I have previously posted number puzzles. This time it's letter puzzles. Some clues have a Harry Potter theme.

12 puzzles


The answer is a 6-letter word

Hint 2:


Hint 3:


  • $\begingroup$ Just a comment - if the “hints” are more or less mandatory then include them in the puzzle and not as a hint (the second hint seems pretty close to this); if they’re NOT mandatory, it’s recommended you wait a while and let folks have a chance at solving it without them. See also puzzling.meta.stackexchange.com/questions/5005/… and its answers. $\endgroup$
    – Rubio
    Mar 2, 2020 at 3:22

2 Answers 2



Answers provided for all 12 clues, including 4 found by @Randal'Thor...

1 = X (In A1Z26 each row sums to 24; the 24th letter is X)

2 = L (Alternate letters spell the words WIZARD anti-clockwise and MUGGLE (JK Rowling's word for a non-magical person) clockwise, with the ‘L’ missing)

3 = W (Bottom letters are initials of the 4 ‘Marauders’ who created the "Marauders' Map" first seen in the third HP book), and the top letter the initial of their nickname: James Potter = Prongs, Remus Lupin = Moony, Sirius Black = Padfoot, and Peter Pettigrew = WORMTAIL, which begins with W)

4 = V (Sequence snakes down a column, up the next, etc., moving 3 letters on each time) (also found by @Randal'Thor)

5 = R (Sequence is +2, +3, +4 and then +5)

6 = H (Each triangle represents the initials of one of the 3 main characters of the HP books: Hermione Jean Granger, Ronald Bilius Weasley, and HARRY James Potter, which yields H)

7 = H (Sought-for letter is midway between the pair of letters) (found by @Randal'Thor)

8 = K (Sequence is alternate terms of +7, -4)

9 = I (Opposing pairs of letters add to the central S (19) in A1Z26) (method found by @Randal'Thor, though his answer is incorrect!)

10 = L (Arithmetic with Roman numerals) (found by @Randal'Thor)

11 = F (Letter on left is 5 on from letter on right, moving from Z to A when required) (found by @Randal'Thor)

12 = G (The circles hide four 4-letter creatures from the HP books - find them by choosing a letter from the first circle, then choosing the next clockwise from each subsequent circle. This produces OGRE, TOAD, STAG and FROG if the missing letter is a G.)

These give us the letters:


This letter sequence:

has been encoded with a substitution cipher where each of the first 13 letters of the alphabet are reflected about the halfway point (M/N) and substituted with the letter in the second half of the alphabet that is the same distance from Z as the original letter was from A.

i.e. A = Z, B = Y, C = X, etc.

Translated, this reveals:

CODE IS SPROUT ('Sprout' being the name of the Head of Hufflepuff House and professor of the magical plant science Herbology in the Harry Potter series).

  • $\begingroup$ All are correct except #12. It is a good solution but not the intended one. $\endgroup$ Mar 1, 2020 at 17:21
  • $\begingroup$ @eyl327 Is it a Harry Potter one? $\endgroup$
    – Stiv
    Mar 1, 2020 at 18:08
  • $\begingroup$ In a way, it is Harry Potter themed. Does not require any knowledge of Harry Potter, though. $\endgroup$ Mar 1, 2020 at 18:33
  • 1
    $\begingroup$ @eyl327 Aha, got that one now! Nice puzzle clue that one :) $\endgroup$
    – Stiv
    Mar 1, 2020 at 19:04
  • $\begingroup$ You got it. I added hints for the final step. $\endgroup$ Mar 1, 2020 at 20:10

Partial answer for the clues:

  1. $V$ (in each column the letters are advancing by 3 places through the alphabet as you go up or down)

  2. $H$ (the letter on the right is exactly in between, the midpoint, of the two on the left)

  3. I wonder if this might be something to do with

    each letter representing one of the seven Harry Potter books, somehow. Just a hunch.

  4. $G$ (treating letters as numbers, the sum of each opposite pair is the one in the middle)

  5. $L$ (arithmetic with Roman numerals)

  6. $F$ (the letter on the right is five before the one on the left in the alphabet, cycling if necessary)

Partially filled grid:

? ? ? V
? ? H ?
G L F ?

  • $\begingroup$ The ones you have filled in (4, 7, 9, 10, 11) are correct. $\endgroup$ Mar 1, 2020 at 17:35

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