## Saturday, December 27, 2008

### Year end something or other

## Sunday, December 21, 2008

### Needs more cinnamon

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.

## Thursday, December 18, 2008

### Two for one fun

## Saturday, December 13, 2008

### Time to get your grades

This past week I gave you two SAT problems. You can find the first one here. The second one required a diagram, and you can view that one here.

I also gave you a riddle of sorts, though it involved a lot of math. I appreciated all the complaints about the wording, but the fact remains that I got in one try and so did Eric.

Finally, this brain teaser comes to you compliments of my friend, Aithan. You'll find another question from Victor in the comments. I personally agree with Eric's answer to that one. Maybe Victor will tell us if he's right.

And what good are all these fun questions without answers?

SAT question one:

Answer: 105

For this problem there are a number of strategies you can use. One is to recognize that if the average of ten numbers is 100, then the numbers must add up to 1000. If we then remove the numbers 90 and 70 from the group we have eight numbers that add up to 840. Dividing this by eight gives us a new average of 105.

SAT question two:

Answer: A) 21

This problem simply requires you to know the pythagorean theorem which states that in a right right traingle with legs of length a and b and hypotenuse of length c, a^2 + b^2 = c^2. Using this on all the triangles eventually yields x^2 + y^2 = 21.

Mathy riddle question:

Answer: 3 minutes fast

The initial trip was suppose to take an hour, but was five minutes late, so it actually took 65 minutes. The fact that Mr. Moody's watch shows the train to be three minutes early indicates that he has no idea the train is actually late. Therefore, when he sets his watch forward he is actually setting to match an hour-long trip when the trip was actually 65 minutes. At this point his watch is off by five minutes. The return trip is 25% faster. This refers to the speed, not the time. To figure out what this means for the speed we can use the face that distance = speed x time. For both trips the distance remains the same, so speedtrip1 x timetrip1 = speedtrip2 x timetrip2. We also know that speedtrip2 = 1.25 x speedtrip1. This means that speedtrip1 x 65 = 1.25 x speedtrip1 x timetrip2. Dividing both sides of the equation by 1.25 and speedtrip1 gives us timetrip2 = 52. This is a difference of eight minutes on a trip that should take one hour. However, because Mr. Moody's watch is off by five minutes, the station is actually only fast by three minutes.

People on a bus question:

Answer: 4 people get off the bus

There is a pattern to this problem. One way to figure it out is to consider a group of four people. Let's assume one has a blue hat. The person with this hat will see all white hats on the other people. Since they know someone has to have a blue hat, this person will immediately know he has the blue hat and get off at the first stop. Now consider the case where two people have blue hats. They will both see one blue hat and two white hats. If the person that they see with the blue hat was the only one wearing a blue hat, that person would immediately know it and would have gotten off at the first stop. Because he did not get off, then there must be another blue hat. Since these people can only see one blue hat, they know they must be wearing the other one, and they will then get off at the second stop. Assume now that there are three blue hats. These three people see two blue hats and one white hat. They know that if there were only the two blue hats they see then those two people would have gotten off at the second stop. When they don't get off they know there must be another blue hat. Since they only see two, they must be the one's wearing it. They will then get off at the third stop. Now think back and realize it doesn't matter how many white hats there are in any of these situations, so we don't need to start with four people for this to work. The pattern will continue, so if people get off at the fourth stop, there must be four blue hats.

PS. I advise you to watch this.

## Tuesday, December 9, 2008

### Mid-week special

Here's some boring wood I got for the future:

I also purchased a new gouge and a veiner. A veiner has a really small U-shaped cutting edge which I'm pretty sure you can't make out in this photo. Next to that is a photo of the finger guards I got.

The finger guards basically guard the hand you hold the knife with. They allow you to cut toward your thumb without fearing for your life as shown below.

And finally, I have another SAT question for you. You need the diagram for this one, so I couldn't just put it in a caption. HAVE FUN:

## Thursday, December 4, 2008

### The first one

Ok, so first a little tour. On the right you'll find links to the main album page as well as links to current, completed, and future projects. You will also find an entirely unenlightening profile page and a followers thingy. I don't know exactly how that works, but if you check out my stuff regularly be a follower. I know most of the people that check stuff daily, but I'm always curious to see if there are any random people out there. My minimal research suggests you have to have a google account to use that, though, so maybe it's not going to be so useful.

Now some stuff you may have missed.

Did you know my carving has prompted Palestinian and Israeli cooperation? Unfortunately, peace talks broke down rather quickly. If you have anything to say about either group, I invite you to leave your comments.

If anyone else has any Plaxico Burress comments, this is the place to put them. At least he doesn't drop every pass or bullet like some receivers.

Try to answer this if you haven't yet. Mike, Victor, and Eric have given it a shot. They all got it, but none on the first try.

I asked for some suggestions here, but no one actually submitted any. Unless whatever Az wrote was a suggestion? Anyway, I'm not joking. Leg pictures have to be getting boring.

Eric is trying to insult me here. I hope you'll help him out.

There are some pictures from a recent visit to Eric and Chrissy in Hartford. Oh, thinking about Chrissy gave me an idea. One sec. Ok, I added a link to her stuff over there on the right. She makes good stuff. Ok, Mike and I were visiting and we revived an old tradition. Start there and keep going. You can't miss it.

I think that's good for now.

The lion is finished, so for now enjoy these pictures. You can see the full set with all your comments in the finished projects section.

Completed lion:

The lion was always for my sister, but thanks to some fortunate timing and manipulation of the rules it was also able to fulfill my secret santa obligation. Of course, I have to make it extra entertaining, so I carefully boxed her gift and wrapped it with some nice Christmas wrapping paper. When she opened it she saw this:

She tends to cry when she's laughing hard. I didn't hurt her this time. She is also insisted on keeping the full block of wood that cost me $14. Couple more of her with the lion(s).

My brother also cheated to get me as his gift receivulator. He got me a sharpening stone for my knives.

The oil is freshly squeezed from my hair. I think it's olive oil.

On the fun riddle front there were really only a couple things this week. Mike sumbitted this guy, which no one actually answered. Are you going to let him win like that?

I gave you these two as well.

As usual, here are some answers.

For the triangle problem, the key is to realize that the two things aren't actually triangles. The red and green triangles in each picture have different slopes, so they don't actually make a straight line when joined up. One line is slightly indented and one bulges slightly. This accounts for the one missing block.

For the pool problem start by figuring out how many pools each pipe can fill or empty in one hour. This comes out to 1/3 and 1/6 filled and 1/9 emptied. Together they fill 1/3 + 1/6 - 1/9 pools in one hour. Multiplying by the .64 hours we want we get .25.

For the pyramid problem the fact that both must have the same mass means that the one with 1/7 density must have 7 times greater volume. While it helps to know the formula for volume of a pyramid, V = (1/3)Bh, it's enough to know that volume has units of some measurement cubed. Since all measurements in a scale model must increase or decrease proportionally, height will increase as a cube root of volume. This means the original height of 6 gets mulitplied my the cube root of 7, giving a final height of 11.48.

I'll be going to Florida for a week on Sunday, so nothing new for at least that long. When I get back I'll update the preview section and start something new. In the mean time feel free to entertain each other in my absence. By that I mean fill this thing up with racial slurs and jokes about moms and faces and what she said.