## How To Study For Math Placement Test?

What should I study for a college math placement test? – You should study the following topics to make sure that your college math placement test prep covers everything you need to know.

## Is math placement test hard?

What Will a Math Placement Test Day Be Like? – In many colleges, you will take the placement test on a computer in a selected room or lab. Study the regulations for the math test day and the venue. Ask what you are and are not around to take into the examination venue.

Find out whether there is a time limit; typically, there is no time limit for the college math placement test. In some instances, a learning institution can permit you to go to a test site with a calculator, while others forbid calculators. In order to get the most meaningful grades from your exam, prepare well for your mathematics skills through study and practice, and ensure that you get enough rest and take care of your individual needs before taking the math placement test.

We will assist you in understanding more about college math placement tests. Contact us today at +18882424262,

### How do you pass math placement?

Download Article Download Article While you can’t pass or fail a placement test, your score will determine which mathematics class you should be placed in based on your knowledge and skills. Before you take the test, review basic math concepts, geometry, algebra, and calculus.

1. 1 Review division, multiplication, and the order of operations. Basic math concepts are still fair game on a math placement test. Go over long division, multiplication, and how to follow the order of operations. Make sure you know how to apply those to different math concepts and can do them long form as well as with a calculator.
• You will most likely be allowed to use a calculator on a math placement test, but some problems might ask you to show your work.
2. 2 Go over conversions between fractions and decimals. Fractions and decimals can describe the same number, but they display it in a different way. Make sure you know how to convert from fractions to decimals and vice versa. Look out for fractions that are written as percentages.
• For example, 24% indicates a fraction of 24/100.
• You may be asked to convert 24% to a decimal. Divide 24/100 to get 0.24.

3. 3 Make sure you know how to work with inequalities. Most often, these problems are given as 2 numbers and you are asked to put a ” ” or “=” in between them to signify if they are less than, greater than, or equal to each other.
• For example, a problem might read -3 _ 6
4. 4 Look over scientific notation rules. Scientific notation is the standard for which long numbers, either whole or decimal, are written. Significant digits are the numbers that are significant. These usually exclude 0s.
• For example, when writing 0.0000001, you would instead notate 1 x 10^-7. This signifies that to get the decimal number, you have to move the decimal point 7 times to the left.

1. 1 Reexamine intermediate algebra. Most of intermediate algebra consists of solving equations that have letters substituted for numbers. Brush up on how you would solve for “x” in any given equation. Be prepared to solve for multiple letters in 1 problem. Some examples of intermediate algebra are:
• Factoring polynomials, setting up equations, linear equations, exponents and radicals, simplifying radical expressions, and the AC method of factoring.
2. 2 Go over geometry concepts. Geometry is a large field of study, and there will most likely be a dedicated section to it on your placement test. Review concepts that you learned from high school and be prepared to use your graphing calculator to assist you. Some concepts from the geometry portion of the test include:
• Nonlinear points, intersecting lines, perpendicular lines, angle measurements, and line distance measurements.
3. 3 Prepare for the calculus portion of the test. Calculus includes graphing, functions, and trigonometry. Review these concepts and prepare to use the graphing function on your calculator to solve these problems. Some examples of the calculus portion of the test include:
• Functions, exponents and radicals, complex numbers, matrices, and factorials.

1. 1 Look over as many topics as you can before the test. Most math placement tests will give you credit for proving that you understand most of each topic. Study a wide range of what you have learned in the past: algebra, statistics, geometry, and calculus are all fair game when it comes to a math placement test. Tip: This is especially true for the ALEKS math placement test, which is the one most commonly used for colleges.
2. 2 Go over the concepts before you try any practice questions. You will have a much easier time studying for your placement test if you review the math concepts you have learned throughout your school years. Take a look at math that you learned at the beginning of high school all the way up to the highest level that you have learned. Look through your old notes or find example questions online of basic algebra, fractions, and even word problems.
3. 3 Avoid trying to learn concepts you aren’t familiar with. Everyone’s math level is different. If you didn’t quite make it to calculus in high school or you are still stuck on algebra, that’s okay! Don’t try to teach yourself new concepts. Instead, review the ones that you are knowledgeable in so that you do well on those sections of the test. Focus on reexamining rather than learning.
• You can’t fail a math placement test. Just do the best that you can on the levels that you do know.
4. 4 Take practice tests and answer practice problems online. The best way to prepare for a placement test is to answer questions that are similar to the ones on the real test. Check to see if your college website has a practice test that you can print out and answer.
• If you can find out which specific placement test you will be taking, search practice tests for that specific test. Otherwise, just search for general math placement test questions.

1. 1 Take your time when working through the problems. Don’t rush through your placement test, even if you think you are an expert. Work slowly through the problems and make sure to read each one carefully. Double check your work if you have time to make sure you are answering correctly.
• Some math placement tests are timed, but others are not. If yours is timed, do the questions that take you the least time first and then move onto harder ones.
2. 2 Complete the sample questions if your test gives them to you. Some standardized placement tests, like Accuplacer, give you 1 to 2 sample questions to prepare you for the section you are about to take. Use these questions as an example of what is to come and prepare yourself for questions that are similar.
• Use these questions to think back on what you have reviewed and go over the general rules in your head.
3. 3 Skip the questions you don’t know and come back to them. If you get stuck on a question, leave it until the very end and focus on the problems that you do know how to answer. Be sure to mark your page or remember to come back to it somehow so you don’t leave it blank. Leave yourself enough time at the end of the test to complete the problems you were unsure about.
4. 4 Review your answers before you turn in your test. Sometimes the stress of doing well on a test can make you make simple mistakes. If you have time, go over your answers right before you turn in your test. Double-check your work with a calculator to make sure you didn’t accidentally subtract when you meant to add, or multiply instead of divide. Tip: If you don’t have time to go over all of your answers, that’s okay. Check the ones that you had the most trouble with to make sure you did them right.
5. 5 Try not to be disappointed if you place into a lower-level math class. A placement test is a way to gauge what level of math you are proficient in and what you need to learn still. Don’t be discouraged if you get your results back and they are lower than you expected. All that means is you might need a few more classes to boost your math skills.
• Some colleges will not give you credit for your math class until you reach a certain level. Check with your course advisor to see how many math classes you need to take for your particular department or degree.
• Often, your score on each section of the test correlates to what math class you are placed into. If you did great on the geometry section but poorly on the calculus one, you will likely be placed into a lower-level calculus class.
• The level of math that you need to take in college varies based on your major. An English major might not need to take as many math classes as a chemistry student would. Check with your advisor if you are confused about how many credits you need.

• Question How do I study for a college placement test? Arash Fayz is the Co-Founder and Executive Director of LA Tutors 123, an academic consulting and private tutoring company based in Los Angeles, California. Arash has over 10 years of educational consulting experience, managing the tutoring of students of all ages, abilities, and backgrounds to score higher on standardized tests and gain admission to their target schools. Test Prep Tutor Expert Answer
• Question What kind of math is on a placement test? Arash Fayz is the Co-Founder and Executive Director of LA Tutors 123, an academic consulting and private tutoring company based in Los Angeles, California. Arash has over 10 years of educational consulting experience, managing the tutoring of students of all ages, abilities, and backgrounds to score higher on standardized tests and gain admission to their target schools. Test Prep Tutor Expert Answer

## What is the hardest math test in the world?

What is the William Lowell Putnam Mathematical Competition? – The William Lowell Putnam Mathematical Competition, known to many as the World’s Hardest Math Competition, is a prestigious mathematics competition for undergraduate college students in the United States and Canada.

It was founded in 1927 by Elizabeth Lowell Putnam in memory of her husband William Lowell Putnam, who was an advocate of intercollegiate intellectual competition. The competition is administered by the Mathematical Association of America and takes place annually on the first Saturday of December. The competition consists of two 3-hour sessions, one in the morning and one in the afternoon.

During each session, participants work individually on 6 challenging mathematical problems that cover a range of advanced material in undergraduate mathematics, including concepts from group theory, set theory, graph theory, lattice theory, and number theory.

Each problem is worth 10 points, and partial credit is given for incomplete or nearly complete solutions. The competition is very difficult: the median score is usually zero or one point out of 120 possible, and there have been only five perfect scores as of 2021. The competition awards scholarships and cash prizes to the top individual scorers and the top teams.

The top five individual scorers are designated as Putnam Fellows and receive a scholarship of up to \$12,000 plus tuition at Harvard University (Putnam Fellow Prize Fellowship). The next highest-ranking individuals receive cash prizes ranging from \$250 to \$2,500.

The top 100 individual scorers have their names mentioned in the American Mathematical Monthly (alphabetically ordered within rank), and the names and addresses of the top 500 contestants are mailed to all participating institutions. The top teams receive cash prizes ranging from \$5,000 to \$25,000 for their departments of mathematics.

A team’s score is the sum of the scores of its three team members. The competition aims to stimulate interest in mathematics among college students and to recognize their mathematical talent and achievement. It also provides an opportunity for students to challenge themselves and to interact with other students who share their passion for mathematics.

The competition has grown to be the leading university-level mathematics examination in the world and attracts thousands of participants every year. Thank you for taking the time to read this story. If you enjoyed it, please show your appreciation by clicking the clap icon as many times as you wish. If you appreciate my writing and would like to support me, you can or treat me to a,

Stay tuned for more content! : 3 Quick Questions from the Hardest Math Test in the World

#### Which math exam is the hardest?

The William Lowell Putnam Mathematical Competition – Mathematical Association of America (MAA). Image source: https://www.cmu.edu/news/stories/archives/2016/april/putnam-2015.html M athematics is one of my favorite subjects. I love it because it is beautiful, it is logical and we can use it to explain any physical phenomena of the universe.

#### What are good questions for a placement?

What do you hope to learn here? What strengths to you bring to this field? What are your areas of professional growth? Can you tell me about your previous work, volunteer and/or field placement experiences?

### How long does it take to study for ALEKS?

At the other extreme, it is not unusual for students to take 100 hours or more to complete a course; these are mostly cases where students were placed into a course that is too advanced. As a general rule, most students should plan to spend at least 40 hours in ALEKS to complete a course.

### What does math placement test score mean?

The mathematics placement test is a prerequisite for only the introductory courses above. All other mathematics, statistics and computer science courses require previous coursework. The mathematics placement test score indicates a student’s appropriate level of introductory course and does not earn a student credit for any course. It does not satisfy any requirement for graduation. Math credits are awarded for the successful completion of VCU math courses, transfer credits or satisfactory AP/IB test scores. Math courses taken in high school that are not dual enrollment courses with an accredited university cannot be used to place into a higher introductory math course.

## What is a 30 on ALEKS math placement test?

Understand Your Score ALEKS scores cannot be interpreted in the same way as exam grades. An ALEKS Assessment Score of 30 or higher indicates adequate preparation for most general education mathematics or statistics courses. The below chart shows the recommended NKU course placements for various ALEKS scores.

ALEKS Assessment Score Recommended Course Placement
0-29
• MAT 101 (Preparation for College Algebra)
• MAT 115R (Math for Liberal Arts with Recitation)
• STA 205R (Statistical Methods with Recitation)
30-45
1. MAT 102 (Introductory College Algebra)
2. MAT/STA 110 (Introductory Probability)
3. MAT 115 (Mathematics for Liberal Arts – QR)

STA 113 (Prob. And Stat. with El. Ed. Applications – QR) STA 205 (Introduction to Statistical Methods – QR) STA 212 (Statistics for Business Applications I – QR)

46-60 MAT 103 (Intermediate College Algebra) MAT 114 (Finite Mathematics – QR)
61-75
• MAT 112 (Applied Calculus)
• MAT 119 (Pre-calculus Mathematics)
• MAT 185 (Introductory Discrete Mathematics – QR)
76 or greater MAT 128 (Calculus A – QR) MAT 129 (Calculus I – QR)

QR denotes a course which fulfills the Mathematics & Statistics requirement for general education Students may enroll in courses at or below their ALEKS score level. While alternative placements are possible for some courses, NKU strongly recommends adherence to the placements indicated in this chart.

## Can ALEKS math placement test detect cheating?

The Perfect Book to Ace the ALEKS Math Test – Rated 4.45 out of 5 based on 225 customer ratings Satisfied 213 Students ALEKS stands for Assessment and LEarning in Knowledge Spaces. It has been widely used by students and universities to make learning fun and orderly. This tool has served as an assessment tool to measure students’ learning of different topics, contexts, and subjects.

1. ALEKS avoids multiple-choice questions and instead utilizes flexible, easy-to-use input tools that mimic paper and pencil techniques.
2. ALEKS assesses your current content knowledge in a short amount of time (30-45 minutes for most courses) by asking several questions (usually 20-30).
3. Direct cheating when attending the assessment or test at ALEKS is difficult.

Even survey results display that cheating in ALEKS is harder than cheating in real classes. Because ALEKS is an electronic web page that easily detects if you are cheating. You can get help with questions or problems with homework, but you cannot cheat directly while you are being for the assessment.

1. Respondus LockDown Browser is one of the tools developed to combat cheating on ALEKS,
2. This browser acts as a special browser that restricts the testing environment.
3. When launching the browser, your computer’s webcam and microphone must be set to run throughout the test session.
4. Therefore, the browser relies heavily on your computer microphone and webcam to detect cheating,

The important thing about cheating is that cheating in the placement assessments is useless – your reward will be for taking a very difficult class for your current level of math knowledge. Looking for the best resources to help you or your student succeed on the ALEKS Math test?

#### Can you use a calculator for ALEKS math placement?

May I use a calculator while using ALEKS? ALEKS will provide an on-screen calculator if you need one to complete a particular problem. Otherwise, please do not use a calculator.

#### Which country has toughest maths?

Which country has the hardest math? The United Kingdom, The United States of America, etc are the countries having one of the best education systems. But when it comes to having the hardest math, China and South Korea top the list.

### Has 3X 1 been solved?

The Analysis of Convergence for the 3 X + 1 Problem and Crandall Conjecture for the a X + 1 Problem Abstract The 3 X + 1 problem (Collatz conjecture) has been proposed for many years, however no major breakthrough has been made so far. As we know, the Crandall conjecture is a well-known generalization of the 3 X + 1 problem. It is worth noting that, both conjectures are infamous for their simplicity in stating but intractability in solving. In this paper, I aim to provide a clear explanation about the reason why these two problems are difficult to handle and have very different characteristics on convergence of the series via creatively applying the probability theory and global expectancy value E ( n ) of energy contraction index. The corresponding convergence analysis explicitly shows that a = 3 leads to a difficult problem, while a > 3 leads to a divergent series. To the best of my knowledge, this is the first work to point out the difference between these cases. The corresponding results not only propose a new angle to analyze the 3 X + 1 problem, but also shed some light on the future research. Share and Cite: Hu, Z. (2021) The Analysis of Convergence for the 3 X + 1 Problem and Crandall Conjecture for the a X + 1 Problem. Advances in Pure Mathematics, 11, 400-407. doi: 10.4236/apm.2021.115027,1. Introduction The 3 X + 1 problem is known as the Collatz problem. It focuses on the behavior of the iteration of the function which takes odd integers n to 3 n + 1 and even integers n to n /2. The Collatz conjecture declares that no matter starting from any positive integer n, if we repeat the iteration of this function, eventually we will converge to the value 1. Its corresponding mathematical statement is as follows. If S ( X ) is equal to 3 X + 1 ( X is a positive odd number) or X /2 n ( X is a positive even number, n is an unidentified value), starting from any positive integer X, iterating the function repeatedly to get a sequence of S ( 0 ) ( X ), S ( 1 ) ( X ), S ( 2 ) ( X ), ⋯, where S ( 0 ) ( X ) = X, S ( n + 1 ) ( X ) = S ( S ( n ) ( X ) ) ( n = 0, 1, 2, ⋯ ), there is a positive integer I, which makes S ( i ) ( X ) = 1, This problem is traditionally proposed by Lothar Collatz during his student days in the 1930’s. At that time, he was interested in graph theory and studied the behavior of iterations of mummer-theoretic functions as directed graphs. Although Collatz never published his iteration problems, he presented them at the International Congress of Mathematicians in 1950 in Cambridge. Then the original 3 X + 1 problem appeared in print, After that, the 3 X + 1 problem has appeared in various forms. It is one of the most infamous unsolved puzzles in the word. Prizes have been offered for its solution for more than forty years, but no one has completely and successfully solved it,1.1. Selecting a Template 3 X + 1 Problem the Specified Calculation and Verification 1.1.1. The Specified Calculations The 3 X + 1 problem has been numerically checked for a large range of values on n, In 1992, Leavens and Vermeulen proved that the conjecture is true for positive integers less than 5.6 × 10 13, In 2006, Distributed computing has proved for positive integers less than 510 × 10 15, Yoneda, the University of Tokyo (Japan) has proved that the conjecture is true for positive integers less than 2 40 ≈ 1.1 × 10 12, Fraenkel has check that all positive integers less than 2 50 have a finite stopping time and the conjecture is still not erroneous, All these numerical experiments verify the correctness of the conjecture in a very large range of positive integers. Hence, most of the mathematicians believe that the conjecture is true and aim to find a theoretic proof for the conjecture. It is worth pointing out that, due to the large development of computer and software, the upper bound of positive integers that has been verified to be true is keeping increasing very quickly nowadays.1.1.2. Find the Upper Bound of S ( i ) ( X ) In 1976, Terras proved that almost all positive integers n (in the sense of natural density), S ( n ) < n exists, In 1979, Allouche proved that for any a > 0.869, almost all positive integers n (in the sense of natural density), S ( n ) < n a exists, In 1994, Korec proved that for any a > ln3/ln4 ≈ 0.7924, almost all positive integers n (in the sense of natural density), S ( n ) < n a exists. In 2019, Fields Prize winner Terence Chi-Shen Tao published his results on arXiv.org, attempted to prove that as long as is a sequence of real numbers that tends to be positive infinity, for almost all positive integers n (in the sense of logarithmic density), S ( n ) < f ( n ) exists. In his conclusion, f ( n ) could be a sequence that growing slowly, such as f ( n ) = lnlnlnln( n ), All these works established certain skew random walks on cyclic groups at high frequencies and estimated the certain renew processes which interact with a union of triangles. Though these works provide the closest conclusions to the original conjecture, they still can't get on the final peak.1.2. Crandall Conjecture The 3 X + 1 problem also has various extensions, the most famous extension is the aX + b problem proposed by Crandall in 1978. Generally, let a and b positive integers, a > 1, and b is odd integer, this so-called aX + b problem is specified as whether 1 can be obtained after a finite number of iterations for any positive odd number x? It is obvious that a = 3, b = 1 is the 3 X + 1 problem. After research, Crandall proposed the following conjecture: for all the other positive integers ( a > 1), we can find a positive odd number r, which makes S ( i ) ( r ) is not equal to 1 for all positive integers except a = 3, b = 1 (3 x + 1 problem). Crandall proved that if b > 1, the conjecture is correct. In the case of aX + 1, he only proves that the conjecture is correct when a = 5, 181 and 1093, In 1995, Franco and Pom-erance proved that the Crandall conjecture about the aX + 1 problem is correct for almost all positive odd numbers a > 3, under the definition of asymptotic density. However, both of the 3 X + 1 problem and Crandall conjecture have not been solved yet. And to the best of my knowledge, the convergence analysis of these two problems is still blank.1.3. Probabilistic Argument Instead of directly solving the 3 X + 1 problem, many researchers also proposed some heuristic probabilistic arguments to support the conjecture. For example, Olliveira and Silva proposed an empirical verification of the conjecture ; Sinai extends the 3 X + 1 problem to a statistical (3 X + 1) problem and studied its properties ; Thomas showed a non-uniform distribution property of most orbits ; Tao recently claimed that almost all orbits of the Collatz map attain almost bounded values, Note that, these probabilistic approaches not only extend the original problem, but also provide some new ideas to the future study of the problem.2. Energy Method, Expectancy Theory, Weak 3 X + 1 Problem and Crandall Conjecture In order to solve the 3 X + 1 problem and the Crandall conjecture for the aX + 1 problem, the key is to prove that the global convergence (diffusion) of the iteration on the corresponding number field is consistent with the individual convergence (diffusion) and global convergence (diffusion) of the number field. Some researchers have pointed out that, it is necessary to prove the uniqueness of the “4-1” cycle in the 3 X + 1 loop. Otherwise, it is definite as the weak 3 X + 1 problem which omits the loop factor, Expected Value of Energy Contraction Index For each iteration of an arbitrary positive odd number X, based on the description of the problem, the constructor T ( X ) = ( 3 X + 1 ) / 2 n ( X is 1, 3, 5, and n is an undefined value). By using the energy analysis method, defined 3 and 1 on the molecule as the expansion energy factor X, 2 n on the denominator as the average contraction energy factor, when the contraction energy is greater than the expansion energy, the T ( X ) converges, so as expected value E ( n ) of energy contraction index, the larger the relative expansion energy, the faster speed of the convergence rate T ( X ), Now, I use some numerical instances to explicitly illustrate how to calculate E ( n ). For example: X = 7, based on the description of the problem, 7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1. There are 16 steps for 7 to converge to 1. The calculation of E ( n ) = 1 × 2 / 5 + 2 × 1 / 5 + 3 × 1 / 5 + 4 × 1 / 5 = 11 / 5 = 2.2 X = 11, based on the description of the problem, 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1. There are 14 steps for 11 to converge to 1.The calculation of E ( n ) = 1 × 1 / 4 + 2 × 1 / 4 + 3 × 1 / 4 + 4 × 1 / 4 = 10 / 4 = 2.25 X = 27, based on the description of the problem, 27 → 82 → 41 → 124 → 62 → 31 → 94 → 47 → 142 → 71 → 214 → 107 → 322 → 161 → 484 → 242 → 121 → 364 → 182 → 91 → 274 → 137 → 412 → 206 → 103 → 310 → 155 → 466 → 233 → 700 → 350 → 175 → 526 → 263 → 790 → 395 → 1186 → 593 → 1780 → 890 → 445 → 1336 → 668 → 334 → 167 → 502 → 251 → 754 → 377 → 1132 → 566 → 283 → 850 → 425 → 1276 → 638 → 319 → 958 → 479 → 1438 → 719 → 2158 → 1079 → 3238 → 1619 → 4858 → 2429 → 7288 → 3644 → 1822 → 911 → 2734 → 1367 → 4102 → 2051 → 6154 → 3077 → 9232 → 4616 → 2308 → 1154 → 577 → 1732 → 866 → 433 → 1300 → 650 → 325 → 976 → 488 → 244 → 122 → 61 → 184 → 92 → 46 → 23 → 70 → 35 → 106 → 53 → 160 → 80 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1. There are 111 steps for 27 to converge to 1. The calculation of E ( n ) = 1 × 24 / 41 + 2 × 10 / 41 + 3 × 3 / 41 + 4 × 3 / 41 + 5 × 1 / 41 = 70 / 41 ≈ 1.71 Since 2 ln(3)/ln(2) = 3, 1.71 is larger than the energy equilibrium point ≈ ln(3)/ln(2) ≈ 1.585, compared with the above two equations, it is closer to the equilibrium point. Although the initial values are different, the relative speed of convergence to 1 is slower. From these numerical examples we can clearly see that when n is sufficiently large and the corresponding X is sufficiently large too, the contribution of 1 on the molecule to the energy expansion is negligible, which can be regarded as an even transformation factor. So whether T’ ( X ) = 3 X /2 n, X converges or not depends on the expansion-contraction ratio coefficient λ, that is 3/2 n, Since the contraction energy index n is an undefined value, its global expected value E ( n ) is determined by the distribution of two factors in the even number of 3 X + 1 ( X is an odd positive integer). By using the law of distribution, I state the above result as the following theory. Theorem 2.1 The mathematically expected expression for the 2-factor distribution of 3 X 1-type even numbers ( x is positive odd) is 2 − ( 2 + n ) / 2 n Proof S ( X ) = 3 X + 1, X is an odd positive number, which makes X = 2 K + 1 ( K = 0, 1, 2, ⋯ ) 1) S ( X ) = 3 X + 1 = 3 ( 2 K + 1 ) + 1 = 6 K + 4 = 2 ( 3 K + 2 ) That is 3 X + 1 even number ( X is an odd positive number) has at least one factor of 2.2) S ( K ) = 2 ( 3 K + 2 ) ( K = 0, 1, 2, ⋯ ) K = 2 K 1 (50%) or 2 K 1 + 1 (50%) Let K = 2 K 1 S ( K 1 ) = 2 ( 3 × 2 K 1 + 2 ) = 2 ( 6 K 1 + 2 ) = 2 2 ( 3 K 1 + 1 ) That is the even number of 3 X + 1 ( X is an odd positive number) has at least two factors of 2 in proportion of 50% (That’s 1/2), and the remaining 50% (That’s 1/2) has only one factor of 2.3) S ( K 1 ) = 2 2 ( 3 K 1 + 1 ) K 1 = 2 K 2 (50%) or 2 K 2 + 1 (50%) Let K 1 = 2 K 2 + 1 S ( K 2 ) = 2 2 = 2 2 ( 6 K 2 + 4 ) = 2 3 ( 3 K 2 + 2 ) That is the even number of 3 X + 1 ( X is positive odd) has at least three factors of 2 in proportion of 25% (That’s 1/4), and the remaining 25% (That’s 1/4) has only two factors of 2.4) If S ( K n ) = 2 n + 1 ( 3 K n + 2 ) ( n is an even positive number) So S ( K n + 1 ) is K n = 2 K n + 1 + 1 (50%) S ( K n + 1 ) = 2 n + 1 ( 3 × 2 K n + 1 + 2 ) = 2 n + 2 ( 3 K n + 1 + 1 ) If S ( K n ) = 2 n + 1 ( 3 K n + 1 ) ( n is an odd positive number) So S ( K n + 1 ) is K n = 2 K n + 1 + 1 (50%) S ( K n + 1 ) = 2 n + 1 = 2 n + 1 ( 6 K n + 1 + 4 ) = 2 n + 2 ( 3 K n + 1 + 2 ) That is to say, the factor ratio of 3 X + 1 even number ( X is an odd positive number) with at least 2 n +2 of 2 in the proportion of (1/2) n +1, and the remaining (1/2) n +1 has only 2 n +1 factors of 2. Therefore, in the even number of 3 X + 1 type ( X is an odd positive number), only one factor of 2 accounts for 1/2, only two factors of 2 account for 1/4, only three factors of 2 account for 1/8. Only n factors of 2 account for (1/2) n, The mathematical expectation expression of its 2-factor distribution: E ( n ) = 1 × 1 / 2 + 2 × 1 / 4 + 3 × 1 / 8 + ⋯ + n × ( 1 / 2 ) n (2.1) 2 × E ( n ) = 1 × 1 + 2 × 1 / 2 + 3 × 1 / 4 + ⋯ + n × ( 1 / 2 ) n − 1 (2.2) Equation (2.2) minus Equation (2.1) E ( n ) = 1 × 1 / 2 + 1 / 4 + ⋯ + ( 1 / 2 ) n − 1 − n × ( 1 / 2 ) n E ( n ) = 2 − 2 × ( 1 / 2 ) n − n × ( 1 / 2 ) n = 2 − ( 2 + n ) / 2 n (2.3) Theorem 2.2 When n tends to infinity, From Equation (2.3), the global expectancy value E ( n ) = 2, that is T ′ ( X ) = λ X = ( 3 / 2 E ( n ) ) X = ( 3 / 4 ) X, λ = 3 / 4 < 1, T ′ ( X ) converges globally. In the same way, this distribution law is applicable to 5 X + 1, 7 X + 1, 9 X + 1, the even number of type A ( X is a positive odd number greater than 3) proves its global divergence. Proof S ( X ) = N X + 1, N and X are both odd positive numbers, which make X = 2 K + 1 ( K = 0, 1, 2, ⋯ ) 1) S ( X ) = N X + 1 = N ( 2 K + 1 ) + 1 = 2 N K + N + 1 = 2 Since N and X are both odd positive number, NX + 1 is an even number and has at least one factor of 2.2) S ( K ) = 2 ( K = 0, 1, 2, ⋯ ) K = 2 K 1 (50%) or 2 K 1 + 1 (50%) If ( N + 1 ) / 2 is a positive even number, let K = 2 K 1 S ( K 1 ) = 2 2 If ( N + 1 ) / 2 is a positive odd number, let K = 2 K 1 + 1 S ( K 1 ) = 2 2 That is the even number of NX + 1 ( N and X are both odd positive number) has at least two factors of 2 in proportion of 50% (That's 1/2), and the remaining 50% (That's 1/2) has only one factor of 2.3) If S ( K n ) = 2 n + 1 ( N is a positive odd number, Z 1 is a positive even number) Then S ( K n + 1 ) is when K n = 2 K n + 1 (50%) S ( K n + 1 ) = 2 n + 2 If S ( K n ) = 2 n + 1 ( N and Z 2 are both odd positive number) Then S ( K n + 1 ) is when K n = 2 K n + 1 (50%) S ( K n + 1 ) = 2 n + 1 = 2 n + 2 That is to say, the factor ratio of NX + 1 even number ( N and X are both odd positive number) with at least 2 n + 2 of 2 in the proportion of (1/2) n +1, and the remaining (1/2) n +1 has only 2 n + 1 factors of 2. Therefore, in the even number of NX + 1 type ( N and X are both odd positive number), only one factor of 2 accounts for 1/2, only two factors of 2 account for 1/4, only three factors of 2 account for 1/8. Only n factors of 2 account for (1/2) n, The mathematical expectation expression of the 2-factor distribution is the same as Equation (2.3), That is E ( n ) = 1 × 1 / 2 + 2 × 1 / 4 + 3 × 1 / 8 + ⋯ + n × ( 1 / 2 ) n = 2 − ( 2 + n ) / 2 n The above proof and theorem 2.2 can lead to the following important result: The global expected value E ( N ) of 2-factor distribution is always 2. On the other hand, if the following conditions are met: 1) After infinite iterations, it approaches infinity; 2) Loops do not occur during iterations. X i will finally diverge (calculation verification shows that the smaller number converges to 1) with a sufficiently large positive integer. Theorem 2.3 Since in the infinite acyclic iteration, the even number of 3 X + 1 type of the large number X i is infinite and does not repeat, According to the law of large numbers, the local expected value of the contraction energy index must get close to the global expected value 2, and the expansion-contraction ratio coefficient λ is less than 1, X i convergences, so the assumption is not true, and the individual convergence is consistent with the global convergence, that is, In a statistical sense, the weak 3 X + 1 problem established. For the Crandall conjecture of the ax + 1 problem, λ = 5 / 4, 7 / 4, 9 / 4, ⋯ > 1, In the same way, it can prove that it spreads at 5 X + 1, 7 X + 1, 9 X + 1,, The S ( i ) ( r ) is always not 1 when the even number of type ( X is a positive odd number greater than 3) does not intersect singularity 2 n,3. Conclusions This article analyzes the convergence for the 3 X + 1 problem and Crandall conjecture. It creatively applies the global expectance value E ( n ) of energy contraction index, and proposes a new angle to check the function iteration process. Moreover, it establishes a relationship between sequence process analysis and static mean via probability and statistics. The corresponding analysis clearly shows the orbit of iterations on the 3 X + 1 problem is on the imbalanced point, hence it is difficult to depict the orbit. Besides, this analysis also shows the reason why ( a > 3) leads to a divergent series. It is worth pointing out that, this manuscript is the first work to point out the difference between these cases by creatively applying the probability analysis and global expectancy value E ( n ) of energy contraction index. Since a lot of researchers have tried various ways to approach the 3 X + 1 problem but it has not been solved yet for many years, this manuscript indeed provides a totally new idea to address the problem. Hence, it may play a substantial role in the future study and shed some light on the research of this area. Conflicts of Interest The author declares no conflicts of interest regarding the publication of this paper.

## What’s the answer to x3 y3 z3 K?

After cracking the “sum of cubes” puzzle for 42, mathematicians discover a new solution for 3 What do you do after solving the answer to life, the universe, and everything? If you’re mathematicians Drew Sutherland and Andy Booker, you go for the harder problem.

• In 2019, Booker, at the University of Bristol, and Sutherland, principal research scientist at MIT, were the first to find the answer to 42.
• The number has pop culture significance as the fictional answer to “the ultimate question of life, the universe, and everything,” as Douglas Adams famously penned in his novel “The Hitchhiker’s Guide to the Galaxy.” The question that begets 42, at least in the novel, is frustratingly, hilariously unknown.

In mathematics, entirely by coincidence, there exists a polynomial equation for which the answer, 42, had similarly eluded mathematicians for decades. The equation x 3 +y 3 +z 3 =k is known as the sum of cubes problem. While seemingly straightforward, the equation becomes exponentially difficult to solve when framed as a “Diophantine equation” — a problem that stipulates that, for any value of k, the values for x, y, and z must each be whole numbers.

When the sum of cubes equation is framed in this way, for certain values of k, the integer solutions for x, y, and z can grow to enormous numbers. The number space that mathematicians must search across for these numbers is larger still, requiring intricate and massive computations. Over the years, mathematicians had managed through various means to solve the equation, either finding a solution or determining that a solution must not exist, for every value of k between 1 and 100 — except for 42.

In September 2019, Booker and Sutherland, harnessing the combined power of half a million home computers around the world, for the first time found, The widely reported breakthrough spurred the team to tackle an even harder, and in some ways more universal problem: finding the next solution for 3.

Booker and Sutherland have now published the solutions for 42 and 3, along with several other numbers greater than 100, this week in the Proceedings of the National Academy of Sciences, Picking up the gauntlet The first two solutions for the equation x 3 + y 3 + z 3 = 3 might be obvious to any high school algebra student, where x, y, and z can be either 1, 1, and 1, or 4, 4, and -5.

Finding a third solution, however, has stumped expert number theorists for decades, and in 1953 the puzzle prompted pioneering mathematician Louis Mordell to ask the question: Is it even possible to know whether other solutions for 3 exist? “This was sort of like Mordell throwing down the gauntlet,” says Sutherland.

The interest in solving this question is not so much for the particular solution, but to better understand how hard these equations are to solve. It’s a benchmark against which we can measure ourselves.” As decades went by with no new solutions for 3, many began to believe there were none to be found.

But soon after finding the answer to 42, Booker and Sutherland’s method, in a surprisingly short time, turned up the next solution for 3: 569936821221962380720 3 + (−569936821113563493509) 3 + (−472715493453327032) 3 = 3 The discovery was a direct answer to Mordell’s question: Yes, it is possible to find the next solution to 3, and what’s more, here is that solution.

And perhaps more universally, the solution, involving gigantic, 21-digit numbers that were not possible to sift out until now, suggests that there are more solutions out there, for 3, and other values of k. “There had been some serious doubt in the mathematical and computational communities, because is very hard to test,” Sutherland says.

“The numbers get so big so fast. You’re never going to find more than the first few solutions. But what I can say is, having found this one solution, I’m convinced there are infinitely many more out there.”

• A solution’s twist
• To find the solutions for both 42 and 3, the team started with an existing algorithm, or a twisting of the sum of cubes equation into a form they believed would be more manageable to solve:
• k − z 3 = x 3 + y 3 = ( x + y )( x 2 − xy + y 2 )

This approach was first proposed by mathematician Roger Heath-Brown, who conjectured that there should be infinitely many solutions for every suitable k. The team further modified the algorithm by representing x+y as a single parameter, d. They then reduced the equation by dividing both sides by d and keeping only the remainder — an operation in mathematics termed “modulo d” — leaving a simplified representation of the problem.

“You can now think of k as a cube root of z, modulo d,” Sutherland explains. “So imagine working in a system of arithmetic where you only care about the remainder modulo d, and we’re trying to compute a cube root of k.” With this sleeker version of the equation, the researchers would only need to look for values of d and z that would guarantee finding the ultimate solutions to x, y, and z, for k=3.

But still, the space of numbers that they would have to search through would be infinitely large.

1. So, the researchers optimized the algorithm by using mathematical “sieving” techniques to dramatically cut down the space of possible solutions for d.
2. “This involves some fairly advanced number theory, using the structure of what we know about number fields to avoid looking in places we don’t need to look,” Sutherland says.

The team also developed ways to efficiently split the algorithm’s search into hundreds of thousands of parallel processing streams. If the algorithm were run on just one computer, it would have taken hundreds of years to find a solution to k=3. By dividing the job into millions of smaller tasks, each independently run on a separate computer, the team could further speed up their search.

In September 2019, the researchers put their plan in play through Charity Engine, a project that can be downloaded as a free app by any personal computer, and which is designed to harness any spare home computing power to collectively solve hard mathematical problems. At the time, Charity Engine’s grid comprised over 400,000 computers around the world, and Booker and Sutherland were able to run their algorithm on the network as a test of Charity Engine’s new software platform.

“For each computer in the network, they are told, ‘your job is to look for d’s whose prime factor falls within this range, subject to some other conditions,'” Sutherland says. “And we had to figure out how to divide the job up into roughly 4 million tasks that would each take about three hours for a computer to complete.” Very quickly, the global grid returned the very first solution to k=42, and just two weeks later, the researchers confirmed they had found the third solution for k=3 — a milestone that they marked, in part, by printing the equation on t-shirts.

The fact that a third solution to k=3 exists suggests that Heath-Brown’s original conjecture was right and that there are infinitely more solutions beyond this newest one. Heath-Brown also predicts the space between solutions will grow exponentially, along with their searches. For instance, rather than the third solution’s 21-digit values, the fourth solution for x, y, and z will likely involve numbers with a mind-boggling 28 digits.

“The amount of work you have to do for each new solution grows by a factor of more than 10 million, so the next solution for 3 will need 10 million times 400,000 computers to find, and there’s no guarantee that’s even enough,” Sutherland says. “I don’t know if we’ll ever know the fourth solution.

#### What is the easiest to hardest math?

Which math classes are the easiest? – According to a large group of high-schoolers, the easiest math class is Algebra 1. That is the reason why most of the students in their freshman year end up taking Algebra 1. Following Algebra 1, Geometry is the second easiest math course in high school. Apart from these, Calculus and Statistics are considered to be the hardest math courses in high school.

## What type of math is on a math placement test?

What math is on a placement test? – The math on a placement test falls into three categories: arithmetic, algebra, and advanced math. Mastering the arithmetic on the test will place you out of one remedial math course. Acing the algebra will allow you skip remedial math entirely. Conquer the advanced math, and you might even place out of College Algebra.

## What is the hardest math taught in school?

What is the Hardest Math Class in High School? – In most cases, you’ll find that AP Calculus BC or IB Math HL is the most difficult math course your school offers. Note that AP Calculus BC covers the material in AP Calculus AB but also continues the curriculum, addressing more challenging and advanced concepts.