Mathematics Olympiad 2013
Year round begins again the National Math Olympiad 2013. In view of the problems and the solutions Mathematics Olympiad preparation techniques are discussed.
The strategy to solve the problem :
Math Olympiad problems, our principal difference between the exercise book, exercise book, we already know the solution options.
The problem for us is the way to apply. However, most of the National Mathematics Olympiad multiple step problems. The problem will be the beginning of some of our plans to pursue. Ekekajanera the different types of steps may resolve the issue. But in general, the problem may be four steps:
1. The problem is properly understood, we should first understand the problem better. It is necessary to solve the problem have all the information. It is all part of the problem again and again to read and clear idea of the problem.
2. Working on the problem, the problem could be the first thing to see, it is a method or formula can be used to solve this problem or how to solve the problem egole may be, it will be out. The problem is that information can be viewed using different values. Many times, especially in the case of an image amkale geometry is easy to fix. But to an end, not the beginning of understanding can easily work out how to fix it. So I know all the tricks to try to be a useful strategy.
3. Properly designed, the way to solve the problem and fix it properly plan and the steps will solve the problem.
4. Properly written and tested solution: the problem is the inability to write a large part of the problem. So the solution is to present the solution and not the result of the test will be to see.
Olympiad regular practice to prepare for mathematics and education options. To prepare a good way to try to resolve the problem over the last year. Clear idea of the type of problem can be found, although the common examination board has no chance to read it.
A different category of questions to discuss and resolve the problem. Book List for help with the preparation
Primary section are to do is to learn arithmetic, Number theory, algebra, geometry. Some of the puzzle – the problem is a national problem and treatment. Pathyabaiyera, as well as (1), (ii) the number of practices can be. Here are some of the problems in the National Mathematics Olympiad are:
# Problem: the product of two odd numbers 143. Number two was a friend.
Solution: The solution to the problem of our numbers will be a little play! Since the product of two numbers, so we can see the number utpadake analysis; 143 = 11 × 13. Iureka! We have two odd numbers were detected. Number two is 11 and 13.
# Problem: Abul, Badrul and Jashim started running at the same time a circular path. Babul per minute 1/3 rotation, Badrul 1/5 rotation, Jashim 1/6 rotation can start running. Minimum of circling race terminate the daurale they reach you?
Solution: This problem surely know what you think of the book issue hours. I also know a lot of you. Observe, we need a number, which is 3, 5, 6 divisible by. The smallest number is 30. (All the numbers lasagu 30) Babul 10 of the 30 minutes, Badrul 6 and Jashim 5 wheel after they run the same race last with land boundaries.
# Problem: prove that the large number of fundamental differences between the two can not be other than a prime number.
Solution: Since the large prime number, so they are seamless. We know the difference between two odd numbers is always an even number. The basic difference between the two numbers is the number of basic, but it will be an even prime number, which means it will be.
In addition to the provisions of this section text book problem, Polynomial, structure equations, geometry and number Advertisers may be a good idea. (1), (ii), (5) number of exercises can be done from.
Problems: A triangular three-arm length, respectively, a, b, c. If a2 + b2 + c2 = ab + bc + ca is to prove that the triangle is equilateral.
Solution: The problem is a technique that can be used!
a2 + b2 + c2 = ab + bc + ca
2a2 +2 b2 +2 c2-2ab-2bc-2ca = 0
(a2-2ab + b2) + (b2-2bc + c2) + (c2-2ca + a2) = 0
(a-b) 2 + (b-c) 2 + (c-a) 2 = 0
We know that some of the real family the sum of the number of 0, they are each independently 0.
As a result, (a-b) 2 = (b-c) 2 = (c-a) 2 = 0. From here, you can easily tell that a = b = c. The equilateral triangle.
Problems: p, 33 – a prime number greater than. How much will the remaining 12 percent by p2-out?
Solution: We’ll use the division algorithm. (You do not get anything, we learned that many of the dividend divisor = quotient + remainder , the Division Algorithm) 12 with the remainder may be divided into any number of 12. (0, …, 11), so we can write the number,
p = 12k +0, p = 12k +1, p = 12k +3, p = 12k +4, …, p = 12k +11. I noticed that it was 12k +1, 12k +5, 12k +7, 12k +11 exception of a number of other basic can not. [Because, 12k-'s multiple manufacturers have, p = 12k +2 = 6 (6k +1) - than at least two manufacturers have, check the rest of the way.]
Now, the values derived class caratike the screenshot we can get through that 12 per share for the remainder of 1.
By showing: p2 = (12k +11) 2 = 144k2 +2.12.11 k +121 = 12 (12k2 +22 k +10) +1
Secondary and Higher Secondary :
These two types of errors are almost the same. There Geometry, Statistics, Algebra, kambinetariksa the problem is. The hayara sekendarite Calculus and analysis from the two – there may be a problem. Geometry, the geometry of iuklidiyana more emphasis should be. 9 th -10 geometry class, as well as (7) and the number (6) number if it is reading the book. As well as (3), (4), (5) the number of problem can be resolved. Here are some of the problems in the National Mathematics Olympiad are:
problems: a currency back to 1, two different currencies at back, three different currencies back 3, … … … …, 49 different currencies and 50 different currencies back to back 49 to 50 are written. All currency in a black bag from the one for the currency. Choosing a minimum number of coins to ensure that at least 10 of any one type of coin has been taken?
Solution: It seems like a lot of feasibility problems, but it will get you to think a little differently. It may be our fate It’s too bad! We think our fate is worse than it is, but what can be. Since no one is writing mudragulora 10 mudrai (maximum nine), we assume that the first tulechi mudraguloi. I think we all kind of exchange of the currencies of the nine tulechi. Therefore, (1 +9 +9 +9 + in + … + … 9) = 45 +9.41 = 414 does not give us any kind of coin after coin drop of 10 no. So if you add another currency, that is to say, a total of 415 coins to recommend any one of 10 types of currency we’ll get.
Problem: ABCDE pancabhuje triangle ABC, BCD, CDE, DEA, EAB-’s the same area. BE-straight lines AC and AD, respectively, M and N as the point of intersection. Evidence that BM = EN
Solution: (This is a picture amki) BCD and CDE triangle of the same area from CD | | BE; same area of the triangle ABC BCD and BC | | AD, similar to CDE, DEA triangle from DE | | CA. So we can say that parellelogram BNDC and MCDE. The BN = CD = ME and BM = EM.
Problem: Suppose, a is a whole number. The m = 4a +3 and m, 11 -’s multiplier. a4 is much left to be divided by 11?
Solution: mod 11 of the pie,
4a +3 0mod11 4a -3 8mod11
a 2mod11 a4 16 5mod11
The remainder 5.
Problem: integers solutions Logs: x2 / 2 +5 / y = 7
Solution: Observe, x2 +10 / y = 14, the 10, y divisible by. It’s the value of 10 -’s gen (1, 2, 5, 10, -1, -2, -5, -10) putting it, y = 1, 2, -5 – x-value for the integer. Not all solutions are (x, y) = (1, 2), (1, -2), (2, 3), (2, -3), (-5, 4), (-5, -4) .
Problem: xy +9 (x + y) = 2006 equations have solutions that are integer number?
Solution: xy +9 (x + y) = 2006 xy +9 x +9 y +81 = (x +9) (y +9) = 2087
The 087 is a prime number, so (x +9) = 1, 2087, or -1, -2087. Thus, positive – negative overall total of 4 possible solutions.
There is no specific syllabus for the preparation of national Olympiad different category. The first text book Advertisers must have a good possession. The Olympiad questions are designed so that most problems can be resolved without text book are based on different concepts. However, any problems can be achieved. In some cases, new techniques will be useful if you know how to fix the problem. However, for the Secondary and Secondary hayara text book knowledge all the time is not enough. The kambinetariksa or number of issues in our school – college, as is taught. There have been numerous English books. It will be good if you want to build the reading habit.
How do you solve the problem, it is the only way to learn a lot of problems. The National Olympiad preparation will be good to you throughout the year. The last moments of preparation, as well as a list of books in the past to solve problems from a variety of practices and techniques can be learned. For good luck.
Some useful websites :
Bangladesh mathematics alimipayadera official website. There is a variety of information related to Math Olympiad and questions can be found in the past year.
www.matholympiad.org.bd / forum
This is the best place online to prepare for the Math Olympiad. Mathematics Olympiad hundreds of students throughout the year to discuss the issue here. The problem last year and the related discussions can be found here and a little khumjalei.
www.kmcbd.org / Home / documents
There are a variety of useful articles prastutibisayaka Mathematics Olympiad, problem sets, and notes can be found.
Book List :
1. Mathematics and further mathematics
It Zafar Iqbal and Zakari Swapan
The. Our math festival
3. The Art and Craft of Problem Solving
4. Problem Solving Strategies
5. 500 Mathematical Challenges
6. 104 Number Theory Problems
7. Geometry Revisited