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Competitions » MATHS OLYMPIAD (RMO, INMO, IMO) » Maths Olympiad Previous Papers

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Free Download RMO 2017 Solved Paper

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Free Download Pre-RMO 2016 Paper Chandigarh

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SOME SAMPLE QUESTIONS

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
         = (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

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