## Wednesday, March 17, 2010

### The 2010 Bernoulli Trials

Here's a copy of the 2010 Bernoulli trials, a contest held annually at the University of Waterloo and open to all undergraduates. The students were given 13 statements and had to decide the truth or falsity of each one. Two incorrect answers and you're out. (For some previous years, see here.)

I like many of these questions, and a couple of them made me think for a while. How many can you do? Go ahead and post solutions in the comments, if you feel like it.

Jeffrey Shallit said...

Hint for number 11: seven primes in arithmetic progression must have a common difference divisible by what?

hpgross said...

For #4 it seems to be false as if you take the set of p_n for n such that n=(2*10^k)-1 you get a sequence 1/1,11/19,111/199 which is decreasing but cannot go below 5/9. When n=10^k-1, the sequence equals exactly 1/9. Since after each of the tens powers the probability gets larger, it will oscillate between going above 5/9 and at 1/9 so a limit will not exist.

For #11 the common difference must be divisible by 2,3,4,5,6,7, and the lowest common multiple is 2*5*6*7 which is 420, and thus the progression cannot exist with all primes less than 900.

I'm feeling rusty right now so I'm not sure if these are right but I'm just throwing them out here.

Miranda said...

The last question: "A rectangular box with sides of length 1, 2 and 3 hovers above the
flat ground. The
maximum possible area of its shadow on the ground is equal to 7"

Nope, it's infinity.
The light source isn't directly overhead.

Heh, just kidding. I hope to work on this one.

IvanM said...

Harrison: That's not quite right for #11. You only need the common difference to be divisible by a primorial (i.e., 2*3*5*7*11*13*... in general). Also, which primorial is needed depends on the starting term.

Since we don't have spoiler tags here, I'll give my answers in sorta-cryptic form:

The problem numbers whose answers are true are exactly those which are congruent modulo ten to either the additive identity or one of the first two primes.

Let me know if I made any embarrassing mistakes.

IvanM said...

I just discovered an error in the 2006 Bernoulli Trials solutions, in the last problem, #16. The limit there is -1/30, not 2/15, and thus the correct answer is FALSE, not TRUE. (It looks like the only mistake in the solution is that b_3 - a_3 is given as -1/2 when it's actually +1/2.)

Does this mean that the winner of that contest was awarded wrongly? Yikes!

Anonymous said...

ivanm: double check that. There's a second negative in the numerator of the last term which you missed.

Anonymous said...

ivanm - I take it back, I see the error you point out.