The Museum of HP Calculators

HP Forum Archive 19

 University of Houston Calculator ContestMessage #1 Posted by David Hayden on 6 Nov 2009, 3:38 p.m. The following appeared on comp.sys.hp48 recently. I thought this community might enjoy it. Please note that I was not the orignal poster. ---------------------------------------------------------- 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: 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

 Re: University of Houston Calculator ContestMessage #2 Posted by Paul Dale on 6 Nov 2009, 5:12 p.m.,in response to message #1 by David Hayden Eeek, that test would be quite a challenge in an hour. A high end graphing calculator would help a lot evaluating the relatively long series. - Pauli

 Re: University of Houston Calculator ContestMessage #3 Posted by Marcus von Cube, Germany on 6 Nov 2009, 5:24 p.m.,in response to message #2 by Paul Dale If you look at the chapter names of the PDF (I viewed it with Preview on my Mac) a TI-83 is recommended.

 Re: Some of my answersMessage #5 Posted by Crawl on 6 Nov 2009, 7:07 p.m.,in response to message #4 by Crawl Quote:7. 7351344 = 2^4 * 3 ^ 3 * 7 * 11 * 13 * 17 11 factors You can include or not include any of them. (2^11 values of i) so 2^22 points Answer: 4194304 I already see that I did that totally wrong. It's 5 * 4 * 2 * 2 * 2 * 2 (The prime exponent plus 1) so 320 factors. And 10240 points.

 One moreMessage #6 Posted by Crawl on 6 Nov 2009, 8:26 p.m.,in response to message #5 by Crawl 19. This is much easier than it first seems. The square root of 1100 is 33.1... The 25th prime is 98. That means if any number less than 1100 is not divisible by the first 25 primes then it itself must be prime. << 0 -> X << 2 1100 FOR J J ISPRIME? IF THEN X 1 + 'X' STO END NEXT X >> >> That gives 184 primes less than 1100 So the answer is 159. It would be harder if they asked if they weren't divisible by, say, the first 5 primes (up to 11). Then you'd have to include numbers like 13^2=169, 17^2=289, 13*17=221, etc.

 Most of the restMessage #7 Posted by Crawl on 7 Nov 2009, 10:56 a.m.,in response to message #6 by Crawl 13. I really hated this one. It's not a math problem; it's a "can you penetrate our impenetrable writing" problem. Too bad those being judged can't sometimes turn around and judge the judges. Anyway, I guess the idea is that to 6 pm she has gone 67 mph for 2 hours, so she's gone 134 miles. For the next 40 minutes she's able to drive the full 75 mph, so that gets her another 50 miles ahead, for 184 total. I presume that for 25 minutes she just drove in a big circle, so at 7:33 she's still 184 miles from Houston (and 55 from Dallas). Then for 20 minutes she drove for 64 mph, getting her another 21.333... miles ahead. So, it's now 7:53 and she has 33.666... miles to go. She can go that distance at 75 mph, so it takes her 0.448888... hours = 26.93333.... minutes to go the rest of the way. She arrives at about 8:20. 17. I already answered this one, but just as a note, they never specified that the trial numbers would be chosen with uniform random distribution in the interval. 18. We could do this with a financial function trivially if not for the alternating deposits. There are 126 6-month periods in 21 years. << 0 1 126 FOR X 1 .04 12 / + * 600 + 1 .04 12 / + * 900 + NEXT >> Gives 295356.71 as the answer. A refinement could be to round down fractions of a penny. If it wasn't for the compound interest, the total would have been 189000.

 Re: One moreMessage #8 Posted by David Hayden on 9 Nov 2009, 1:39 p.m.,in response to message #6 by Crawl There's an evil twist to problem 19, which is probably intentional: Quote: 19. Give the number of positive integers that are no larger than 1,100 and are not divisible by any of the first 25 prime numbers. The number 1 is positive, less than 1,100 and not divisible by any of the first 25 primes, so it counts too. Therefore the correct answer is 160, not 159.

 Re: Some of my answersMessage #9 Posted by David Hayden on 6 Nov 2009, 10:36 p.m.,in response to message #5 by Crawl Quote: I already see that I did that totally wrong. It's 5 * 4 * 2 * 2 * 2 * 2 (The prime exponent plus 1) so 320 factors. And 10240 points. 102400 actually. You missed a zero. But thanks for the solution. I did it completely wrong, in a complete different way from either your right answer or your wrong one. I thought they were plotting only the prime factors.

 Re: 2009 UH calc test: Some "old-school" answersMessage #10 Posted by Karl Schneider on 7 Nov 2009, 2:31 p.m.,in response to message #4 by Crawl "Crawl" -- Thanks for your answers. Certainly, a modern calc with built-in graphing, fast processing, and mathematical functions such as prime number and maybe LCM, LCD, etc. would helpful for this exercise. Still, even a venerable HP-34C, HP-11C, or HP-15C is useful for some of the problems requiring more than straightforward transcedental functions: Question 3 Find the smallest natural number k so that 1 + 1/2 + 1/3 + 1/4 + ... + 1/k > 7.2 I'd forgotten all about the Euler-Mascheroni Constant, which had been discussed in the Forum, and was surely was intended to be utilized for analyzing this harmonic series. As "Crawl" pointed out, the "rate of divergence" equation will give a very accurate estimate of k. However, ignoring the error 'epsilon' in the equation will lead the user to assume that the correct answer is 753, not 752 (for which 'epsilon' is about 0.00072418). Directly-calculated solution -- the sum is not too difficult to automate using ISG: ```Program: Execution: LBL A 1.99901 RCL I STO I INT 7.2 1/x GSB A - ... x < 0? RCL I RTN INT ISG (or ISG I) (752.) GTO A RTN ``` Question 5 Give the sum of the x coordinates for the points of intersection of the graphs of f(x) = sin(x) + 2x g(x) = 5 - x2 SOLVE in the HP-34C and HP-15C will tackle this, but they will find only one root at a time, so the guesses that define initial search ranges must be selected intelligently. Don't forget RADian mode: ```Program: Execution: LBL B 0 RAD ENTER 2 2 + SOLVE B * (1.247614363) 5 STO 0 - -3 x<>y ENTER SIN -4 + SOLVE B RTN (-3.397303973) RCL+ 0 (-2.149689610) ``` Question 6: Approximate the largest value of f(x) = -x8 + 787x4 + 673x3 + 521x2 + 840x + 12. -x8 is a symmetrical term dominant for larger-magnitude values of x, reducing the sum of the remaining terms (except at x = 0). Since all the coefficients for the lower-order terms are positive, it's reasonably clear that, when x > 0, the lower-order terms all contribute positively in a monotonically-increasing sum, providing the maximum function value. Therefore, there is indeed only one root for the derivative of the complete function for x > 0, indicating a local maximum; the largest value is the function evaluated at that point. Using SOLVE between 0 and 10 to find the root of the derivative yields 4.619292584, consistent with what "Crawl" found. Programming the derivative using Horner's Method will speed things up: ```Program: Execution: LBL 0 0 * ENTER * 10 * SOLVE 0 -8 ... * (4.619292584) 3148 ENTER + ENTER * ENTER 2019 * + * * * 1042 CHS + 787 * + 840 * + 673 RTN + * 521 + * 840 + * 12 + (232,366.5024) (coefficients merged for clarity) ``` Question 10 Give the y coordinate of the solution to the system ``` 31x - 29y = 43 -51x + 19y = 16 ``` in reduced fraction form. The approach, of course, is to utilize ``` [a b] 1 [ d -b] inv | | = ------- * | | [c d] ad - bc [-c a] ``` and multiply to solve for x and y. Answers can be checked with the HP-15C determinant and matrix solutions. -- KS Edited: 9 Nov 2009, 1:17 a.m.