# The missing page of Adrian Puzzlinger

“Knock, knock, Adrian Puzzlinger. It’s the puzzled police. Are you home?”

“Unless I’m mistaken.”

“We just received a tip. Is one of your puzzle preparation pages missing, perchance?”

“I need my missing page for the puzzle I’m presently preparing.”

“What does it look like?”

“Exactly like this.”

 39 1 2 3       35 1 2 3 4       30 1 2 3 4 5       24 1 2 3 4 5 6         17 1 2 3 4 5 6 7
38 1 2 4       34 1 2 3 5       29 1 2 3 4 6       23 1 2 3 4 5 7         16 1 2 3 4 5 6 8
37 1           33 1 2           28 1 2 3           22 1 2 3 4             15 1 2 3 4 5
32 1             27 1 2             21 1 2 3               14 1 2 3 4   7
26 1               20 1 2                 13 1 2 3
19 1                   12 1 2     6
11 1
6)             5)               4)                3)                   2)  10       5
9             9
11           9          8     4     8 9
14         9       10         8 9          7         7 8 9
18       9       13       8 9        9       7 8 9          6   3   6 7 8 9
23     9       17     8 9       12     7 8 9        8     6 7 8 9          5     5 6 7 8 9
22 6 8 9       16 5 7 8 9       11 4 6 7 8 9        7 3 5 6 7 8 9          4 2 4 5 6 7 8 9
21 7 8 9       15 6 7 8 9       10 5 6 7 8 9        6 4 5 6 7 8 9          3 3 4 5 6 7 8 9


“Why do the subsequences in the top two and bottom two rows add up to 45?
And why do other subsequences add up to less?”

“Because this is my missing page.”

“But you plainly have it, so just what is missing?”

Just what, indeed, is missing?Why?What type of puzzle is Adrian Puzzlinger preparing?

• Related :P Sep 4 '16 at 9:11
• You're missing an 'a' in 'human'. :P Sep 4 '16 at 10:32
• Lost my 'a' in a bet to a user named 'rdvark', who's still looking for a nother sucker. Word to the wise, @Mithrandir, you have an 'a' too, watch it closely!
– humn
Sep 4 '16 at 10:39
• Yeah, be careful @Mithrndir or you'll end up in this puzzle being Mthrndr! Sep 4 '16 at 11:01
• @randal'thor I was so tempted to change my username at that, but I learned my lesson. Sep 4 '16 at 11:12

The puzzle Adrian is preparing is

a Killer Sudoku, or Kakuro, or any other puzzle which involves solving for a bunch of unknown numbers some of whose sums are known.

I'll try to flesh out the reasoning here (it's quite hard to explain in words!), but the basic inspiration for Adrian's mysterious page is that

as all Kakuro or Killer Sudoku solvers know, there are certain given sums which can tell us a lot about the summands. For instance, if we know two numbers sum to $17$, then they must be $8$ and $9$ in some order. If we know four numbers sum to $11$, then they must be $1,2,3,5$ in some order. If we know three numbers sum to $8$, then they must be either $1,2,5$ or $1,3,4$.

all possible sums which are restricted in this way: all $n$ such that when $k$ distinct digits have to sum to $n$, we have some restrictions on what those $k$ numbers can be.

• The columns are separated into groups labelled $6),5),4),3),2)$, each of which corresponds to

a value of $k$, this value being the label with the bracket. These are the interesting values of $k$, since $k=1$ is trivial and $k=7,8,9$ are rarely encountered in actual puzzles.

• Each of these column groups contains several rows, each of which corresponds to

a value of $m$, this value being the leftmost number in the row.

• The other numbers in the row correspond to

the numbers which cannot be included in a set of $k$ digits which sum to $m$.

As a random example, let us take group $3)$, the row beginning with $21$. What does this row tell us?

Here $k=3$ and $m=21$, so we're looking for a set of three distinct digits summing to twenty-one. What we know about these six numbers, thanks to Adrian's table, is that none of them can be $1,2,3$. Why? Because if any of them was that small, their sum would be at most $9+8+3=20$, which is just too small.

Another example: column-group $2)$, row beginning $8$.

Here we're looking for two distinct digits summing to eight. Clearly neither of them can be $8$ or $9$, and (since they are distinct) nor can either of them be $4$.

The significance of the number $45$ is of course that

it's the sum of all the digits from $1$ to $9$.

• The explanation would be simpler and more in the spirit of "missing," though no more correct overall, if you take 6), 5), ...2) at face value as $k$ and, in a sense, shake the bush by its roots instead of its leaves. No rush, and I'd be glad to add the original perspective as a note eventually if you don't.
– humn
Sep 4 '16 at 11:15
• @humn OK, gotcha. Then the numbers appearing in each row are those that can't appear in the collection summing to $m$ rather than those that must appear. Will edit. Sep 4 '16 at 11:43
• @humn How does it look now? Sep 4 '16 at 13:28
• Beautiful! And wish I could ^vote extra for explaining the half-number inclusions/exclusions.
– humn
Sep 4 '16 at 18:57
• @humn No worries :-) And I'm impressed at your ability to do those nice green ticks in comments! Sep 4 '16 at 19:04