**The foundation of the Maths 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 olympaid every year. NBHM coordinates and support maths Olympaid 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.**

**Many Students from the country participate in this olympiad which itself is one of the best and toughest exam in the country. The selected talents from the country ultimately competes on the global international level.**

**Following are the different stages of the exam:**

**Stage-I: RMO (Regional Mathematics Olympiad), is the first step in selecting the Indian Team for IMO. **

**Stage-II:**INMO (Indian National Mathematics Olympiad)

**Stage-III:**IMOTC (International Mathematics Olympiad Training Camp)

**Stage-IV:**IMO (International Mathematics Olympiad)

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**Pioneer mathematics suppports you in the preperation of Math Olympiad by providing the study material, online tests, I.Q.sharpeners, etc. **

**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]**

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**Divisibility and Congruence**

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**Case 1: What is the remainder when each of the numbers 1, 7, 16, 91, is divided by 3? The answer is obviously 1.**

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**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.

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.

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**For example, 91 ≡ 1 (mod 3). Therefore 91 may be replaced by 1. Let us learn to use this concept.**

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**Problem 1: Prove that 2 ^{20} – 1 is divisible by 41.**

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**Sol: We know that 2 ^{20 }– 1 =_{ }(2^{5})^{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 2^{20} – 1 leaves remainder zero when divided by 41

Hence 2^{20} – 1 is divisible by 41

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**Problem 2: Prove that (53) ^{103} (103)^{53} is divisible by 39.**

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**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} = (14)^{103} – (14)^{53} = 14^{53} (14^{50} - 1) Since 14^{53} and 39 are co-prime, therefore 14^{53} cannot be multiple of 39. So let us consider 14^{50} – 1 only 14^{50}- 1 = (14^{2})^{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 14^{50} – 1 is divided by 39) Thus 14^{50} – 1 is divisible by 39. Hence (53)^{103} (103)^{53} is divisible by 39**

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