http://mathcontest.uh.edu/Exams09/calculator.pdf

It was interesting.  The 20 questions range from reasonably easy to
downright hard to completely ridiculously and confusing.

The contest is supposed to be one hour, but I took closer to two.
Unfortunately, after I finished, I realized that there is no answer
key given, so I couldn't check my results, but it was still a fun
challenge.

Thanks for this pointer.  I am always looking for good calculator
problems, and these were good ones.  It took me around five hours, but
I wasn't racing, just enjoying.  I will compare answers, if you
promise not to laugh.

Regards,

I found that this test seemed to particularly emphasize strengths of
the HP-50g.  The ability to cob together quick little programs to
solve a problem is just what I think the designers of RPL had in
mind.  Also, the NEXTPRIME & PREVPRIME commands proved invaluable
several times.

Interestingly, the Univ of Houston contest info states: "Any
calculator can be used on the Calculator Exam."  However, the contest
results pages shows a "TI-83/84" catagory of winners and a "TI-89/92"
catagory of winners, so maybe "any calculator" really meant "any TI
calculator."  Or perhaps it really mean "CAS" and "non-CAS"
catagories.  I'll have to try it again using an 89 to see how it
compares.  I've got a nephew in Texas who has been active on his high
school's math team and he told me that some on his team use hp's, so I
know it's not unheard of for high school students to use them in these
contests.

Here are my answers below, mistakes and all.  The instructions said to
"round your answers to 8 decimal places" but perhaps it meant 8
significant digits as some of the questions could not be written to 8
decimal places with any standard calculator.  I just ignored the this
and listed the full precision.

Wes's Answers to the 2009 contest:

 1. 950753

 2. 552426

 3. 752

 4. 3.42871301499E-6

 5. -2.14968960968

 6. 232366.502395

 7. 102400

 8. 15.4097986727

 9. 0.321750554396

10. -2689/890

11. $310,000.00 (did I miss something here?)

12. 937 (not 941)

13. This question serves no purpose other than to see if a student can
follow an intentionally confusing question.  I refuse to answer it on
principle.  (ie, I didn't have the gumption follow it either.)

14. 17610

15. 0.634247256824, 0.695933022105, 0.634830752548

16. 0.451499600416

17. 0.494190956873

18. $295356.71857 (nearest penny: $295,356.72)

19. 160

20. 1/74 = 0.0135135135135 (Assuming that "prime numbers in increasing
order" meant "consecutive prime numbers in increasing order."
Otherwise, it's a _much_ harder question.)

Any corrections?

17) 8.68856271971E-4

The question stated: "10 randomly chosen values in the interval" so I
think a uniform random distribution seems reasonable.

Thirteen men and nineteen women are placed in line so that 3 women are
at the front of the line, 4 women are at the end of the line, and men
and women alternate in the line from the fourth position to the 29th
position, with a man standing in the 4th position. Each of the first
31 people choose a random number in the set {2, 3, … , 100}. You don’t
know their choices, but you know that the 32nd person knows their
choices, and you hear her say that all of the men have chosen prime
numbers in increasing order, and all of the women have chosen
composite numbers in increasing order. What is the probability that
she can choose a composite number in the set {2, 3, … , 100} that is
larger than any of the numbers chosen by the women and forces the sum
of all chosen numbers to be an integer multiple of 103?

I interpreted each term in the product as (1+1/3^k)^(k+1) with k
running between 1 and 14.  Oh, I think you summed the terms instead of
multiplying.

When I took the test, I thought we might be looking at a place where
the function had a periodicity of 2, but if you make a bifurcation
plot, you see that r=3 is right where the plot bifurcates, so s_n
actually converges, albeit slowly.  It would have been of more
historical significance to use a value like r=3.83 where the
periodicity is 3.  ("Period Three Implies Chaos"
http://pb.math.univ.gda.pl/chaos/pdf/li-yorke.pdf )

Fortunately, they did not choose a value of r that is chaotic.  Using
r=3.80 on the 12 digit 50g, s_53 is accurate to only 2 decimal
places.  The 14 digit TI's fair a bit better, accurate to 5 digits (as
does the 50g using 15 digit extended reals).

Thanks for the corrections.

On Nov 23, 12:24 am, Crawl <crawland1...@lycos.com> wrote:

There are so many ways to interpret this question.  The ambiguity is
that the given is unclear.  For instance:

1) Given that she chooses a composite number the set {2, ... , 100},
what's the probability that it has the following properties:
   a) it is larger than any of the others chosen by the women
   b) it forces the the sum of all chosen numbers to be an integer
multiple of 103

2) Given that she chooses a composite number the set {2, ... , 100}
that is larger than any of the others chosen by the women, what's the
probability that it has the following properties:
   a) it forces the the sum of all chosen numbers to be an integer
multiple of 103

3) My original (probably incorrect) interpretation was that it meant
consecutive primes and composites and that she was randomly selecting
from 1 of the 74 composites from 2 to 100.  I answered P=1/74 since
there was only one such composite that was larger than the other
women's and made the sum a multiple of 103.  However, since the
question asked for the probability that she "can" choose such a
number, my answer should have then been 100%.  Since the woman knows
what everyone else has chosen, she CAN choose a number that works.
But like I said, this interpretation is probably not what they
intended.

It looks like I'm not the only one seeing different possibilities
since the two other suggests answers are P~0.6 and P~0.0286

Here's the text again:

Thirteen men and nineteen women are placed in line so that 3 women are
at the front of the line, 4 women are at the end of the line, and men
and women alternate in the line from the fourth position to the 29th
position, with a man standing in the 4th position. Each of the first
31 people choose a random number in the set {2, 3, … , 100}. You don’t
know their choices, but you know that the 32nd person knows their
choices, and you hear her say that all of the men have chosen prime
numbers in increasing order, and all of the women have chosen
composite numbers in increasing order. What is the probability that
she can choose a composite number in the set {2, 3, … , 100} that is
larger than any of the numbers chosen by the women and forces the sum
of all chosen numbers to be an integer multiple of 103?

Obviously, it is very doubtful that I will ever get into the
University of Houston.

But help me here, get an education.  On number 12, I wrote a recursive
function, Q, to come up with Q(48) = 980, less 39, gives me 941. Why
not 941?
Do you happen to know this family and know that 3 of them didn't show
up to the 2008 meeting?

Cheers,

George Polya's Steps for Solving a Problem:
   1. Understanding the problem.
   2. Devising a plan.
   3. Carrying out the plan.
   4. Looking back.

One down, three to go.

  2. Devise a plan that works for small numbers of men and women, but
  would require too much memory and/or CPU time to work for the
  numbers in the problem.

  3. Carry out the plan, and check the results versus Monte Carlo
  simulation.

  4. Look at ways to reduce memory and/or CPU usage, and go back to 2
  with a better plan.

If I say too much about what I did, I run the risk of guiding you to
getting the same (possibly wrong) answer that I did, but I can give
you some hints if you are interested.

In some problems, you have to be careful that the events that you
count are equal probability.  For example, if you are calculating the
probabilites of rolling two dice totaling 2,...,12 you can't assign
those 11 events equal probabilities if you want to get the correct
answer.  (I know you know this, but this is sort of what I meant when
I said you have to be careful when you change the problem.)

In the original problem, the first 31 people generate 99^31 equal
probability events.  We are not concered with the ones where the men
choose composite numbers, the women choose prime numbers, or where
someone chooses the same number as someone else in his group.  So it
doesn't matter that your method doesn't even generate these kind of
events.  We are also not concerned with events where the members of a
group choose numbers out of order.  Instead discarding these event,
however, you treat the numbers as if they had been chosen in order.
This works because for each of the events we care about, your method
generates exactly 18! permutations (for the women, or 13! permutations
for the men) with equal probability.

(partial spoiler)

Out of the total 99^31 events, the ones we are interested total
C(74,18)*C(25,13) = 377893220518123106330800.  This number is too big
to fit in a 64-bit integer, so depending on your hardware you might
have to either use multiple precision, or settle for an approximate
answer with floating point arithmetic.  My method grouped events
together based on the only two things we need to know about them: the
number that the last person chose, and the total mod 103.  I kept a
two dimensional array with the total number of events that fell into
each of these categories, which I built up person-by-person.  I did
the men first, followed by the women.

Dan Doherty

On Nov 28, 1:27 am, Wes <wjltemp...@yahoo.com> wrote: