of 2
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

 Posted by Charlie on 2020-08-26 15:49:59 Please log in:

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