Guilds Puzzle League!

[worldwideweb3]

Nice work, oty's. On to the next challenge!

[Submachine]

I know that this challenge is long over, but I'd like to present a new logic task for those who are interested.

The following table is given:

0	1	2	3	4	5	6	7	8	9
?	?	?	?	?	?	?	?	?	?

The task is to fill in the questions marks with numbers from 0 to 9. The rules are the following:

• The numbers in the bottom row are showing an amount. This amount refers to another number that is in the top row of the same column.
• The indicated amount of numbers must be placed in the bottom row.
• To complete the task, all ten question marks must be filled with a number from 0 to 9.

For example:

0	1	2	3	4	5	6	7	8	9
				2

This table would mean that there is an amount of 2 of the 'number 4' in the bottom row.

Spoiler for Hard Mode:

Solve the puzzle for an n-column table.

0	1	2	3	...	i-1	i	i+1	...	n
?	?	?	?	?	?	?	?	?	?

Based on the single solution I found, it might be possible to solve the puzzle for n = imaginary 1-digit numbers over 9.

[Zawadx]

Your wording for hard mode was a bit confusing. What you want is an n-column table, where the numbers in top row run from 0 to n-1. The numbers in bottom row can be from 0 to n-1, reflecting the number of times the number above it appears in the bottom row. n is any natural number.

Using i in the table and putting n = imaginary number was confusing, because i refers to the standard imaginary number and those don't interact with normal numbers easily.

Spoiler for solution:

There are three solutions if n<7:

0	1	2	3
1	2	1	0

0	1	2	3
2	0	2	0

0	1	2	3	4
2	1	2	0	0

For n greater than or equal to 7, a solution is:

0	1	2	...	n-4	n-3	n-2	n-1
n-4	2	1	0	1	0	0	0

Easy to check why, but the how is a bit tough. First, we define the numbers in the bottom row as a(k) where k is the number in the same column above it:

0	1	2	...	n-3	n-2	n-1
a(0)	a(1)	a(2)	...	a(n-3)	a(n-2)	a(n-1)

Now, all the a(k) must add up to n because you only have n numbers in the bottom row (and each of the a(k) is the no. of times k appears on the bottom row). So a(0) + a(1) + a(2) + ... + a(n-2) + a(n-1) = n

Again, because a(k) is the no. of times k appears in the bottom row, you can calculate its sum as the sum of k*a(k) for all k. Therefore, 0*a(0) + 1*a(1) + 2*a(2) + ... + (n-2)*a(n-2) + (n-1)*a(n-1) = n

If you put in the value of n from first equation in second and take all the terms to left side, you get:

(-1)*a(0) + 0*a(1) + 1*a(2) + ... + (n-3)*a(n-2) + (n-2)*a(n-1) = 0
or, 1*a(2) + ... + (n-3)*a(n-2) + (n-2)*a(n-1) = a(0)

Now, let a(0) = X for a solution to the problem. Then a(X) > 0. If a(X) > 1, then (X-1)*a(X) > 2(X-1) > X for all X>2 (and we can prove that X>2 by using the fact that n<7, which can be used to prove that a(n-1), a(n-2) and a(n-3) are all 0).

So, if a(X) = 1, then 1*a(2) + ... + (X-2)*a(X-1) + (X-1)*a(X) + (X)*a(X+1) + ... + (n-3)*a(n-2) + (n-2)*a(n-1) = a(0) leads to:

1*a(2) + ... + (X-2)*a(X-1) + X - 1 + (X)*a(X+1) + ... + (n-3)*a(n-2) + (n-2)*a(n-1) = X
or, 1*a(2) + ... + (X-2)*a(X-1) + (X)*a(X+1) + ... + (n-3)*a(n-2) + (n-2)*a(n-1) = 1

So if any a(k)>0 for k>2 and k=/=X, the left hand side becomes greater than one. So, a(k) = 0 for all k > 2 and k=/=X

or, 1*a(2) + 0 = 1
or, a(2) = 1

Now since a(2) = 1, a(1) > 0. So a(0) is the number of numbers from 3 to n-1, minus 1.

So, X = n-1 - 3 + 1 - 1 = n-4

And so the solution I found before is legit, and it is a unique solution ^^

Edit: Clarified solution for more rigor, thanks Sub for catching my mistake and showing me the other solutions!

[Zawadx]

So I awarded points based on n(n+1)*5/2, where n is the last round a team participated in. Tbolts count up to round 5. Oty counts to R7 as they won R6. So points are:

OtyUnity - 140
Bunchie Peace Embassy - 105
Thunderbolts - 75
AbomiNation - 15
RainBros -15
The Graveyard - 15

[Submachine]

Here is a short puzzle for fun.

There are 9 letters: A, B, C, D, E, F, G, H, I. These letters represent different numbers from 1 to 9. Two letters cannot represent the same number and all numbers from 1 to 9 are used. The task is to determine which letter represents which number. The following equations will help solving the puzzle:

• A + B + C = 18
• B + C + D = 21
• C + D + E = 16
• D + E + F = 15
• E + F + G = 10
• F + G + H = 16

There is only one good solution.

[Linkcat]

Spoiler for Solution:

I  = 1
G = 2
E = 3
A = 4
F = 5
C = 6
D = 7
B = 8
H = 9

[Ginyu]

Spoiler for Hidden:

a = {4}
b = {8}
c = {6}
d = {7}
e = {3}
f = {5}
g = {2}
h = {9}
i = {1}

[Zawadx]

Spoiler for How I solved:

I just subtracted each equation from the next, obtaining some shapes for a, d, g; b, e, h and c, f. Then considering the possible arrangements of numbers according to these shapes gives the solution.

Here's another puzzle... of sorts. It was a problem in the IMO this year, but the math involved is elementary. The puzzle like factor makes it challenging:

[Submachine]

Lately, someone has been writing puzzles on the cafeteria board. Last month's puzzle was something like "2, 6, 30, 210, 1890, ?". I could solve that one, but I still don't know the solution to this month's puzzle.

It is as follows:

23, 27, 29, 34, 37, 38,      43, 46, 47, 49, ?

I don't know if the big space between 38 and 43 is intentional, but it's there. The tasker wrote a Hint#1 below the puzzle last week, and by this week, somebody already solved it.

Try to solve this puzzle. If you send me a PM, I can give you Hint#1.

[dragtom]

Spoiler for probably an incorrect attempt:

53
difference between 2 consecutive numbers have been
4, 2, 5, 3, 1,    5, 3, 1, 2.
suggesting the new difference would be 4, as that 'finishes' the set.
Doesn't allow me to give the next set though, so it is most likely incorrect.

[Submachine]

I finally convinced the puzzler to give the method of the solution. Without giving it away, the order of the numbers is not relevant in this one. It's just that between 0 and 49, these are the only numbers that share this "trait". The gap is apparently irrelevant too.

With this new information revealed, I am still only giving Hint#1 to those who ask privately. At least until later.

Now without the gap:

23, 27, 29, 34, 37, 38, 43, 46, 47, 49, ?

[Submachine]

Enough time passed, so I'm posting the details.

Spoiler for Hint#1:

None of the numbers have 1 in them.

If the hint is not enough, you can read the solution below.

Spoiler for Solution:

This is a list of numbers that cannot be divided by their own digits, in ascending order from 0. So the next step, and also the solution, is 53.

Kaempfer and dragtom got to the correct solution but not with the intended method.