HP50g or other: U. of Houston calc exams

[Peter Murphy]

Back in November 2009, David Hayden linked to some calculator exams from the University of Houston's annual high school math contest. Now I find that many more of those exams are available, at

<http://mathcontest.uh.edu/>

I've learned a lot about using the 50g in solving problems in these exams, which at one time were explicitly intended for users of the TI-89/92. Perhaps others will profit from them as well.

In the 2015 exam, I solved problem 14 quickly on the 50g using the LAGRANGE function to do the fitting. I found problem 20 tedious, because the 50g provides no direct way (I think) to create a vector in polar coordinates from its radial and angular coordinates.

I'd be interested in seeing solutions to 2015 problem 23, and to any other problems whose solutions you think might help users of the 50g get better at using it.

[Brad Barton]

(08-05-2015 04:58 AM)Peter Murphy Wrote:  I found problem 20 tedious, because the 50g provides no direct way (I think) to create a vector in polar coordinates from its radial and angular coordinates.

What about MTH>VECTR> ->V2 ?

Of course you have to have Polar or Spherical coordinates selected in the Mode menu, but it works fine for composing 2D vectors. There are also commands for composing 3D vectors, and decomposing vectors.

Also, thanks for the link.

Brad

[Peter Murphy]

Hi Brad,

Thanks for the tip on composing vectors in polar form; it works fine.

Even with this help, solving 2015 problem 20 on the 50g still takes some thinking, because each leg of the motion is specified with respect to the direction of the previous leg and not in absolute terms.

As to problem 14, somewhere I learned that one can create a polynomial in its symbolic form by specifying all its numerical coefficients, in the form [a b c] for example, and applying the program << 'X' PEVAL EXPAND >>. This is a faster way to create symbolic polynomials of any modest degree than using EQW, for example.

With a longer process using the command DIV2, it's possible to undo this, going from the symbolic form to the array of coefficients. The two programs are handy when a lot of polynomial algebra is involved.

I'm still hoping that one or more members of the Forum will find problems worth discussing.

Peter

[Gilles]

(08-05-2015 04:58 AM)Peter Murphy Wrote:  I'd be interested in seeing solutions to 2015 problem 23,

Here is my idea for problem 23 (but i dont know what is a " vertice", perhaps I misundertsood the problem), exact mode, rigorous on

['-1-X' '2-X²'] ['2-X' '7-X²'] CROSS ABS 2 / (here we have the aera of the triangle, function of X)
ENTER (dup the result )
DERVX (to get the extremum , then delete the sign part of the derivate because we search the root)
SOLVEX (here we get -> X=5/6 )
SUBST EVAL

-> 157/24

[Peter Murphy]

Hello Gilles,

Vertices = sommets (and vertex = sommet), so you were entirely right. I solved problem 23 in that way, more or less; every step you made replaced several of mine. I learned a lot from your solution. Thanks!

Peter

[Joe Horn]

(08-05-2015 06:40 PM)Brad Barton Wrote:  

(08-05-2015 04:58 AM)Peter Murphy Wrote:  I found problem 20 tedious, because the 50g provides no direct way (I think) to create a vector in polar coordinates from its radial and angular coordinates.

What about MTH>VECTR> ->V2 ?

You can also type polar vectors directly into the command line, in either parentheses or square brackets:

(2 ∡3)
[2 ∡3]

The angle symbol is obtained by pressing ALPHA left-shift 6. When you press ENTER, the angle will be interpreted according to the current DEG/RAD/GRAD setting, and the vector will be displayed according to the current RECT/CYLIN/SPHERE setting. 3D vectors of course can only be typed in square brackets, not parentheses:

[1 ∡2 ∡3] or [1 ∡2 3]

[Peter Murphy]

Hi Joe,

And thanks for the ALPHA-left shift-6 sequence for inserting the angle symbol. It's an annoyance, but it works.

That reminds me of the factorial !, accessed with ALPHA-left shft-2 and occasionally very helpful.

Joe, you are (I believe) one of very few Forum participants who use the 50g in the classroom. I'd be very interested to hear your thoughts on the U of H calculator exam problems, and even more interested to know details of your approach to teaching the use of the 50g. Actual exercises would be nice.

Thanks,

Peter

[Jlouis]

(08-05-2015 04:58 AM)Peter Murphy Wrote:  Back in November 2009, David Hayden linked to some calculator exams from the University of Houston's annual high school math contest. Now I find that many more of those exams are available, at

<http://mathcontest.uh.edu/>

I've learned a lot about using the 50g in solving problems in these exams, which at one time were explicitly intended for users of the TI-89/92. Perhaps others will profit from them as well.

In the 2015 exam, I solved problem 14 quickly on the 50g using the LAGRANGE function to do the fitting. I found problem 20 tedious, because the 50g provides no direct way (I think) to create a vector in polar coordinates from its radial and angular coordinates.

I'd be interested in seeing solutions to 2015 problem 23, and to any other problems whose solutions you think might help users of the 50g get better at using it.

Thank you very much indeed Peter.

I'm sure I will learn very much with the tests and it will help me in my studies.

Warm Regards

JL

[Peter Murphy]

Hello JL,

I have also found that Forum members can solve these problems much more cleanly than I can, so that although I learn a lot solving them myself there is still more to learn. Ask questions.

What can Forum participants make of this problem, from 2014?

9. Give the sum of the prime factors (including repetition) for 224551579230.

After FACTOR, the solution seems clumsy but straightforward until you get to repetition, at which point I'm at a loss.

Peter

[Hlib]

9. Give the sum of the prime factors (including repetition) for 224551579230.

224551579230 FACTORS /access LS 1 F6, press RS ▼ to see full menu list /
OBJ→ DROP *
UNROT * + ..... UNROT * + ..... till we will get the answer 301
A little clumsily.
Edit: flags -3, -105 clear, -117 set

[Gilles]

(08-08-2015 10:19 PM)Hlib Wrote:  9. Give the sum of the prime factors (including repetition) for 224551579230.

224551579230 FACTORS /access LS 1 F6, press RS ▼ to see full menu list /
DROP *
UNROT * + ..... UNROT * + ..... till we will get the answer 301
A little clumsily.

Hi, I proceeded also with FACTORS (and not FACTOR) like this :

FACTORS
here, you can see there is only one repetition (3), so edit the list (down arrow) and delete repetition and just change 3 with a 6 ENTER and SigmaLIST

[DavidM]

(08-08-2015 06:00 PM)Peter Murphy Wrote:  9. Give the sum of the prime factors (including repetition) for 224551579230.

After FACTOR, the solution seems clumsy but straightforward until you get to repetition, at which point I'm at a loss.

Like Hlib and Gilles, I thought of using FACTORS instead of FACTOR. Then, my brain got stuck on using DOT to compute the sum of the products of the factors. The problem with this method is getting the result of FACTORS split into two vectors, which I haven't found a good way to do. This is what I ended up with for a "general form" of the problem:

Code:

FACTORS
OBJ-> 2 / 2 DUP ->LIST ->ARRY 1 COL- SWAP 1 COL- SWAP DROP
DOT

I'm sure there's simpler methods.

[Didier Lachieze]

Here is what I did:

Code:

FACTORS
OBJ-> 2 / {2} + ->ARRY ->COL DROP
DOT

EDIT: another way to do the same thing:

Code:

FACTORS
2 << * NSUB 2 MOD * >> DOSUBS
ΣLIST

[Peter Murphy]

In thread-2771, Gilles (presumably asleep as I write this, or he'd be writing this himself) very nicely shows how to split the output of FACTORS into two lists, one of primes and the other of their corresponding multiplicities. (He makes use of the seemingly gratuitous difference between the object types of the two quantities; maybe he has thus shown us that the difference was intentional.)

Multiplying those lists together and summing the resulting single list yields the 301 result here.

With those split lists, by the way, one can do other things (things often asked of competitors in the USA MATHCOUNTS program, for example): calculate the number of distinct positive integer factors of the original integer, the sum of those factors, and the sum of the inverses of those factors.

Thread-2771 is here:

<http://www.hpmuseum.org/forum/archive/index.php?thread-2771.html>

Peter

[Hlib]

(08-05-2015 04:58 AM)Peter Murphy Wrote:  .........
I'd be interested in seeing solutions to 2015 problem 23, and to any other problems whose solutions you think might help users of the 50g get better at using it.

19. Give the value of "a" for which the system below has a solution (Calculator Exam – 2015)
2x -3y = 6 - 7a
-3x + 5y = 7+ 2a
7x +11y = 4 - 6a

Solution was based on the Kronecker-Capelli theorem and its consequences.

Code:

 flags: -2,-3,-55,-105 clear -117 set

2 -3 -3 5 7 11 {3 2} →ARRY DUP RANK SWAP
[′6-7*a′ ′7+2*a′ ′4-6*a′] 3 COL+ DET = ′a′ ZEROS

a=707/384

If I wasn't wrong.
Edit 1: sorry, this is incorrect. I will think

[CR Haeger]

Here is how I worked this problem:

19. Give the value of "a" for which the system below has a solution (Calculator Exam – 2015)
2x -3y = 6 - 7a
-3x + 5y = 7+ 2a
7x +11y = 4 - 6a

Rearranged to a 3x3 linear system

2x -3y +7a = 6
-3x + 5y -2a = 7
7x +11y +6a= 4

I used a TI36x [sys solv] and obtained

x = -287/128 (corrected the sign)
y = 101/128
a = 235/128 shown as z in output

To check, used [matrix] to create [3x3 coeff] and [3x1, x-z coeff] matrices then multiplied them, giving [6 7 4] 1x3 array.

These exam questions are interesting to try to solve on the TI36x.

[Gilles]

(08-09-2015 07:14 PM)CR Haeger Wrote:  Here is how I worked this problem:

19. Give the value of "a" for which the system below has a solution (Calculator Exam – 2015)
2x -3y = 6 - 7a
-3x + 5y = 7+ 2a
7x +11y = 4 - 6a

It's 'native' with the 50G (exact mode)

Code:

['2.x-3.y=6-7.a' '-3.x+5.y=7+2.a' '7.x +11.y=4-6.a' ]
['x' 'y' 'a' ]
LINSOLVE

There is a typo with x above : x = -287/128

[CR Haeger]

(08-09-2015 07:46 PM)Gilles Wrote:  

(08-09-2015 07:14 PM)CR Haeger Wrote:  Here is how I worked this problem:

19. Give the value of "a" for which the system below has a solution (Calculator Exam – 2015)
2x -3y = 6 - 7a
-3x + 5y = 7+ 2a
7x +11y = 4 - 6a

It's 'native' with the 50G (exact mode)

Code:

['2.x-3.y=6-7.a' '-3.x+5.y=7+2.a' '7.x +11.y=4-6.a' ]
['x' 'y' 'a' ]
LINSOLVE

There is a typo with x above : x = -287/128

Thanks for catching my typing error and for reminding me of the HP50G LINSOLVE command. Although the TI36x has far less functionality, its pretty efficient and required maybe half the keystrokes of the 50G on this problem.

[Hlib]

Thanks, CR Haeger and Gilles, you rescued me from a failure at exam :-)
On a statement of the problem we have to find parameter "a" without solving system of the linear equations. You simply solved ordinary system, having replaced "a" with "z":
2x -3y = 6 - 7z
-3x + 5y = 7+ 2z
7x +11y = 4 - 6z
I have the same result. It seems to me, it is not a proved trick. We must have:
rank[[2 -3][-3 5][7 11]]=rank[[2 -3 6-7a][-3 5 7+2a][7 11 4-6a]]<=2,
therefore det[[2 -3 6-7a][-3 5 7+2a][7 11 4-6a]]=0.

Code:

[[2 -3 ′6 - 7a′ ] [-3 5 ′7+ 2a′] [7 11 ′4 - 6a′]] DET 0 = ′a′ ZEROS

a=235/128.
Initially I was going to solve quite so.
††††What would you do, if:
2x -3y = 6 - 7*SQ(a)
-3x + 5y = 7+ 2a
7x +11y = 4 - 6a ?

[CR Haeger]

Its been a while, but I think you can reduce to the a terms as follows:

A. 2x -3y = 6 - 7a^2 Parabolic surface?
B. -3x + 5y = 7+ 2a Plane
C. 7x +11y = 4 - 6a Plane

B*7/3 +C:
-7x + (35/3)y = (49/3) + (14/3)a
+7x +11y = 4 - 6a
D. 0x +(68/3)y = (61/3) - (4/3)a

A + B*⅔::
+2x -3y = 6 - 7a^2
-2x (10/3)y = (14/3) +(4/3)a
E. 0x +(⅓)y = (32/3) -7a^2 +(4/3)a

D - 68*E:
(68/3)y = (61/3) - (4/3)a
(-68/3)y = (-68*32/3) -68*(-7a^2) +(-68*4/3)a
F. 0y = -705 +476a^2 -92a

Ti36x [poly solv] for the quadratic (F) yielded a = -1.123... or +1.317.... Back solving for x = 0.0214 and y = 0.963 for a=-1.123... If this is correct, Ill have a drink and if not, two.

I expect the 50g or Prime could make quick work of this.