LHIRT POD #3

It's time for a little number theory for the POD:

Let _nC_r be the number of ways to choose r items from n objects (or "n choose r"). Assume p is prime. Determine what the remainder of _2pC_p divided by p is, and prove it. For extra credit, is the converse true?

This one may be a little tough if you don't see it off the bat, so I'll leave it up for a while. Besides, I'm not entirely sure about the converse part myself. :)

LHIRT POD #2 solution

Solution for POD #2 below (white on white below):

Let the point on the inner circle be T. Because PQ is perpendicular to OT, OTP is a right triangle. Therefore OP²=OT²+PT². Now, the true area of the doughnut is (π/2)(OP²-OT²)= (π/2)(OT²+PT²-OT²)=(π/2)(PQ/2)².

Amazing, huh? Of course the smart-ass answer is that because a formula must exist for this area based on the length of that tangent alone (or else the problem wouldn't have an answer). Since the radius of the inner circle does not matter, make it radius 0, and the area will stay the same or just the area of the remaining circle. In a non-authoritative setting this form of reasoning is invalid, because you can't be sure if the questioner is being honest in implying the existence of that formula.

Another problem will be posted tomorrow.

Ha ha

It seems that I can't keep up with a daily puzzle, it is a lot of work and my schedule prevents it. I will be instead doing a couple of PODs (Problems of Difficulty :P) a week. This will also allow me to vary my posts more. There is too much in my life going on just to flood it with daily puzzles, no matter how much fun they are.

There's a lot of cleaning in my life recently. Recently I sold my PS2 and all games. I've vastly reduced my diet. My activities are paired down to affordable basics. My theory is that the less I'm invested in, the more invested I am in those activities. Maybe it'll pan out, maybe not, but it's something to try.

LHIRT's POD #2

The problem derives itself from an old puzzle book that solves this problem in a very clever way. I am looking for the real solution, which isn't more difficult than the original method to solve it.

Take a pair of concentric circles, that share a center O. Take a point on the inner circle, and extend its tangent to its intersection with the outer circle at points P and Q. Given only the length of this chord PQ of the outer circle, calculate the area between the two concentric circles.

I will give both solutions in the morning, but I am looking for the one that does not assume that a formula exists in the first place.

Answer to POD #1

The answer is in white on white below, highlight to verify your answer. Surprisingly this didn't take me much time, so I'll continue with my questions tomorrow. I will have to qualify by stating that I may pose a problem that I don't know the answer to, and may pass on the problem until I can research it further. This is all but an experimental exercise. :)
Here's how you figure out the expected number of complete rows: there are five possibilities to consider, because you can leave anywhere from 0 to 4 complete rows. You cannot leave 3 or 4 complete rows because of pigeonhole principle: 4 is impossible after removing one, and 3 is impossible after removing 4 (because you can remove an entire row and have one more to remove, destroying another row). So, in all C(12,4) possibilities there are 3 outcomes: leaving 0 rows, leaving 1 row, and leaving 2 rows.

Case 1: leaving 0 rows... you can do this by choosing 1 from each row, or 3*3*3*3 of 81 possibilities.

Case 2: leaving 2 rows... you can do this by first choosing the rows that are complete ( C(4,2) ) then finishing by choosing what to take within the rows that aren't complete ( C(6,4) ) leaving C(4,2)*C(6,4)=6*15=90 possibilities

Case 3: leaving 1 row. First, choose the complete row C(4,1) then choose which 4 of the 9 remaining to remove. But, be careful: of those C(9,4) you must exclude the ones that leave a full row which is [choose first the row remaining then the 4 from the 6 remaining, or C(3,1)*C(6,4)=3*15 or 45] so C(4,1)*(C(9,4)-45)=4*81=324.

To confirm: 81+90+324=495=11*5*9=12*11*10*9/4*3*2*1=C(12,4)

So, now that we have all the possibilities, let's calculate the expected value E=(81*0+90*1+324*2)/495=738/495=1.49090909... so the expected number of rows is about 1 1/2 rows. More likely than not, you'll have 1 complete row remaining.

I had a really fun time tonight, scoring 46 points in the brainmaster's competition at Tigin. I must look up this website; it sounds like a lot of fun. But, it makes for a long night, and your dear author is too tired to continue. Will write again tomorrow.

LHIRT's POD #1

As promised, here is the first problem of the day. This one was brought to me by Kyle, my sister's friend's brother whom I met at my sister's graduation last week:

Take 12 pennies, arranged in a rectangle of 4 rows and 3 columns. Next, randomly remove 4 of those pennies. What is the expected number of complete rows that remain?

I will post the answer to this puzzle later tonight. I may ROT13 it so that I don't spoil it for any readers.

(PS -- LHIRT is the acronym of my pseudonym)