Math IOQM

Q1. A DNA chain is composed of basic building blocks in the form of four chemicals, known by the symbols A,C,T,G. Consider three-letter chains consisting of these symbols (with or without repetitions). How many such chains are there? S1. 4*4*4 = 64 Q2. Consider the set {a,b,c,d,e}. How many three-letter words can be made out of them, with or without meaning? How many of these will have at least one vowel in them? Answer these questions both for the case when repetitions of letters are allowed and for the case when repetitions of letters are not allowed. S2. 2.1 With repetition: 5*5*5 = 125 Without: 5*4*3 = 60 2.2 With repetition: Total words: 5^3 = 125 Without any vowels = 3^3 = 27 Answer = 125 - 27 = 98 Without: Total: 5*4*3 = 60 Without vowels = 3*2*1 Answer = 60 - 6 = 54 Q3. A permutation (a1,a2,a3,a4,a5) of (1,2,3,4,5) is called heavy-tailed if a1 + a2 < a4 + a5. What is the number of heavy-tailed permutations? S3. Total such: 5! = 120 Let's fix 1,2 to a1,a2 => 2!*3! = 12...

1. Find the total number of words formed by the letters a,a,b,b,c. S1. 5!/2!.2! 2. In how many ways can you arrange 2 beads of red colour, 2 beads of blue colour and 3 beads of black colour in a ring? Note that beads with the same colour are identical and arrangements that can be obtained by rotating (or) flipping the ring are identical. S2. First attempt(wrong): (7-1)! circular permutations Divide by 2!2!3! to account for repeated elements. Divide by 2 to account for flipping. Answer = 15 This is wrong, correct answer is 18. Why? Let's see using a smaller example. 2 blue, 3 black. (5-1)! circular permutations Divide by 2!3!. So far so good. Mistake is when we divide by 2 for flipping. Why do we even divide by 2 for flipping. Let's say we have 4 distinct beads. Here is one necklace: 1 --- 2 |       | 4 -- 3 and flipping it we get: 2 --- 1 |       | 3 -- 4 So 1-2-3-4 was clockwise and it became counter clockwise. Flipping gives a distinct perm...

1. 5 chairs, 4 people? How many ways can they sit? 2. 10 mangoes, 12 apples, 16 guava, 15 bananas. How many ways to eat 3 fruits? Case 1: Every single fruit is distinct. We have 53 fruits in total. So 53C3. That is if the order of eating doesn't matter. If the order of eating matters, then: 53.52.51 Case 2: Fruits of one type are all same.  Case 2.1 - Order of eating doesn't matter. Then this is simply stars and bars. (n+k-1)C(n) where 'n' is the number of items and 'k' is the number of buckets. Here,  n = 3 k = 4 So: All 3 fruits are same: 4C1 = 4 2 Same 1 distinct: 4C2 * 2C1 = 12 All 3 distinct: 4C3 = 4 Total: 20 Case 2.2 - Order of eating does matter. 4 * 4 * 4 = 64 Case 3: One fruit of each type, i.e. I eat 1 mango, 1 apple, 1 banana for e.g. Case 3.1 Fruits of same type are also distinct We have to consider 4 different cases(4C3): Mango, banana, apple = 10*15*12 .... If order of eating matters, multiply them by 6.(3!) Case 3.2 They are not distinct. 4C3 if ...

  Problem 1. Let (a) be a positive real number. Prove that one of the positive real numbers [ a,;2a,;3a,...;(n-1)a ] is at most (1/n) distant from a positive integer. S1. Let's test for n = 2 The list is simply: 'a'. And any real positive number is at most 1/2 away from a positive integer. So it works for n = 2. Let 1 <= k <= n-1. We want to show that at least one k.a is at most 1/n away from a positive integer. Let {k.a} be the fractional part of k.a. Distance of {k.a} from a positive integer is min( {ka}, 1 - {ka}). To understand, see the number line: 0______{ka}_______1 Its distance from 0 is {ka} and 1 - {ka} from 1. Let's prove by contradiction. Let's say for each 'k', 1 <= k <= n-1, {k.a}is more than 1/n away from the nearest integer. Like this: 0___1/n___{ka}___1 - 1/n____1 So {ka} is always between 1/n and (n-1)/n. Distance between 1/n and (n-1)/n is (n-2)/n. So there are (n-2) gaps each of length 1/n. Using pigeonhole argument, fraction...

1) P(x) = x^4- 4x^3 + 6x^2 - 4x - 2005. Find the integer part of the product of all non-real roots of (P). S1) P(x) = 0 => (x-1)^4 = 2006 => (x-1) = K.(2006)^(1/4) where K = -+1, -+i So non real roots are x = 1 +-i.(2006)^(1/4) Multiply: 1 + 2006^(1/2) 45^2 = 2025 45^2 - 44^2 = 89 So 2006^(1/2) = 44.something Answer: 45 2) First we will prove this: In the incircle trisects the median then the sides are in the ratio 5:10:13 where 10 maps to the base where median is landing. Once we do that, since BC = 20, other sides will become 10 and 26. Then apply Heron's formula. s = 28 s - a = 28 - 20 = 8 s - b = 28 - 10 = 18 s - c = 28 - 26 = 2 [ABC] = sqrt(28.18.8.2) = sqrt(64.7.2.9) = 24.sqrt(14) m + n =38 Ok coming to the proof of the first part. Using power of the point theorem: If DF is the tangent from D, then DF^2 = DY.DX = d/3 * 2d/3 = 2d^2/9 Similarly Tangent from A = AE then AE^2 = AX.AY = DF^2 => AE^2 = DF^2 But AE = s - a from incircle properties. And DF = (s-c) - a/2 = a+...

  1) Find the largest natural number 'm' such that:  (m^2) is the difference of two consecutive perfect cubes. AND  (2m + 79) is a perfect square. S1) m^2 = (n+1)^3 - n^3 = 3n^2 + 3n + 1 Since m,n are both natural numbers, RHS will grow much faster than LHS and at some point it will irreversibly exceed it. m^2 - 1 = 3n(n+1) So RHS is a multiple of 6 => m^2 = 1 mod 6 m^2 = 25,49 satisfy this but 24 = 3n(n+1) => 8 = n(n+1) no go. 48 = 3n(n+1) => 16 = n(n+1) no go. what next? So let's try some other approach. Once you complete the square in 'n', you will get: 4m^2 = 3(2n+1)^2 - 1 (2m-1)(2m+1) = 3(2n+1)^2 LHS has 2 consecutive odd numbers which are co prime. Why? gcd(2m-1,2m+1) = gcd(2,2m-1) = 1 So RHS should split cleanly into 2 factors which don't share any prime factors. Case 1: 2m-1 = a^2 2m + 1 = 3b^2 Case 2: 2m + 1 = a^2 2m - 1 = 3b^2 Case 2: 2 = a^2 - 3b^2 => a^2 = 3b^2 + 2 => a^2 mod 3 = 2 which is not possible since it can be only 0,1 Case 1...