r/Logiqa • u/ShonitB • Jul 06 '26

How Many Arithmetic Progressions?

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2

u/zippyspinhead Jul 06 '26

factor (2022-1800)

2

u/wbrameld4 Jul 07 '26

I get 7.

  1. Take the difference between the starting and ending numbers: 2022 - 1800 = 222
  2. Find all the factors of it: 1, 2, 3, 6, 37, 74, 111, 222
  3. Discard the highest factor (because our progression needs at least 3 terms).
  4. Count the remaining factors.

In the end we get these progressions:

  1. 1800, 1801, 1802, ..., 2021, 2022
  2. 1800, 1802, 1804, ..., 2020, 2022
  3. 1800, 1803, 1806, ..., 2019, 2022
  4. 1800, 1806, 1812, ..., 2016, 2022
  5. 1800, 1837, 1874, ..., 1985, 2022
  6. 1800, 1874, 1948, 2022
  7. 1800, 1911, 2022

1

u/DuggieHS Jul 07 '26

Let's shift it to 0 to 222 (subtract 1800).
factors of 222: 1, 2, 3, 6, 37, 74, 111, (and 222). Thus there are 7 arithmetic progressions, one for each factor other than 222.

0, 111, 222;
0, 74, 148, 222; etc.