## HMNT 2009

I love Harvard-MIT Math Tournaments. I like the mini-events, especially when I learn a new game. I also like the guts round, where I enjoy the adrenaline rush of watching the progress in real time. I also like the fact that I know many of the kids from different teams: my current students, my former students, the members of my club, my Sergei’s friends.

The problems for the competitions are designed by undergraduate students at MIT and Harvard. Kudos to them. Still, I was somewhat disappointed with the November 2009 problems. Most problems are variations of standard problems with different parameters. It is not easy to design a problem, but I was hoping for something fresh.

My favorite problem from the HMNT 2009 tournament was in the theme round:

There are five guys named Alan, Bob, Casey, Dan, and Eric. Each one either always tells the truth or always lies. You overhear the following discussion between them:

• Alan: “All of us are truth-tellers.”
• Bob: “No, only Alan and I are truth-tellers.”
• Casey: “You are both liars.”
• Dan: “If Casey is a truth-teller, then Eric is too.”
• Eric: “An odd number of us are liars.”

Who are the liars?

My second favorite problem was in the guts round:

Six men and their wives are sitting at a round table with 12 seats. These men and women are very jealous — no man will allow his wife to sit next to any man except for himself, and no woman will allow her husband to sit next to any woman except for herself. In how many distinct ways can these 12 people be seated such that these conditions are satisfied? (Rotations of a valid seating are considered distinct.)

This was the funniest problem:

You are trapped in ancient Japan, and a giant enemy crab is approaching! You must defeat it by cutting off its two claws and six legs and attacking its weak point for massive damage. You cannot cut off any of its claws until you cut off at least three of its legs, and you cannot attack its weak point until you have cut off all of its claws and legs. In how many ways can you defeat the giant enemy crab? (Note that the legs are distinguishable, as are the claws.)

It is difficult to arrange so many problems for four rounds without mistakes. The error in the following problem is not a typo and it bothers me that no one caught it:

Pick a random digit in the decimal expansion of 1/99999. What is the probability that it is 0?

Hey, there is no uniform distribution on an infinite set of integers: picking a random digit is not defined.

Share:

1. #### Anonymous Rex:

I think the last problem is just fine, but that there would be an issue if it were a fraction like 1/6.

2. #### Jonathan:

These are ok, but I agree, not amazing. But yes, it must be very hard to create interesting, challenging problems.

I think I will borrow the seating one. Think before counting. Sounds like a good lesson.

3. #### Bill:

I’ll take a shot at the first one. I think Casey is the only truth teller. The other four are liars.

4. #### Norm:

Regarding the last question, there’s a lot of precedent for treating a periodic sequence as a finite sequence.

5. #### Qiaochu Yuan:

Picking random digits is perfectly well-defined for rational numbers: one ignores the nonperiodic part of the decimal expansion and picks a random digit in the periodic part.

6. #### colorblind:

I noted the first problem doesn’t mention to whom Casey was speaking. Then I realized it doesn’t matter. Cute.

I wish there were more interesting math in the crab problem. Maybe 2 crabs and one regrows a limb when the other is attacked twice in a row.

The seating problem brings back memories of grade school when I first saw one of Martin Gardner’s circle problems: As I remember, “There was a ring of 12 people. They counted off clockwise by 3’s and each person left the circle as they said 3. Then they did it again counting off by a different number. Miraculously, they ended up leaving in the same order. What was the smallest number they could have counted off by?”

7. #### Bill:

Wow, that’s an elegant problem.

The answer is 1. When the people returned to the circle, they were arranged in a different order, which happened to be the same order they had originally left.

Before I was able to post this, one of Hassan’s horses called to tell me to assume that the same people are in the same positions the second time around, and that the counting starts from the same person. If he is incorrect, then my answer may be wrong, but hey, I got it from the horse’s mouth.

Okay, let’s call the correct answer X. So the first time around the circle, X has to go around the circle a number of times and end 3 spots ahead of the starting point. So X, when divided by 12, yields a remainder of 3. With one person missing, X now goes around the circle and yields a remainder of 3 when divided by 11. X must also yield a remainder of 3 when divided by any number from 10 down to 4. As this number yields a remainder of 3 when divided by 9, it will be already be divisible by 3, and as it yields a remainder of 3 when divided by 8, it will already be an odd number. So the answer is the LCM of the numbers from 4 to 12, plus 3. Starting from 4 and adding prime factors as needed, we find the following:

2 x 2 x 5 x 3 x 7 x 2 x 3 x 11 = 27,720

X=27,723

Is this right?

8. #### Anonymous:

Picking random digits is perfectly well-defined for rational numbers

Other than terminating decimals, of course (due to the 999… vs. 000… ambiguity).

9. #### colorblind:

*DOH* This is why I didn’t go to grad school….

10. #### Qiaochu Yuan:

In a sensible definition of “decimal expansion” one disallows repeating 9s; this ensures that the corresponding representation is actually unique. Anyway, even without the rational hypothesis one can still sensibly define asymptotic densities and this is clearly the intended meaning.

11. #### Andrew MW:

Is the answer to the crab question 10? If so it seems very easy… if not, then I’m clearly missing something!

12. #### Andrew MW:

I came up with 48 for the table question, again probably missing something here…. cheers PS I got Casey as the only truth-teller as well.

13. #### Bill:

I came up with 2,880 for the original table question, but that’s assuming that all married couples are seated together. If other arrangements are allowed (such as all women seated together and all men seated together, with just two married couples seated together where the two groups meet), then there would obviously be many more possibilities.

14. #### Natalia Demina:

Sorry for off-top. Just for the linguistical purposes. When you wrote “Kudos to them”, what did you mean? How to translate it into Russian?

15. #### DeepButi:

Am I absolutely wrong or the crab can be defeated in
P6*P2 + V6,5*(P2 + V2,1) + V6,4*(P2*P2 + V2,1*P2 + V2,1*V2,1) + V6,3*(P2*P3 + V2,1*P3 +V3,2 + V2,1*V3,1*P2)
diferent ways?
Uff … sure it exists a simpler formula. Will try to find the final number!

PS. Vx,y Variations of x elements in groups of y; Px Permutations of x elements

16. #### DeepButi:

Sorry typo mistake:
P6*P2 + V6,5*(P2 + V2,1) + V6,4*(P2*P2 + V2,1*P2 + V2,1*V2,1) + V6,3*(P2*P3 + V2,1*P3 + V2,1*V3,2 + V2,1*V3,1*P2)

17. #### DeepButi:

WHile I was making the calculus (well, in fact excell did them for me, so it would be more acurate: while I was entering the formula) I thought that the other way would be easier. Which are the incorrect ways to attack the crab?

1: claw – anything
2: leg – claw – anything
3: leg -leg – claw – anything

so the correct ones woule be:

P8 – (P2*P7 + V6,1*P2*P6 – V6,2*P2*P5)

And oh, yes, both formulas give the same nice number: 14,400 = 10**2 * 12**2

Anyone knows if I’m right? 🙂

18. #### Lauren:

Just thought I’d put in a word about the HMNT – the problems are not meant to be as challenging as the February tournament, since it is mainly intended to give a chance to local schools (particularly weaker schools) which find the February tournament too difficult/don’t get much out of it. So, that might be why the problems seemed less interesting than the normal HMMT-fare.