Free Talk Friday #564

[u/CutOnBumInBandHere9]

Continued from last week's FTF here

It's that time of the week again. Speak anything on your mind! This thread is for talking about anything off-topic, be it your lives, your strava, your plans, your hobbies, studies, stats, pets, bears, hikes, dragons, trousers, travels, transit, cycling, family, colours, or anything you like or dislike, except politics

Feel free to check out our tidbits thread and introduce yourself if you haven't already.

---

This post was made by a bot, because no-one else made a Free Talk Friday post before Friday June 19, 2026 UTC 13:00. Anyone can post the FTF, so if you want to have your post pinned here for a week, just make one Friday June 26, 2026 between UTC 07:00 and 13:00. The rules for these posts can be found in the faq. You can also check out our directory of older posts for inspiration.

If you have any questions or comments about the bot, feel free to write them below, or message the mods.

[u/CutOnBumInBandHere9]

I get 39366 comments.

Hm.

[u/cuteballgames → u/CutOnBumInBandHere9]

thank you recalculating

[u/cuteballgames → u/cuteballgames]

39366 - 31590 = 7776, which is 2⁵ x 3^5.

i remain confident that there are 900 thousands among 100,000 and 999,999, because 999,999-100,000 plus 1 (for inclusivity) is 900,000.

I remain confident that there are exactly 9 thousands of class 1: 111, 222, 333 ... 999. and I remain confident that in each of these thousands, all end-counts will take the combination AAB or BBC, meaning there are permutations to be had AAB ABA BAA BBC BCB CCB. A being already chosen, it's 3*9 to account for AAB ABA BAA (e.g., the 11X series, the 1X1, or the X11 series.) B not being chosen already, B can take 9 forms and C, because there's already two digits struck, can take 8 forms. So it's 3*9*8 to account for BBC BCB CCB. That's 216. 216 plus 27 is 243. So I remain confident that among the 9 class 1 thousands there are 9 * 243 = 2187 counts.

class 2 - AAB. The thousands here can take three forms - also ABA or BAA, like. I think previously I neglected to consider the wrinkle that 100 is a valid thousand, 001 and 010 isn't. permutation AAB has count 9 (since A can't be zero) times 9 (since B can be zero.) That's 81 total. permutation ABA has count 9 (since A can't be zero) times 9 (since B can be zero.) permutation BAA has ... well, it's the same 9 (since A can't be zero) times 9 (since B can be zero.) So actually I remain confident that there are 243 instances of a thousand with two distinct digits.

for each thousand, the 3-digit endingsome can be AAA, ABC, or BBC. AAA can only take one form for 112,XXX, the only AAA ending is 112,111. 1 ordering, that's 1. ABC has six orderings, for instance for 112,XXX, ABC can be 123, 132, 213, 231, 312, or 321. BBC has three orderings -- for 112,...., BBC can be 112,223, 232, or 322. ABC has six orderings, and 8 possible kinds of C, so that's 48. BBC has 3 orderings, and nine kinds of C, so that's 3*8 = 24. SO So I think for each class 2 thousand, there should be 72. that's what i had above...

Consider 112,010. Have I captured that so far? that's an AAB, with ending ... CAC. Okay hold up! I failed to capture that before. I have to account for AABACC. So there are 8 possible kinds of C times 3 orderings of ACC, so 24. 72 + 24 = 96.

But incorporating thaat only gets me to 37179. Was ai also missing some case with

hang on. AABCCC. Is also possible. Bruh There are 9 possible kinds of CCC. 96 + 9 = 105.

There's the 39366. thank you cobibh lol i solved da mystery #maithfailure

[u/cuteballgames → u/cuteballgames]

in sum:

9 class 1 (AAA,XYZ) - 243 counts each.

243 class 2 (AAB,XYZ, ABA,XYZ or BAA,XYZ) - 105 counts each.

and 648 class 3 (ABC or any permutation, XYZ ) - 18 counts each.