Needs more cinnamon

It was another week filled with riddles and puzzles. I shifted away from math heavy questions to more logic based stuff. Some of you spent up to three hours answering some of these, while others spent up to three hours writing computer programs that would answer for you. At least you didn't spend that time working at your jobs where you get paid to work. +/-

You'll find the first question here. There's still some math involved in this, but I think everyone was able to get it. Mike also posted another problem in the comments. That one is entirely logic based.

For whatever reason people decided to be mean in this picture and in the next one. At least keep the criticism focused on my I-talianess like everyone else does, Mike. When you make fun of my wood I cry.

I kept up the logic theme with this question. Victor was able to get it in roughly three hours, while Mike spent that much time relearning Prolog and having a computer solve it. He was thoughtful enough to include his code for everyone else to enjoy.

I gave you two puzzles near the end of the week, which you can find below in "Two for one fun." Again Mike was able to write a program to solve these, but so far no one else has solved them.

Mike has also provided something for you to think about this week. The picture below says it all.

AND FINALLY, AND THIS IS IN ALL CAPS BECAUSE OF HOW IMPORTANT IT IS, I LEAVE YOU WITH A VERY SPECIAL VIDEO. ENJOY.

Solutions to last week's questions are here.

Address question:

The address of Mr. Street is 14 Main Street.

There are 8 possible combinations of answers on the three questions:

Case: Which block? Even? Square? Conclusion
A first yes yes 4 or 16
B first yes no many
C first no yes 1 or 9
D first no no many
E second yes yes 36
F second yes no many
G second no yes 25 or 49
H second no no many

Now it turns out that we only shoud consider cases A and G (because then the question about whether there is a 4 in the number makes sense). In case A the real number should be larger than 20, odd and no square. But then it would be impossible for Mr. House to know the correct number.
So we're left with case G. Mr. Street lives in the first block, with an even number, which is not a square. There is only one number which has a 4 in it, number 14.
Mr. House was told by the passer-by to go to number 25. But Mr. Street lives on 14 Main Street.


The fish question:

The answer is the German, but the process takes too long, so I'm not going to go through it all.


Long division problem:

The letters and numbers match up like this:

0 1 2 3 4 5 6 7 8 9
F L O W S U N D E R

This might take too long to explain, but I'll give it a shot. F has to be 0 because of the last subraction where D-D=F. U is 1 more than S and N is 1 more than U because both subtract to nothing, meaning in each case one was borrowed before the subraction. Similarly, E is one more than D because in the second subraction E-F=D. Since F is 0 E must have had one borrowed first. L is the letter that needed to borrow this one, so it must be less than D. From the last subraction D-N=L didn't need to borrow from the U, so D > N. L did have to borrow from N in the second subraction though, so L is also less than N. From all of this we know that so far the letters SUN and DE must be next to each other with DE being higher than SUN. We can write those borrowed subractions involving L as the equations 10+L-D=S and 10+L-N=U. Some rearranging tells us that D+S=N+U. With the fact that SUN must be together, this means DE must follow right after the N. The only numbers that now work for these letters are S=4, U=5, N=6, D=7, and E=8. From here it's relatively simple subractions to fine the other letters. There are also probably other ways to figure this out.


Two for one fun puzzles:

For the first puzzle it works out this way:

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

There are only two ways to add to 33. Either 98763 or 98754. G must be 4 though, because the other 8 spaces total 21 and 20 and the numbers 1-9 total 45. So the 33 is made by the numbers 98754. This leaves a total of 12 for the other spots, made by the numbers 1236. The only possible combination for 17 that works with the above facts is 467, so H must be 7 and I must be 6. This leaves 7 more in C and D to complete the 20. 3+4 is not possible because the 4 is used, so these must be 5+2, with the 5 in C and the 2 in D. Now the only remaining numbers that work for 15 are 483, so E is 3 and F is 8. This leave a for A and 9 for B.


Second puzzle:

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

This one is a little tougher.The 22 and 15 include everything except H, so H must be 8. There is only one way to sum to 34 and that's with 98764. This leaves 1235 for the other spots. 4 can't be part of 22 or 24 because it is too small given the constraints that A, B, E, and F are some combination of 1, 2, 3, and 5. This means the only place 4 can go is I. The 24 also can't have both 6 and 7 for the same reason. This means 9 must be part of 24. The 9 can't be part of the 15 because this would require 123 to be the other 3 spots for 15, but the one of those spots is shared with 34 which can't have 1, 2, or 3. This puts 9 at C. The only way to reach 24 is 9753, so D is 7 and G is 6. To reach 15 F must 1 and E must be 3 and to reach 22 A must be 2 and B must be 5.