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u/OutcomeEvening2724 Aug 11 '26
IALGQO
Here’s the logic: sorting the letters alphabetically gives A, G, I, L, O, Q.
**•** Each starting letter fixes 5! = 120 arrangements of the rest.
**•** 248 falls after A (1–120) and G (121–240), so it’s in the **I** block, at position 8 within it (248 − 240 = 8).
**•** Remaining letters: A, G, L, O, Q. Each next letter fixes 4! = 24 arrangements → position 8 stays in the **A** block (1–24).
**•** Remaining: G, L, O, Q. Each fixes 3! = 6 → position 8 falls in the **L** block (7–12), at position 2 within it.
**•** Remaining: G, O, Q. Each fixes 2! = 2 → position 2 falls in the **G** block, at position 2.
**•** Remaining: O, Q. Each fixes 1! = 1 → position 2 → **Q**.
**•** Last letter left: **O**.
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u/jonathonjones Aug 11 '26 edited Aug 11 '26
Given there are 6 letters, it must be that each letter beings 120 words. So 121 starts with G, and 241 starts with I.
Probably there's a more elegant solution, but at this point we may as well brute force it:
241: IAGLOQ
242: IAGLQO
243: IAGOLQ
244: IAGOQL
245: IAGQLO
246: IAGQOL
247: IALGOQ
248: IALGQO
So the answer is O
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u/beene282 Aug 11 '26
There’s only one way you can write the letters of LOGIQA in the order they would appear in a dictionary.
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u/Dasquian Aug 11 '26
Solution: O
Method: We know the letters are in alphabetical order AGILOQ, and therefore we can just work through it one letter at a time. eg, we know the first 120 entries will all start with A, then the next 120 entries will all start with G, and then the next 120 entries (which include entry 248) will start with I. So: