Classic Exponent Challenge for SATs, Algebra 2, Math Contests...

Don't forget to register for the upcoming 2nd MathNotations Free Online Contest for secondary students. Click here for more info.



The first 4 terms of a sequence are 2, 6, 18, and 54.
Each term after that is three times the preceding term.
If the sum of the 49th, 50th and 51st terms of this sequence is expressed as k⋅3⁴⁹, then k = ?

Click Read more to see the answer, solution, discussion...


Answer: 26/3

Suggested Solution
The first three terms can be written as
2(3⁰), 2(3¹) and 2(3²). (***)
In general, the nth term is 2(3^n-1).
The sum of the 3 desired terms would then be 2(3⁴⁸) + 2(3⁴⁹) +2(3⁵⁰). Factoring out 3⁴⁹, we obtain 3⁴⁹(2/3 + 2 + 6) = (26/3)(3⁴⁹), so k = 26/3.

Comments
(1) Too hard for SATs? Similar (but slightly easier) problems have appeared on the test.
(2) Could students use the "Make it simpler strategy" here to reduce the problem to the sum of just the first three terms? But this is the essence of geometric sequences (or exponential functions):
From (***) above, this sum would be
2(3⁰) + 2(3¹) + 2(3²) = 2 + 6 + 18 = 3(2/3 + 2 +6) = 3(26/3). The coefficient 26/3 would be the same for any three consecutive terms! Is this concept/technique worth developing?

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MathNotations Second Math Contest (FREE) Announced!

Silly Joke For Today
Have you heard the very sad "fractioned" fairy tale?
Six out of seven dwarfs are not 'happy.'

Announcing MathNotation's Second Free "Online" Team Math Contest!

If interested in participating, please send an email as soon as possible to "dmarain at gmail dot com." I will then email a registration form. Your initial email expresses only your interest. You are under no obligation.

There will be several changes from the first contest.

1) Team advisers may administer the contest at any time during the week of May 18th-22nd.
2) Advisers must email the registration form no later than Fri 5-15-09.
3) The contest is designed for secondary students who have completed Algebra 2. Some trigonometry may be necessary. I would not recommend middle schoolers take the contest unless they have completed or are completing Algebra 2.
4) Questions include topics from geometry, algebra, trigonometry, discrete math, etc. No calculus...
5) Teams must consist of from two to six members. Homeschooling and international teams are welcome!

For more details, click Read more.




FORMAT OF TEST
Questions types include
(a) Short constructed response (students enter only numerical answers)
(b) Open-ended requiring detailed work and explanations
(c) Multi-part questions


ADMINISTRATION OF CONTEST
1) Team members must complete the contest within 45 minutes on the same day.
2) Advisers must email the official answer form the same day the contest is administered. Scanned solutions will be accepted.
3) Any scientific or graphing calculator is allowed.

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Number Theory, Logic, Proofs and Patterns for Middle School and Beyond...

HAPPY HOLIDAYS!

The following is a series of apparently straightforward arithmetic problems for middle schoolers. However, the objective is to have students justify their reasoning beyond "guess and test" methods. Proving there is only one solution or none requires more careful logic using algebra as needed. Students will need some basic algebra for the "proofs." For the younger student, modify these questions to have them find the squares in questions 1,2,3 and 5. Take this as far as you wish...


In the following, square refers to the square of an integer. Justify your reasoning or prove each of the following.

(1) There is only one square which is 1 more than a prime.

(2) There is only one square which is 4 more than a prime.

(3) There is only one square which is 9 more than a prime.

(4) There is no square which is 16 more than a prime.

(5) There is only one square which is 25 more than a prime.

(6) Can one generalize this or not??

Click Read More for selected answers, solutions...



Selected Answers, Solutions

(2) If n² is 4 more than some prime, p, then we can write
p = n² - 4 = (n-2)(n+2). Since p is prime, the smaller factor must be 1, so
n-2 = 1 or n = 3. Thus, there is only one square, 9, which is 4 more than the prime, 5.

(4) p = n² - 16 = (n-4)(n+4). There n would have to equal 5, n² would equal 25 but 25 - 16 = 9 is not prime.

(6) If there were a general rule would that mean we'd have a formula for primes?

Your thoughts about these questions...

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A Unique Math Blog and Website

I'm sure many of my readers out there are aware of Tanya Khovanova. Including her blog in my blogroll is long overdue and was a result of ignorance on my part. I've corrected that oversight and, now that I have discovered and enjoyed some of her essays, I want to share this with the rest of you. Brilliant (winning a Gold Medal in the IMO might suggest that!) and witty, her writing style is original and provocative. BTW, we share a passion for linguistic puzzles although I don't consider that unusual for those with a math bent. I strongly recommend you visit her website, Tanya Khovanova's Home Page as well as her blog. Her site contains a rich mine of creative and challenging problems. One of my favorite 'puzzles' is called Fly Droppings On Your Pizza. Click on Read more for further details...




Her most recent article, Multiple Choice Proofs, is a fascinating discussion of how to improve assessments (math contests in particular) which tend to be multiple choice here in the US and proof-oriented in her native Russia. Tanya suggests a compromise between those two extremes ("balance!") offering some alternatives one of which would involve artificial intelligence (AI) to evaluate proofs. I've always believed it might be possible to evaluate one's writing using an algorithmic approach but the obstacles certainly seem formidable at this time.

I also posted a couple of comments on her "About" page. I included a challenge mental math problem for her talented son, Sergei, to attempt. It may not appear for awhile as her comments are moderated. I may post it here as well...

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A Recurring Problem for SATs (Functions)

SAT "Grid-In" Type
Level of Difficulty: 5 (High)
Content: Algebra 2, precalculus

The function F satisfies the condition
F(N + 6) = F(N) + 8, for all integers N.
If F(7) = -2, what is the value of F(25)?

Click on Read more to see the answer, solution, discussion.


Answer: 22

Suggested Solution:

Replace N by 7 since F(7) is known:
Therefore, F(7 + 6) = F(7) + 8 or F(13) = -2 + 8 = 6

Next, replace N by 13 since F(13) is known:
F(19) = F(13) + 8 = 6 + 8 = 14

Finally, F(25) = F(19) + 8 = 14 + 8 = 22.

Comments
(1) Too difficult for the SATs? Not really! A similar problem recently appeared. There aren't that many "hard" questions (Level 5) on the SAT but, if a student wants to score over 700 they will need exposure to these types in practice.

(2) Consider writing some variations of these function-type problems for additional practice. At first, change the constants, then consider changing the operations (from addition to multiplication for example). One could raise the bar even higher by asking the question in reverse:
If F(25) = 22, what is the value of F(7)?

(3) There is considerable advanced theory in functional equations and recurrence relations underlying these problems. However, the student needs only to feel comfortable with the function symbolism (or should I call it "Math Notations!"). Starting by "plugging in' N = 7 seems simple in retrospect but most students are too intimidated to consider it. Even the precalculus student may be able to get started, but, without experience, they will often get lost. This is all about exposure, but isn't it always?

(4) One could rewrite this problem using sequence notation:
a_N+6 = a_N + 8. By expressing the problem in the context of the Nth term of a sequence, students may grasp it a bit better, but, in the end, it's all about interpreting function notation.

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Math Teachers at Play #4, Krypto, Updates (Odds and Evens),...

1) The latest "biweekly" edition of our new Carnival, Math Teachers at Play, is currently being hosted by Misty over at Homeschool Bytes. I enjoyed the "step by step" approach, progressing from primary math activities like Candy Math through middle school posts like Division of Fractions Conceptually to secondary articles like Ten 16th Century word problems. I contributed a post on Function Questions for the SATs. I'm not including any direct links to these articles. Go to Misty's site to enjoy this carnival!

2) I'm putting a lot of effort into the SAT Math Tips feature in the sidebar (readers of my feed won't get to see this of course unless they visit the site). I realize a good part of the country takes the ACT but the suggestions may be applicable to any standardized test and the math content is valid for anyone who wants to use it.

3) The Read More feature I recently instituted has some bugs as I'm sure you noticed. It doesn't work in RSS or Atom feeds of course and it sometimes doesn't work properly even on this site. I'm a coder at heart but I'm not sure it's worth all the effort. Clearly the Blogger developers are not interested in making things easy for us. You may see Read more... even if there is nothing else to see or it may not work all! Too late for me to migrate to Wordpress at this point... Please be patient with me here as I work through this.

4) The Math Problems of the Day in the sidebar are of good quality and are challenging but I'm seeing more repetition of questions. I will evaluate this and decide if I want to keep this feature.

5) No, I haven't forgotten about the next MathNotations contest I promised for April or May. Stay tuned...

6) Ever hear of Krypto or 24 (the game not the TV show!) or the more well-known "Four Fours Game"? I played many years ago and have always enjoyed the challenge of this tantalizing arithmetic game. I've used it effectively in upper elementary and middle school classrooms when I was a math staff developer to reinforce order of operations and basic fact recall (no calculator allowed!). Ok, so I'm now driving you crazy with another KenKen-like game!!
Click on Read more to learn more. (if this works!).




Here is a sample play of Krypto:

Suppose you're dealt the following five number cards:
5, 9, 4, 11, 1.
A 6th or objective card is turned up, say 2.
Using some or all of the four basic arithmetic operations and the 5 numbers exactly once, produce the objective number. This one is straightforward using only addition and subtraction:
11 + 5 - 9 - 4 - 1 = 2. Krypto! You've Won!

It is possible to be dealt an "impossible" hand for which there is no solution or there could be many solutions for the same hand! The original game did not allow the use of parentheses but you could choose any variation you wish, including reducing the number of cards to 4. An ordinary deck could be used modifying the face cards to be 11,12, etc.

If you want to play the online version from mphgames, go here for instructions and play. Your browser must be java-enabled but most are. For more background on the game and a discussion of the underlying combinatorial mathematics and a discussion of the computer program which generates it, look here.

ENJOY KRYPTO WITH OR WITHOUT YOUR STUDENTS!!

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Two SAT-Type Percent Problems Appropriate for Middle School as well...

Version I (Level of difficulty 3 - medium)

With a special promotion, Al received a 60% discount on a new stereo system and paid $x. Sylvia bought the same system (same original price) but only received a 20% discount. In terms of x, how much (in dollars), did Sylvia pay? Assume x > 0 and disregard sales tax.

(A) 4x (B) 3x (C) 2x (D) 4x/3 (E) x/3


Version II (Level of Difficulty 4 - medium/hard)
Grid-In Type

Maury purchased a new electronic game system with a 25% off coupon. His friend bought the same system (same original price) with a 40% off coupon. If his friend paid $45 less for the system, how much did Maury pay (disregard sales tax)?



For the answers, suggested solutions, strategies and discussion, click Read more...


Level I problem
Answer: (C) 2x

Possible Solutions:
Method I ("Plug-in" SAT Strategy - Student-preferred?)
Let original price = $100.
Then Al's discount was $60, so he paid $40. Thus x = 40.
Sylvia's discount was $20, so she paid $80, which means she paid twice Al's price or 2x.

Method II (conceptual)
Al paid 40% of the original price, Sylvia paid 80%, therefore Sylvia paid twice as much as Al.

Method III (traditional - Algebraic)
Reasoning as in Method I, Al paid 40% and Sylvia paid 80% of the original price.
Let y = original price (before discounts).
Then Al paid 0.4y = x. Solving, y = 2.5x.
Sylvia paid (0.8)(2.5x) = 2x.

Level II Problem
Answer: $225
Methods???
Note: There is a mental math method which will be discussed later.


FOOD FOR THOUGHT

• What are some of the factors which might make the second problem more difficult?
  
• Should these types of percent questions be included in prealgebra texts for middle schoolers?
• The first question appears to be fairly straightforward. Which part of the problem would cause the most difficulty for students in your opinion?
  
• Note that in the ever-popular 'plug in a number method', the ever-popular $100 was chosen for the original price. Can you think of an occasion where you might not want to substitute $100 in a discount problem?
• Students also need to recognize that I plugged in $100 for the original price rather than replace x by $100. They often feel they must substitute a number for the variable given, which, in this case would lead to more work.
• There are other approaches here. Please share different methods you have used!
• How would we assess their learning of the concepts here? Include one of these on the next quiz or test? Use these as warmups occasionally? Develop a worksheet of similar problems for homework?
  

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ADP/Achieve Algebra I Practice Test Now Online! Links, Discussion...

This is not an April Fool's joke! As anticipated and mentioned previously on this blog, a complete practice test is now available for you to download in pdf format from the Achieve web site. Click on the bottom link on the right sidebar.

Student expectations are shown above (you may need to click the image to see a larger version).



From the site:

The ADP Algebra I End-of-Course Exam consists of algebraic topics which will be taken by students across all participating states. These topics are typically taught in an Algebra I course, and fall into four strands: 1) Operations on Numbers and Expressions 2) Linear Relationships 3) Non-linear Relationships and 4) Data, Statistics and Probability.



Composition of Test
47 operational items

• 40 multiple-choice (1 pt ea)

• 5 short answer ( 2 pts ea)

• 2 extended response (4 pts ea)

I'm not permitted to reproduce any of the questions here but I will enumerate some of the topics tested:
Identifying linear functions (y = 2x vs. y = 2^x vs. y = 2x²)
Choosing effective data displays
Absolute value graphs
Rationalizing denominators in radical expressions (square roots only)
Exponential functions (simple)
Graphical interpretation of systems of equations

For further discussion, click Read more below.


and some more...
Solve quadratic equation by factoring
Graph of linear inequality
Probability of independent events
Determine vertex of graph of a quadratic function
Interpretation of slope from problem situation (rate of change)
Solving linear equations
Identifying irrational numbers (square root form)
Multiplication of radical expressions
Solving absolute value equations

Download the entire document to see examples of the multiple-choice, short answer and extended response type questions. This should prove very useful for teachers and students as they prepare for the first administration of this test. From the partial listing of topics above you can see that this is a test reflecting an ambitious first year algebra curriculum which will surely raise the bar for those states and districts participating. I would expect most students will struggle the first time around with the content, format and a level of difficulty they may not have yet experienced. Over time and with experience students should perform at higher and higher levels. This is a learning experience for all of us and it should be viewed as an opportunity to be part of an exciting change in American mathematics education...

Additional Comments

• Even if your district is not participating, this test is an excellent way to measure your Algebra curriculum against a highly regarded world-class standard. I strongly encourage you to include these sample questions in warmups, for classroom discussion or review for other assessments.
  
• As nontraditional or difficult as some of the earlier problems may appear to be, read through the entire sample test. You will definitely see many traditional algebra exercises with which your students should feel comfortable.
• Problem 31 is an interesting application of the Pythagorean Theorem and quadratic equations. Some students have a knack for guessing 'special' right triangles for these however. After discussing the algebra, remind them to consider multiples of 3-4-5. Since the legs differ by 4, it's worth trying sides of 3x4 and 4x4, then checking if the hypotenuse works. It does!
• The test reflects an excellent blend of tradfitional and reform. The extended response and short answer questions should have a definite impact on instruction and assessment in your classes whether or not your district is on board with this test. Ultimately, I predict EVERY state will be on board!
  
  Your comments...
  

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