The erstwhile NBHM (National Board of Higher Mathematics) took the initiative to support the mathematical talent among school students in the country. Hence the foundation of the Mathematics Olympiad was laid in the year 1986 by NBHM. Exhibiting immense courage and commitment NBHM took over the responsibility for selecting and training the Indian team for participation in the International Maths olympiad every year. NBHM coordinates and support maths Olympiad all over the country with the intension to pick the best talent from every nook and corner of the country. It has been divided in 20 regions.
What are the benefits of Maths Olympiad ?
1. If a student happens to clear maths Olympiad, then it gives an added advantage to the candidate in all walks of life e.g. Admissions, jobs and Interviews.
2. Awards & Accolades:
At clearing RMO (Regional Mathematical Olympiad), students get certificate.
At clearing INMO (Indian National Mathematical Olympiad), students get certificate, invitation, expenses, loadging and boarding in camp at Homi Bhabha Centre, Mumbai. After being successful in INMO, students will be eligible for participating in IMO (International Mathematical Olympiad). Then students get International ticket stay from Ministry of HRD.
After clearing IMO (International Mathematical Olympiad), students receive gold, silver and bronze medals with a cash prize of Rs. 50000/-, Rs. 25000/- and Rs. 10000/- respectively.
What are the stages?
Stage-I: RMO (Regional Mathematical Olympiad), is the first step in selecting the Indian Team for IMO.
Eligibility: It is meant primarily for students of Class IX to XII. As special cases, exceptionally talented students from Class VIII may be allowed to appear at the discretion of the Regional Coordinator.
Day of Examination: Second Sunday of October.
Time duration: 3 hours, containing 6 subjective problems.
Stage-II: INMO (Indian National Mathematical Olympiad)
Eligibility: Only students selected on the basis of RMO(Regional Mathematical Olympiad) from different regions are eligible to appear for INMO (Indian National Mathematical Olympiad) . About 30 students from each region are selected in the order of their merit. Approximately 900 students qualify for INMO.
Day of Examination: Third Sunday of January. It takes place at 28 centres across the country.
Time duration: 4 hours.
Stage-III: IMOTC (International Mathematical Olympiad Training Camp)
Eligibility: The top 35-50 performers in INMO receive a certificate of merit. They are invited to a month long training camp (junior batch) conducted in May-June. They undergo Lectures and Problem Solving sessions. In addition, INMO awardees of the previous year that have satisfactorily gone through postal tuition throughout the year are invited again for a second round of training (senior batch).
Stage-IV: IMO (International Mathematical Olympiad):
Eligibility: The team selected (consisting of 6 students) at the end of the camp, a "leader" and a "deputy leader," represent India at the IMO in a different member country of IMO (International Mathematical Olympiad) each year.
Month of Examination: July.
IMO consists of two written tests held on two days with a gap of at least one day.
Time duration: Each test is of 4½ hours duration.
How to apply?
There is no application form. Apply on plain paper giving
Name (In block letters): _______________________________________________
Mailing Address with Phone number: _____________________________________
& e-mail address (if any),: _____________________________________________
Name of the school:___________________________________________________
Postal Address of the School:___________________________________________
Note: RMO awardees of any earlier year do not need any recommendation.
But other students should attach a recommendation letter from the school authorities (Principal/Math teacher) to the following effect:
For students of Class VIII: The candidate is one of the best in his school has seen in last five years.
For students of higher classes: The candidate is among the top 10% of his/her class.
The fee for individual applicants is Rs. 100/-, which includes postage. The amount should be sent as a DD drawn on any nationalized bank in favour of your regional coordinator. Individual applications should be accompanied by a self-addressed unstamped 12" × 6" envelope.
Suggestion: It is preferable that all applicants from the same school send their applications together through the school, with a single DD. Please attach a typed list of all applicants for speedier processing. In this case, the fee per candidate is Rs. 85/-.
Applications should be addressed to the respective regional coordinator. The list of regional coordinators can be downloaded from official website i.e. www.hbcse.tifr.res.in
Olympiad syllabus and methodology for preparation:
1) Geometry: 33%
2) Number Theory: 16%
3) Algebra: 17%
4) Combinatorics : 17%
5) Functional equations: 17%
Student should go through the complete NCERT and NCERT Exemplar books of 6th, 7th, 8th, 9th,10th classes for geometry and collect all the result at one place, then apply them on some good and quality problems. After that you must try unsolved problem banks and sources for them are listed below: Don’t try to jump the tougher problems by saying that you are preparing for Olympiad, instead increase the level of problem gradually. For complete details of study program, Kindly click on the following link.
Following are the best referred books for the preparation of Maths Olympiad.
1. An Excursion in Mathematics
Editors: M. R. Modak, S.A. Katre and V.V. Acharya (Bhaskaracharya Pratishthana, Pune, 2008).
2. Problem Primer for the Olympiads
C.R. Pranesachar, B.J. Venkatachala and C.S. Yogananda (Prism Books Pvt. Ltd., Bangalore, 2008)
3. Challenge and Thrill of Pre-College mathematics
V. Krishnamurthy, C.R. Pranesachar, K.N. Ranganathan and B.J. Venkatachala (New Age International Publishers, New Delhi - 2007).
4. Problem Solving Strategies
Arthur Engel (Springer-Verlag, Germany, 1999).
5. Functional Equations
B.J. Venkatachala (Prism Books Pvt. Ltd., Bangalore, 2008).
6. Mathematical Circles: Russian Experience
Fomin and others (University Press, Hyderabad, 2008).
Sample questions in the "Study Material"
Ques. Five points on a circle are numbered 1, 2, 3, 4, 5 in clockwise order. A frog jumps in the clockwise order from one number to another as follows:
(a) If it is at odd number it moves one place.
(b) If it is at even number it moves two places.
If the frog is initially at 5, where it will be after 2010 jumps? [RMO-2010 Chandigarh]
Ques. A number of bacteria are placed in a glass. One second later each bacterium divides in two, the next second each of the resulting bacteria divides in two again, etc. After one minute the glass is full. When was the glass half full?
[If your ans is 30 sec, then its wrong]
Ques. Jack tore out several successive pages from a book. The number of the first page he tore out was 183, and it is known that the number of the last page is written with the same digits in some order. How many pages did jack tear out of the book?
[Ques is absolutely right]
Ques. A caterpillar crawls up a pole 75 inches high, starting from the ground. Each day it crawls up 5 inches, and each night it slides down 4 inches. When will it first reach the top of the pole?
[75 days is wrong ans]
Divisibility and Congruence
Case 1: What is the remainder when each of the numbers 1, 7, 16, 91, is divided by 3?
The answer is obviously 1.
Case 2: What is the remainder when the number 3, 27, 83, 131 are divided by 8?
The answer to this question is 3.
Let us define the congruence of numbers.
Two numbers a and b are congruent with respect to m if both a and b leave same remainder when divided by m.
Mathematically, a ≡ b(mod m)
Important point in the congruence of numbers is that in problems involving divisibility of numbers, a number may always be replaced by another number congruent to the first.
For example, 91 ≡ 1 (mod 3). Therefore 91 may be replaced by 1.
Let us learn to use this concept.
Problem 1: Prove that 220 – 1 is divisible by 41.
Sol: We know that 220 – 1 = (25)4 – 1 = (32)4 – 1
Let us check number congruent to 32 with respect 41.
-9, 32, 73, 114, 155, 196, etc. are all congruent to 32.
There we may replace 32 by any of these congruent numbers.
Since -9 is smallest (numerically), therefore replacing 32 by -9, we get
(–9)4 – 1 = (81)2 – 1
Now, numbers congruent to 81 with respect to 41 are -1, 40, 81, 122, 163 etc.
Replacing 81 by -1, we get
(-1)2 – 1 = 0
Thus 220 – 1 leaves remainder zero when divided by 41
Hence 220 – 1 is divisible by 41
Problem 2: Prove that (53)103 (103)53 is divisible by 39.
Sol: 53 is congruent to 14 and 103 is congruent to -14 with respect to 39, therefore replacing 53 and 103 by 14 and -14 respectively, we get
= (14)103 – (14)53 = 1453 (1450 - 1)
Since 1453 and 39 are co-prime, therefore 1453 cannot be multiple of 39. So let us consider 1450 – 1 only
1450- 1 = (142)25 – 1 = (196)25 – 1
196 is congruent to 1 with respect to 39, therefore replacing 196 by 1, we get
(1)25 – 1 = 0 (remainder when 1450 – 1 is divided by 39)
Thus 1450 – 1 is divisible by 39.
Hence (53)103 (103)53 is divisible by 39
Problem 3: Prove that 111333 333111 is divisible by 7.
Sol: Here 111 ≡ - 1 (mod 7) [Since 111 is congruent to -1 w.r.t. 7]
and 333 ≡ - 3 (mod 7) [Since 333 is congruent to -1 w.r.t. 7]
Therefore, replacing 111 and 333 by -1 and -3 respectively, we get (-1)333 (–3)111
= (-1) – (33)37
= - 1 – (27)37
Now, 27 ≡ –1 (mod 7),
Replacing 27 by -1,
we get -1 – (-1)37 = 0
Thus 111333 333111 leaves remainder 0 when divided by 7
Hence it is divisible by 7.
Problem 4: Find the remainder when 44444444 is divided by 9.
Sol: 444 ≡ - 2 (mod 9)
Replacing 4444 by -2,
we get (-2)4444 = 24444
= 21 . 24443
= 2. (23)1481
= 2. 81481
Now, 8 ≡ – 1 (mod 9)
Replacing 8 by -1,
we get 2. (-1)1481 = - 2 [Since -2 ≡ 7(mod 9)]
Therefore remainder is 7 when 44444444 is divided by 9.
Problem 5: Find the missing digit x in
(51840). (273581) = 1418243 x 040
Sol: Here 51840 are divisible by 9 and 273581 is divisible by 11, therefore 14182 x 34040 must be divisible by both 9 and 11.
Since 14182 x 34040 is divisible by 9
Hence sum of digits 27 x must be divisible by 9.
i.e. x is either 9 or 0.
Again, Since 1418243 x 040 is divisible by 11
Hence the difference of sum of odd digit and that of even digits is divisible by 11.
(4 8 4 x 4) – (1 1 2 3)
= (20 x) – 7
= 13 x
(13 x) is divisible by 11, therefore x must be 9.