Math IOQM

  Q1(geometry):  AB is a chord in a circle with center O and radius 52 cm. The point M divides the chord AB such that AM = 63 cm and MB = 33 cm. Find the length OM in cm. Solution: Let's say OD is the perpendicular on AB. OD^2 = 52^2 - 48^2 = 400 OM^2 = 15^2 + OD^2 = 625 => OM = 25 Q2(combinatorics,set-theory):  If the number of triplets of set {A, B, C} such that (a) A, B, C ∈ {1, 2, 3, 4, ..., 8} (b) |A ∩ B| = |B ∩ C| = |C ∩ A| = 2 (c) |A| = |B| = |C| = 4 is λ, then find the sum of digits of λ. (where |S| denotes the number of elements in set S) Solution: A can be constructed in 8C4 = 70 ways. B has to pick 2 numbers from A and 2 from A'(A's complement): 4C2*4C2 Let's say A is {1,2,3,4} B is {3,4,5,6} now. C has to pick 4 digits in 3 different ways: 1. It picks 3,4 from A (the same digits which B picked). Now it has to pick 7,8 from A' otherwise (b) given in question won't hold. So C = {3,4,7,8}.  2. It picks 1,2 from A (the digits which B didn't pick)....

Same as PRMO 2012 question . It just asks m + n - 2000.

 Q16 - In a quadrilateral ABCD, it is given that AB = AD = 13, BC = CD = 20, BD = 24. If r is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to r ? Solution  - PRMO 2018 question Answer: 8

Q13. Let ABCD be a trapezium in which AB||CD and AD perpendicular to AB. Suppose ABCD has an incircle which touches AB at Q and CD at P. Given that PC = 36 and QB = 49, Find PQ.  Solution: Let incircle touch BC at R CR = 36, BR = 49 Further let inradius = r So AQ = PD = r and AD = 2r Let perpendicular from C meet AB at S So BS = 13, BC = 85 (CS)^2 = 85^2 - 13^2 = 98*72 = 84^2 CS = 84 = PQ

Q10. Two circles with radii 4 and 9 respectively touch each other externally. Let c be the radius of a circle that touches these two circles as well as a common tangent to the two circles. Find 25c = ? Solution: Use  https://en.wikipedia.org/wiki/Descartes%27_theorem (k1 + k2 + k3 + k4)^2 = 2(k1^2 + k2^2 + k3^2 + k4^2) Here k4 = 0 for the straight line(0 curvature). k1 = 1/4, k2 = 1/9, k3 = 1/c It's a quadratic in k3. You will get k^2 - 26k/36 + 25/36^2 = 0. It has 2 solutions k = 1/36,25/36 => c = 36,36/25  Radius can't be 36 as seen in the diagram. So 25c = 36.

Q7. An infinite geometric progression has sum of 8. The sum of series by squaring each term is 6 times the sum of original series. If common ratio of original series is r. Find 1/r. Answer: 1/7

Q6. Find (q+r) if (x-2)^2 is a factor of x^5 - 5qx + 4r = 0. Solution: If you do the long division eventually you will get this remainder: (-5q+80) x + (4r - 128). This should be 0. If a polynomial is 0, each of its factors has to be 0. 5q = 80 => q = 16. 4r = 128 => r = 32. Answer: 48.

The Principle of Inclusion-Exclusion (PIE) is a counting technique used to find the size of the union of multiple overlapping sets. It corrects for overcounting by systematically adding and subtracting the sizes of intersections. 🧠 Principle of Inclusion-Exclusion (PIE) Let A 1 , A 2 , … , A n A_1, A_2, \dots, A_n be sets. The number of elements in their union is: ✅ PIE Formula: For 2 sets: ∣ A 1 ∪ A 2 ∣ = ∣ A 1 ∣ + ∣ A 2 ∣ − ∣ A 1 ∩ A 2 ∣ |A_1 \cup A_2| = |A_1| + |A_2| - |A_1 \cap A_2| For 3 sets: ∣ A 1 ∪ A 2 ∪ A 3 ∣ = ∣ A 1 ∣ + ∣ A 2 ∣ + ∣ A 3 ∣ − ∣ A 1 ∩ A 2 ∣ − ∣ A 1 ∩ A 3 ∣ − ∣ A 2 ∩ A 3 ∣ + ∣ A 1 ∩ A 2 ∩ A 3 ∣ |A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3| For n n sets: ∣ ⋃ i = 1 n A i ∣ = ∑ ∣ A i ∣ − ∑ ∣ A i ∩ A j ∣ + ∑ ∣ A i ∩ A j ∩ A k ∣ − ⋯ + ( − 1 ) n + 1 ∣ A 1 ∩ A 2 ∩ ⋯ ∩ A n ∣ \left|\bigcup_{i=1}^{n} A_i \right| = \sum |A_i| - \sum |A_i \cap A_j| + \sum |A_i \cap A_j \cap A_k| - \cd...

Q1. Natural solutions of  50 < X + y + z <= 100 x + y + z <= 97 x + y + z + w = 97 Total: 100C3 Subtract those when x + y + z <= 50 x + y + z + w = 47 50C3 Answer: 100C3 - 50C3 Q2. Natural solutions of xyz = 2000 Solution: Factors: 2*2^3*5^3 = 2^4*5^3 It's like distributing 4 apples and 3 oranges among 3 people. Stars and bars twice: 6C2*5C2 = 150 Q3.  Integer solutions of xyz = 3000: Solution: Factorize: 2^3.3^1.5^3 Stars and bars thrice: 5C2*3C2*5C2 = 300 This is when all of x,y,z are positive. Now we can make 2 of  (x,y,z) negative at a time: 3C2 So 4*300 = 1200 is the answer. Q4. How many ways to put 4 numbered-slips in 4 numbered-boxes so that at least one of them is in the correct box. Solution: 4C1.D3 + 4C2.D2 + 4C3.D1 + 4C4.D0  = 4*2 + 6*1 + 4*0 + 1 = 15 Or total - when none is correct = 4! - !4 = 24 - 9 = 15 Q5 . 6 letters in 6 envelopes. How may ways if: a) exactly 2 letters in correct envelopes. 6C2.D4 = 15*9 = 135 b) at least 4 letters go in...

Q1 . A question paper consists of two sections. Section A has 7 questions and section B has 8 questions. A student has to answer 10 questions out of these 15 questions. The number of ways in which he can answer if he must answer at least 3 of section A and at least 4 of the section B is: Solution: One wrong way to solve this: For the mandatory questions: 7C3*8C4 Now 3 more questions need to be answered from 4 of A and 4 of B. Cases: 3A,0B | 2A,1B | 1A,2B | 0A,3B 4C3*4C0 + 4C2*4C1 + 4C1*4C2 + 4C0*4C1 = 4 + 24 + 24 + 4 = 56 Finally: 7C3*8C4*56 = 35*70*56 = 137200 Why is it wrong? Let's say we chose A1,A2,A3 as mandatory from A and then chose A4,A5,A6 when we did 4C3*4C0 for (3A,0B). But we could have done it the other way round also, i.e. choose A4,A5,A6 as mandatory and then choose A1,A2,A3. So there is overcounting. Correct Solution : Simply create the cases for A,B: (3,7),(4,6),(5,5),(6,4) 7C3*8C7 + 7C4*8C6 + 7C5*8C5 + 7C6*8C4 = 2926. Q2 . A kindergarten teacher has 25 kids in her...

(n)C(k) = (n-1)C(k) + (n-1)C(k-1). When n,k are positive integers. It holds even when n < k since in that case nCk = 0. Let's see it using an example: 7C3 = 6C3 + 6C2 => 35 = 20 + 15 => 35 = 35 What is the intuition behind it? Let's say we need to select 3 kids out of 7(A,B,C,D,E,F,G). Now, there are 2 kind of groups: Case 1: 'A' is part of the group. Now we have to select 3 kids from remaining 6. 6C3. Case 2: 'A' is NOT part of the group. Now we select 5 kids from remaining 6. 6C2. So: 7C3 = 6C3 + 6C2

Q1 : How many ways to select 5 kids from 25 kids? Solution: Correct answer: 25C5 = 25!/5!20! But what if I do it like this: First select 1 kid from 25: 25C1 Then select 4 from remaining 24: 24C4 Final answer: 25C1*24C4 = 25!/4!20! What is wrong here? The problem is this: When you do 25C1, let's say you selected 'A'. Then when you select 4 kids from 24, you could choose 'B','C','D','E'. But you could also choose 'B' first and then 'A','C','D','E'. And both would give you same result. So you are overcounting by a factor of 5.

Q1 . How many positive integers 'n' are there such that 'n' is a divisor of one of the numbers 10^40, 20^30? Solution: Factorize 10^40 = 2^40.5^40 Number of divisors = 41.41 = 1681 20^30 = 2^60.5^30 Number of divisors = 61.31 = 1891 What is their gcd? 2^40.5^30 Number of divisors of gcd = 41*31 Total divisors of the given numbers  = 1681 + 1891 But the divisors of gcd would be counted twice so subtract it: 1681 + 1891 - 41*31 = 2301 Q2.  Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7. Before solving this, try solving this similar question: Q. How many 3 digit numbers like aba are divisible by 3? Solution: 100a+10b+a = 0 mod 3 101a + 10b = 0 2a + b = 0 -a + b = 0 => difference of a and b is divisible by 3. => remainder of a and b div 3 is same. a != 0 a = 1,4,7 b = 1,4,7 a = 2,5,8 b = 2,5,8 a = 3,6,9 b = 0,3,6,9 Total 9 + 9 + 12 = 30 Solution to original question: abccba = (1...

Let's say there are 'n' people: P1, P2 ... Pn. And there are 'n' chairs: C1,C2....Cn. And we have to seat the people so that none of them gets the chair with same number. For e.g. for 2 people, there is only 1 way: P2,P1 For 3 people there are 2 ways: P2,P3,P1 P3,P1,P2 And so on.. These are called DeArrangements. For n  = 1 it's D1, for n = 2 it's D2 and in general it's Dn. How to compute Dn? There is a recursive formula: Dn = (n-1)(Dn-1 + Dn-2), D0 = 1, D1 = 0 Using this let's compute: D2 = (2-1)(D0 + D1) = 1.1 = 1 D3 = (3-1)(D1 + D2) = 2.(0+1) = 2 D4 = 3.(2+1) = 9 D5 = 4.(9+2) = 44 ...... There is non recursive formula also: Dn = (n!){1 -1/1! + 1/2! - 1/3! ....(-1)^n.1/n!} D1 = 1!(1-1) = 0 D2 = 2!(0+1/2) = 1 D3 = 3!(1/2-1/6) = 2 .... How do we derive the recursive formula? Let's P1 receives a chair Ci where i != 1. Now the person Pi can receive a chair in 2 ways: 1. Pi receives C1, i.e. P1 and Pi exchange their chairs. Now the remaining (n-2) p...

Prerequisite: Stars and bars method  and Dearrangements . Q1 . In a bakery 8 different brands of biscuits are available. In how many ways 6 biscuits can be selected if biscuits of same brand are identical. Solution: Objects are identical, categories(buckets) are different. Use stars and bars method here. n = 6 k = 8 (n+k-1)C(k-1) = 13C7 = 1716 Q2 . Find the number of solutions of the equation x + y + z + w = 20 under the following restrictions: (i) x,y,z,w are whole numbers. Again stars and bars. n = 20 k = 4 (24-1)C(3) = 23C3 (ii) x,y,z,w are natural numbers. Each bucket would get at least one object. Remaining objects = 16 n = 16 k = 4 19C3 (iii) x and y are odd natural numbers while z & w are even natural numbers. x = 2a + 1, y = 2b + 1, z = 2c, w = 2d a,b >= 0 and c,d >= 1 2a + 1 + 2b + 1 + 2c + 2d = 20 => 2(a+b+c+d) = 18 => a+b+c+d = 9 Now we can apply stars and bars here but the problem is that c,d start from 1 while a,b start from 0. Let c' = c - 1, d' = ...