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Suppose T = 1 + 2 + 3 + ... + 209 . Including the sum, how many sums of two or more consecutive positive integers have value T?

 Oct 29, 2019
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Is 209 the last term? If so, the sum of the sequence =21,945.

 

21,945 has this many consecutive sequences:

 

1  - 10972  to 10973 = 21945
2  - 7314  to 7316 = 21945
3  - 4387  to 4391 = 21945
4  - 3655  to 3660 = 21945
5  - 3132  to 3138 = 21945
6  - 2190  to 2199 = 21945
7  - 1990  to 2000 = 21945
8  - 1561  to 1574 = 21945
9  - 1456  to 1470 = 21945
10  - 1146  to 1164 = 21945
11  - 1035  to 1055 = 21945
12  - 987  to 1008 = 21945
13  - 717  to 746 = 21945
14  - 649  to 681 = 21945
15  - 610  to 644 = 21945
16  - 559  to 596 = 21945
17  - 502  to 543 = 21945
18  - 372  to 426 = 21945
19  - 357  to 413 = 21945
20  - 300  to 365 = 21945
21  - 279  to 348 = 21945
22  - 247  to 323 = 21945
23  - 184  to 278 = 21945
24  - 157  to 261 = 21945
25  - 145  to 254 = 21945
26  - 136  to 249 = 21945
27  - 99  to 231 = 21945
28  - 66  to 219 = 21945
29  - 51  to 215 = 21945
30  - 21  to 210 = 21945
31  - 1  to 209 = 21945

Note: let me know if I misunderstood your question.

 Oct 29, 2019

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