Solution to Mysterious Figures

We're presented with a big table of descriptions, each of which contains a logic clue and describes a person. We start by identifying the people.

After identifying enough of the people, we notice that they all are "Mr." something. This is reinforced by the title, Mysterious Figures.

At some point, we may wonder why the formatting on some of the clues is funny. Or we may wonder why some of them don't have the right personality for the person they clue. Eventually, we may think to look at the Mr. Men series and note that each of the clues refers to a Mr. Man in addition to a "real-life" Mr. person.

While we identify the Misters, we can simultaneously solve the logic puzzle. See the appendix for a logical path.

We reorder the rows by publication order to get a cluephrase: YOU FOUND US! ONE OF OUR COMRADES IS MISSING. WHO IS HE? OMIT MR.. The missing Mr. Man, out of the original 49, is Mr. Topsy-Turvy. Thus, the answer to this puzzle is TOPSY-TURVY.

Appendix: Logical Path

To start solving the logic puzzle part of this puzzle, it's easiest to look at only the logic puzzle clues in a grid, and mark them off as we've used them. Throughout this discussion, we will use R1C2 = row-1-column-2 notation to refer to the grid.

First, we make an observation: R4C3, R6C7 and R6C4 help us understand the rules about neighbors: neighbors are orthogonally adjacent, not wrapping around the grid and not including the center square. Additionally, for the purposes of this logical solution, we will assume that the numbers in the grid are positive integers.

A lot of the clues will be easier to use if we have an upper bound for the grid, so the first thing we do is try to attain one of those. The key here is R3C5, which is tied for the largest number in the grid. We can constrain its value using R3C4 and R3C3, as well as the clue in R1C4. If we aim to maximize the value of R3C5 given the clue in R3C4, we come up with R3C5 = R3C4 + R2C4 + R3C3. R3C4 + R2C4 is upper bounded by 18−3=15 because C4 adds up to 18, and R3C3 is upper bounded by 11−2=9 because of the clue in R4C3. Thus, R3C5 is at most 24. We know that R3C5 is a perfect power based on the R3C6 clue (it can't be a perfect first power because R1C6 must be strictly larger than R2C6). Thus, the largest number in the grid is at most 16.

Now, we can write down the possibilities for each square:

The next thing we can think about is the composition of row 5. R5C2 is the only odd number in the row, so the others must all be even. Thus, R5C4 is 2. Then, R5C6 and R5C7 must both be even and they cannot both be two, because there would be at least three twos in the row. Thus, one of them is two and one of them is four, and the row has 2 fours and 4 twos in it. R5C3 has more holes than any of its neighbors, so it cannot be 2 which has 0 holes. It is thus 4, and then R5C1 and R5C5 are 2.

We can now reapply R4C3 and R5C5 to narrow down some of their neighbors' possibilities.

The next clue to apply is R6C6. There are three sets of four positive integers with harmonic mean 15/7: 1,3,3,5; 1,2,5,6; 1,2,3,30. The third set is impossible given the maximum value in the grid, and the first set is impossible given that there's already a 2/4 adjacent. Thus, the four adjacent numbers are 1/2/5/6.

We can apply the clue in R5C1 to increase the lower bound on R5C2. We can also use this in combination with the R6C2 clue to narrow down the values of R5C2, R4C1 and R6C1.

We reapply the clue in R5C3 to eliminate some possibilities from its neighbors.

Reapply R4C2, R3C3, R4C3 repeatedly to narrow down those three values until only one possibility remains.

R4C1 must be 6−5 or 8−3 or 9−2, and only one of these is 1 or 3.

Now that we know the values of R3C3 and R5C4, we can redo our upper bound calculation. R2C4+R3C4 is now at most 18−2−1−1=14, so R3C5 is at most 14+1=15. Given the perfect power constraint, it's now limited to 8 or 9. Since R4C5 must be unique and even, R4C5 must be 8 and R3C5 9. This also fixes the values of R2C6 and R3C6, by the rule in R3C6.

We can determine the value of R2C5 from the clue in R2C5.

Now, R1C6 is either 7 or 9, since it must be larger than 6 and at most 9, and nothing new can be 8. If it were nine, R1C5 would be greater than nine, so it must be seven. Then R1C5 must be the other 9 in the grid, so everything else is at most 7.

Since we have determined the location of the other nine, we can determine that the only way to satisfy the clue in R3C4 is for R3C4+R2C4 to equal 8. The second half of the clue in R2C4 then limits these to either 1/7 or 3/5.

Now, the only way to make an arithmetic sequence from 2,6,5/7 and 2/3/5/7 is 5,6,7, so R2C3 is 5/7.

Using the clue in R1C4, we can determine that R6C4+R7C4=8. So R7C4 is at most 7, and it must divide 60 so it is not 7. R7C7 must be at least 4 since there are already one 1, two 2s and three 3s in the grid, so R7C4 must be 4, 5, or 6 and must equal R7C7. If these two squares were four, then R6C4 would also be four by rule R1C4, which would put 5 fours in the grid. So they cannot be four. If they were five, R7C2 would also have to be five, since it cannot be 1, and then R6C3 would be two by the rule in R6C2. Then R6C5 would have to be 3 to make the product of 60 work, which violates the harmonic mean rule. Thus, R7C7 and R7C4 must be 6.

Now, the only possibility in R6C7 is 5, because neither 1 nor 6 would have any of the two difference-one neighbors satisfied so far. Then the other harmonic mean numbers are 1 and 6.

Now R6C3 is the only place for the five to come from for the product of 60, so it must be 5. This fixes the last element of the product, R6C5, at 1, and then R7C6 as the last element of the harmonic mean.

Setting R6C3 to 5 also fixes the sum of 16, setting each of the others to their minimum values.

We can now evaluate the rule for R7C3.

We have found all six sixes in the grid, so by R7C7 we know there are no more.

The sum of the grid so far is greater than 49, so the largest proper divisor of the sum is greater than 7 unless the sum is prime. R2C7 cannot be greater than 7, so it must be 1.

By the row constraint, R4C7 must be 1 or 3, so R3C7 must be 4 or 6. It can't be six because we've already found all the sixes, so they're 1 and 4.

We noted earlier that R6C4+R7C4=8, so R6C4=2.

Two already appears more than 8 times in the puzzle, so R7C1 cannot be 2. Thus, since the row sum must be odd by R7C2, R7C6=2.

Rows 4, 5, and 7 cannot add to 30. R2C2 enforces that R2C1 through R2C3 add to a multiple of three, so the row cannot add to 30. R6C5 enforces that R6 cannot add to 30. Then, by R7C5, rows 1 and 3 add to 30.

By this row sum rule, R3C2 is 4 or 6, so it must be 4. This fixes the 5/7 pair by the row sum rule, and it also tells us that there are exactly six fours in the grid. Finally, R2C4 is fixed by the rule in R1C4.

By R1C1 and the row sum of 30, C1 must sum to 15. If R7C1 is 3, then R1C1 and R2C1 must both be 1, but then we have a contradiction in R2C1. So R7C1 is 1 and there are eight ones in the grid.

If R2C1 were even, R2C2 would not be an integer, so R1C1 must be 1 or 3. If it's three, then R1C1 must be 1 by the column sum and R2C2 must be 4 by mean. However, that would put us at 8 ones and 6 fours, which is the number of both we need in the grid; then there would be no possible values for R1C3 to take. Thus, R2C1 is 1.

We can then determine the values of R1C1, R2C2 and R1C2 by their respective rules.

We're in the home stretch! The first row adds to 30, so R1C7 is 2 or 5. There aren't enough fives for 5 to be the most common number in the grid, so the remaining entries in row 1 must be 2 and 4.

Then the remaining two entries must be 2 and 3, so that the row sum satisfies R6C5 and the number of ones and fours in the grid is not too big. Thus, there are twelve twos in the grid, so R6C2 must be three to make R6C3 be satisfied. Then R6C6 is 2, and we've filled the grid.