Assam Academy Of Mathematics RMO Regional Mathematical Olympiad Sample Question Paper : aam.org.in

Name of the Organisation : Assam Academy Of Mathematics
Document Type : Model Question Paper
Exam : Regional Mathematical Olympiad
Subject : Mathematics

Website : http://www.aam.org.in/site/index.html#
Download Sample Question Papers: https://www.pdfquestion.in/uploads/13708-RMO.pdf

RMO Regional Mathematical Olympiad Sample Question Paper :

1. Let ABC be a triangle in which AB = AC and let I be its in-centre. Suppose BC = AB + AI. Find \BAC.
2. Show that there is no integer a such that a2 – 3a – 19 is divisible by 289.

Related : BIOM Brilliant International Olympiad of Mathematics Question Paper : www.pdfquestion.in/13668.html

3. Show that 32008 + 42009 can be written as product of two positive integers each of which is larger than 2009182.
4. Find the sum of all 3-digit natural numbers which contain at least one odd digit and at least one even digit.

5. A convex polygon Z is such that the distance between any two vertices of T does not exceed 1.
(i) Prove that the distance between any two points on the boundary of T does not exceed 1.
(ii) If X and Y are two distinct points inside T, prove that there exists a point Z on the boundary of Z such that XZ + Y Z <= 1.

6. In a book with page numbers from 1 to 100, some pages are torn off. The sum of the numbers on the remaining pages is 4949. How many pages are torn off?

Answer all the Questions :
1. Let ABC be a triangle in which AB = AC and let I be its in-centre. Suppose
BC = AB + AI. Find \BAC.
Solutions :
We observe that \AIB = 90  (C/2). Extend CA to D such that AD = AI. Then CD = CB by the hypothesis. Hence \CDB = \CBD = 90-(C/2). Thus \AIB + \ADB = 90 + (C/2) + 90 – (C/2) = 180. Hence ADBI is a cyclic quadrilateral. This implies that \ADI = \ABI =B/2 . But ADI is isosceles, since AD = AI. This gives \DAI = 180 – 2(\ADI) = 180 – B. Thus \CAI = B and this gives A = 2B. Since C = B, we obtain 4B = 180 and hence B = 45. We thus get A = 2B = 90.

2. We write a2 – 3a – 19 = a2 – 3a – 70 + 51 = (a – 10)(a + 7) + 51.
Solution :
Suppose 289 divides a2 -3a-19 for some integer a. Then 17 divides it and hence 17 divides (a – 10)(a + 7). Since 17 is a prime, it must divide (a – 10) or (a + 7).
But (a+7)-(a-10) = 17. Hence whenever 17 divides one of (a-10) and (a+7), it must divide the other also. Thus 172 = 289 divides (a – 10)(a + 7). It follows that 289 divides 51, which is impossible. Thus, there is no integer a for which 289 divides a2 – 3a – 19.

3. Find the sum of all 3-digit natural numbers which contain at least one odd digit
and at least one even digit.
Solution :
Let X denote the set of all 3-digit natural numbers; let O be those numbers in X having only odd digits; and E be those numbers in X having only even digits. Then X \ (O [ E) is the set of all 3-digit natural numbers having at least one odd digit and at least one even digit.The desired sum is therefore Consider the set O. Each number in O has its digits from the set {1, 3, 5, 7, 9}. Suppose the digit in unit’s place is 1. We can fill the digit in ten’s place in 5 ways and the digit in hundred’s place in 5 ways. Thus there are 25 numbers in the set O each of which has 1 in its unit’s place. Similarly, there are 25 numbers whose digit in unit’s place is 3; 25 having its digit in unit’s place as 5; 25 with 7 and 25 with 9. Thus the sum of the digits in unit’s place of all the numbers in O is 25(1 + 3 + 5 + 7 + 9) = 25 × 25 = 625

A similar argument shows that the sum of digits in ten’s place of all the numbers in O is 625 and that in hundred’s place is also 625. Thus the sum of all the numbers in O is
625(102 + 10 + 1) = 625 × 111 = 69375.

Consider the set E. The digits of numbers in E are from the set {0, 2, 4, 6, 8}, but the digit in hundred’s place is never 0. Suppose the digit in unit’s place is 0. There are 4 × 5 = 20 such numbers. Similarly, 20 numbers each having digits 2,4,6,8 in their unit’s place. Thus the sum of the digits in unit’s place of all the numbers in
E is 20(0 + 2 + 4 + 6 + 8) = 20 × 20 = 400.

A similar reasoning shows that the sum of the digits in ten’s place of all the numbers in E is 400, but the sum of the digits in hundred’s place of all the numbers in E is 25 × 20 = 500. Thus the sum of all the numbers in E is 500 × 102 + 400 × 10 + 400 = 54400.
The required sum is 494550 – 69375 – 54400 = 370775.

5. A convex polygon t is such that the distance between any two vertices of t does not exceed 1.
(i) Prove that the distance between any two points on the boundary of t does not exceed 1.
(ii) If X and Y are two distinct points inside t, prove that there exists a point Z on the boundary of t such that XZ + Y Z < 1.

Solution :
(i) Let S and T be two points on the boundary of T, with S lying on the side AB and T lying on the side PQ of T. (See Fig. 1.) Join TA, TB, TS. Now ST lies between TA and TB in triangle TAB. One of \AST and \BST is at least 90?, say AST < 90. Hence AT <TS. But AT lies inside triangle APQ and one of \ATP and \ATQ is at least 90, say \ATP 90?. Then AP < AT. Thus we get TS < AT > AP < 1.