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Nov 4, 2009, 10:08:46 AM11/4/09

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I came across an interesting site for the annual University of Houston

Mathematics Contest for high school students (http://

mathcontest.uh.edu/). One of the contests is a calculator test. I

tried the 2009 test:

Mathematics Contest for high school students (http://

mathcontest.uh.edu/). One of the contests is a calculator test. I

tried the 2009 test:

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.

Enjoy,

-wes

Nov 8, 2009, 12:27:01 PM11/8/09

to

Wes,

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,

Dan

Nov 22, 2009, 1:22:32 PM11/22/09

to

Sorry it took so long to respond to this. I couldn't find the scrap

paper that I had written all my answers on, but I finally had a chance

to sit down and take the test again. I found at least one that I

misread the first time.

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?

Enjoy,

-wes

Nov 22, 2009, 4:24:40 PM11/22/09

to

> 17. 0.494190956873

>

> 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.)

>

For 17, it looks like you gave the answer if they only sampled one

point. I think the question said they sampled 10 points (with uniform

random distribution?) so it would be your answer to the 10th

power=8.688x10-4.

For 20, I did not think they implied the prime numbers needed to be

consecutive (nor the composites, for that matter), but with that most

general case, I couldn't solve the problem (and would be surprised if

there was a closed form solution that could be computed in realistic

time in that case) Using Monte Carlo simulation, though, I got P~0.6.

Nov 22, 2009, 5:00:16 PM11/22/09

to

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

> > 17. 0.494190956873

>

> > 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.)

>

> For 17, it looks like you gave the answer if they only sampled one

> point. I think the question said they sampled 10 points (with uniform

> random distribution?) so it would be your answer to the 10th

> power=8.688x10-4.

> > 17. 0.494190956873

>

> > 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.)

>

> For 17, it looks like you gave the answer if they only sampled one

> point. I think the question said they sampled 10 points (with uniform

> random distribution?) so it would be your answer to the 10th

> power=8.688x10-4.

Yes, you're right. I remember now doing it correctly the first time I

took the test, but I forgot where I was going with it this time. So

change my answer to:

17) 8.68856271971E-4

The question stated: "10 randomly chosen values in the interval" so I

think a uniform random distribution seems reasonable.

> For 20, I did not think they implied the prime numbers needed to be

> consecutive (nor the composites, for that matter), but with that most

> general case, I couldn't solve the problem (and would be surprised if

> there was a closed form solution that could be computed in realistic

> time in that case) Using Monte Carlo simulation, though, I got P~0.6.

Considering that the test is for high school students and that they

asked for 8 decimal places, I'm guessing that they intended the easier

consecutive case. I'll have to give it some more thought. For

anybody who's interested, the question was:

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?

-wes

Nov 25, 2009, 5:42:40 PM11/25/09

to

Wes <wjlte...@yahoo.com> writes:

> 8. 15.4097986727

I get 3.12155183758.

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.

> 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.)

Yeah, it's confusing. I attempted it and got 8:20. I didn't check my

work.

> 15. 0.634247256824, 0.695933022105, 0.634830752548

We got simmilar answers, but I'm a little surprised. The function is a

well known one from chaos theory. I thought that round off error was

going to cause the errors to explode.

> 17. 0.494190956873

This one was corrected in another thread.

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

I got 295356.84. This should be exact if bankers round to the nearest

penny each month.

> 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.)

Yes, it's hard. I spent three hours on this one problem, and I solved

it with Mathematica, not a pocket calculator. I also wrote a Monte

Carlo implementation to check my work. The answer I get is

approximately 0.0286152273889. I'm withholding the exact fraction

in case someone wants to confirm the result.

Scott

--

Scott Hemphill hemp...@alumni.caltech.edu

"This isn't flying. This is falling, with style." -- Buzz Lightyear

Nov 25, 2009, 5:58:30 PM11/25/09

to

Wes <wjlte...@yahoo.com> writes:

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

[snip]

>> For 20, I did not think they implied the prime numbers needed to be

>> consecutive (nor the composites, for that matter), but with that most

>> general case, I couldn't solve the problem (and would be surprised if

>> there was a closed form solution that could be computed in realistic

>> time in that case) Using Monte Carlo simulation, though, I got P~0.6.

I didn't think they needed to be consecutive either. I did it in

three hours in Mathematica. My anwer was 0.0286152273889, and took

only a few seconds to calculate. (I have an exact fraction, also.) I

checked my method with a Monte Carlo implementation, too. Before I

got both my Mathematica and Monte Carlo code debugged, I had to do the

case with 1 man and 2 women. I also did 2 men and 3 women.

> Considering that the test is for high school students and that they

> asked for 8 decimal places, I'm guessing that they intended the easier

> consecutive case. I'll have to give it some more thought. For

> anybody who's interested, the question was:

It could have been something to keep the students busy until they ran

out of time.

> 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?

Scott

Nov 25, 2009, 6:09:33 PM11/25/09

to

Wes <wjlte...@yahoo.com> writes:

> 12. 937 (not 941)

By the way, that's exactly the way I wrote the answer on my answer

sheet.

Nov 26, 2009, 11:03:18 AM11/26/09

to

On Nov 26, 1:42 am, Scott Hemphill <hemph...@hemphills.net> wrote:

> Oh, I think you summed the terms instead of multiplying.

Yep, I hit SigmaList instead of PiList. I hate it when I make the

silly mistakes that I warn my students against. :-)

> > 15. 0.634247256824, 0.695933022105, 0.634830752548

>

> We got simmilar answers, but I'm a little surprised. The function is a

> well known one from chaos theory. I thought that round off error was

> going to cause the errors to explode.

s -> r*s*(1-s), r=3, initial s=0.5

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).

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

>

> I got 295356.84. This should be exact if bankers round to the nearest

> penny each month.

Do banker's round to the nearest penny each month? I've heard stories

(urban legends?) of bankers skimming a fraction of a penny from

thousands of accounts each day, adding up to a tidy sum.

> > 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.)

> It could have been something to keep the students busy until they ran

> out of time.

I agree. There are clearly more problems than can be done in the

allotted time. Part of the test is to figure out which problems are

doable in a reasonable amount of time. It's important to recognize

what you cannot do.

> Yes, it's hard. I spent three hours on this one problem, and I solved

> it with Mathematica, not a pocket calculator. I also wrote a Monte

> Carlo implementation to check my work. The answer I get is

> approximately 0.0286152273889. I'm withholding the exact fraction

> in case someone wants to confirm the result.

I'll give it some more thought and see what I come up with.

Thanks for the corrections.

-wes

Nov 27, 2009, 10:09:23 AM11/27/09

to

Concerning Question #20

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

> For 20, I did not think they implied the prime numbers needed to be

> consecutive (nor the composites, for that matter), but with that most

> general case, I couldn't solve the problem (and would be surprised if

> there was a closed form solution that could be computed in realistic

> time in that case) Using Monte Carlo simulation, though, I got P~0.6.

On Nov 26, 1:42 am, Scott Hemphill <hemph...@hemphills.net> wrote:

> Yes, it's hard. I spent three hours on this one problem, and I solved

> it with Mathematica, not a pocket calculator. I also wrote a Monte

> Carlo implementation to check my work. The answer I get is

> approximately 0.0286152273889. I'm withholding the exact fraction

> in case someone wants to confirm the result.

The more I look at this problem, the more confused I am about what

it's even asking. After the setup, the actual question is: "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?"

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

-wes

Nov 27, 2009, 1:54:16 PM11/27/09

to

Wes <wjlte...@yahoo.com> writes:

I don't believe that it is either of these two choices. I didn't

think it was so unclear. The woman knows what everyone else has

chosen. Given all the ways that everyone else chosen numbers which

meet the requirements, what is the probability that she CAN choose a

number that meets her special requirements. It is not 100%. If the

woman before her has chosen 100, then there aren't any composites left

for her to pick, since she is limited to 2 to 100 and it must be

greater than that last woman's. Or, if the last woman picked 99, then

there is only one number left that she can choose: 100. But when 100

is added to everyone else's numbers, the result might be a multiple of

103.

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?

Nov 28, 2009, 1:23:59 AM11/28/09

to

Ok, I missed (or didn't try, in the case of #20), 7 out of 20. I feel

good that I made the same mistake as wes on #8 and #17,

but I made stupid errors on some of the others.

good that I made the same mistake as wes on #8 and #17,

but I made stupid errors on some of the others.

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,

Dan Doherty

Nov 28, 2009, 2:27:55 AM11/28/09

to

On Nov 28, 9:23 am, ddoherty03 <ddohert...@gmail.com> wrote:

> I feel good that I made the same mistake as wes on #8 and #17,

> but I made stupid errors on some of the others.

> I feel good that I made the same mistake as wes on #8 and #17,

> but I made stupid errors on some of the others.

I'm glad I could contribute to your "feel good" index. :-)

> 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?

The first time I worked the test, I got 941, but the second time I

noticed that the question said: "each year the family grew ... by an

amount equal to the largest integer less than 6% of the total from the

previous year." The key is that is says "less," NOT "less than or

equal to." So instead of using FLOOR or IP to round down, you can use

CEIL and subtract one. In 1975, Q(15)=150. A 6% increase gives you

exactly 159, which means Q(16)=158. The same thing happens three more

times.

-wes

Nov 28, 2009, 2:04:03 PM11/28/09

to

On Nov 26, 1:42 am, Scott Hemphill <hemph...@hemphills.net> wrote:

> Yes, it's hard. I spent three hours on this one problem, and I solved

> it with Mathematica, not a pocket calculator. I also wrote a Monte

> Carlo implementation to check my work. The answer I get is

> approximately 0.0286152273889. I'm withholding the exact fraction

> in case someone wants to confirm the result.

> it with Mathematica, not a pocket calculator. I also wrote a Monte

> Carlo implementation to check my work. The answer I get is

> approximately 0.0286152273889. I'm withholding the exact fraction

> in case someone wants to confirm the result.

Okay, I think I finally understand the question. I still don't know

how to solve it, but I did write my own Monte Carlo program (in C

using microsoft's rand_s() instead of rand() ). My results are

hovering around P=0.02858 after some 60 million simulations, so I seem

to be in the ballpark.

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.

-wes

Nov 28, 2009, 5:51:50 PM11/28/09

to

Wes <wjlte...@yahoo.com> writes:

> On Nov 26, 1:42 am, Scott Hemphill <hemph...@hemphills.net> wrote:

>> Yes, it's hard. I spent three hours on this one problem, and I solved

>> it with Mathematica, not a pocket calculator. I also wrote a Monte

>> Carlo implementation to check my work. The answer I get is

>> approximately 0.0286152273889. I'm withholding the exact fraction

>> in case someone wants to confirm the result.

>

> Okay, I think I finally understand the question. I still don't know

> how to solve it, but I did write my own Monte Carlo program (in C

> using microsoft's rand_s() instead of rand() ). My results are

> hovering around P=0.02858 after some 60 million simulations, so I seem

> to be in the ballpark.

I consider this to be quite an achievement. Since simulating the

problem as written is very unlikely to produce any of the events which

you are counting, you have to be very careful to change the problem in

ways that don't affect the probability. I confess that I never

simulated the whole problem--I just simulated the results for smaller

numbers of men and women, and verified my answers to several decimal

places. I used the GNU C library's drand48(). By the way, the

expected error in your estimate is sqrt(0.02858*(1-0.02858)/60e6) ~=

2e-5. The actual difference is about 3.5e-5, which is less than two

standard deviations, and wouldn't be considered a significant

difference.

> 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.

That quote is pretty good. In this case I found it useful to repeat

steps 2, 3 and 4 a couple of time.

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.

Nov 29, 2009, 4:17:40 AM11/29/09

to

On Nov 29, 1:51 am, Scott Hemphill <hemph...@hemphills.net> wrote:

> Wes <wjltemp...@yahoo.com> writes:

> > Okay, I think I finally understand the question. I still don't know

> > how to solve it, but I did write my own Monte Carlo program (in C

> > using microsoft's rand_s() instead of rand() ). My results are

> > hovering around P=0.02858 after some 60 million simulations, so I seem

> > to be in the ballpark.

>

> I consider this to be quite an achievement. Since simulating the

> problem as written is very unlikely to produce any of the events which

> you are counting, you have to be very careful to change the problem in

> ways that don't affect the probability.

> Wes <wjltemp...@yahoo.com> writes:

> > Okay, I think I finally understand the question. I still don't know

> > how to solve it, but I did write my own Monte Carlo program (in C

> > using microsoft's rand_s() instead of rand() ). My results are

> > hovering around P=0.02858 after some 60 million simulations, so I seem

> > to be in the ballpark.

>

> I consider this to be quite an achievement. Since simulating the

> problem as written is very unlikely to produce any of the events which

> you are counting, you have to be very careful to change the problem in

> ways that don't affect the probability.

Best I can tell, I didn't change the problem at all. I started with a

list of primes <=100 and a list of composites <=100. I assigned a

random value to each number in the two lists, then sorted each list by

the random values. The first 13 primes were the men's choices and the

first 18 composites were the women's choices. For the 19th woman's

possible choices, starting with the next composite after the largest

of the 18 women's choices, I checked each composite up to 100 to see

whether or not the sum of everybody's choices was a multiple of 103.

If it was a multiple, I stopped looking and mark one down for "yes,

she can choose such a number." Then repeated ad infinitum.

-wes

Nov 29, 2009, 3:35:44 PM11/29/09

to

Wes <wjlte...@yahoo.com> writes:

You changed the problem a little bit, but maybe you consider those

changes trivial.

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.

Nov 29, 2009, 8:01:22 PM11/29/09

to

Ah, now I see. Thanks, Wes.

Dan Doherty

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