Math IOQM

Q1) Solve the cubic equation (9x^3 - 27x^2 + 26x - 8 = 0), given that one of the root of this equation is double the other. Q2) If the product of two roots of the equation (4x^4 - 24x^3 + 31x^2 + 6x - 8 = 0) is 1, find all the roots. Q3 Obtain a polynomial of lowest degree with integral coefficient, whose one of the zeros is sqrt{5} + sqrt{2}. Solution 1: Assume a,2a are the roots. Put the values in the given equation and equate them. You will get a cubic in 'a' with one root as 0. a can't be 0 since that would mean -8 = 0 Other roots will be 2/3,13/21 Putting a = 2/3 in the original equation works out nicely and 13/21 doesn't quite fit it. So roots will be 2/3,4/3,1 = answer. Solution 2: Solution 3: Simplest way to solve such question is to start with x = given root = sqrt(5) + sqrt(2) Square both sides: x^2 = 7 + 2sqrt(10) x^2 - 7 = 2.sqrt(10) Again square to eliminate the sqrt on the right side and now you will have a degree 4 polynomial with all integer coefficients...

Q1). Andy and Bethy are at same point. Andy leaves at 1:30 toward north at a steady 8 miles/hr speed. Bethy leaves at 2:30, toward east at a steady 12 miles/hr speed. At what time they will be exactly the same distance away from their starting point? Q2). A box contains 10 pounds of a nut mix i.e., 50% peanuts, 20% cashews, 30% almonds. A 2nd nut mix containing 20% peanuts, 40% cashews, 40% almonds is added to the box resulting in a new nut mix i.e., 40% peanuts. How many pounds of cashews are now in the box? Q3). How many isosceles triangles are there with positive area whose side lengths are all positive integers and whose longest side has length 2025? Q4) 1, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, ... What is the 2025th term? Q5) Suppose a and b are real numbers. When the polynomial (x^3 + x^2 + ax + b) is divided by (x - 1), the remainder is 4, and when divided by (x - 2), the remainder is 6. b - a = ? Q6) The sequence (1, x, y, ...

Proof: p = a.x q = b.x gcd(p,q) = k Using Bezout's identity: k = s.p + r.q = s.a.x + r.b.x = x(s.a + r.b) => x divides k H.P.

Prove that: If x divides y, then (a^x - 1) divides (a^y - 1) where a,x,y are positive integers and a > 1. Proof: We know that a^x - 1 can be written as (a - 1)(1 + a + a^2 ... a^(x-1)) Let y = kx Then a^y - 1 = a^(kx) - 1 = (a^x)^k - 1 Let u = a^x then a^y - 1 = u^k - 1 = (u - 1)(1 + u + u^2 ... u^(k-1)) = (a^x - 1)(...) H.P. The result above can be used to prove that: gcd(a^m - 1, a^n - 1) = a^(gcd(m,n)) - 1 Proof: Let gcd(m,n) = g Then using Bezout's identity, there exist x,y such that mx + ny = g => a^g = a^mx.a^ny____________[1] Also, since g divides m and n a^g - 1 divides a^m - 1 and a^n - 1 So a^g -1 is a common divisor. Now we need to prove that it is greatest common divisor. Let gcd(a^m - 1, a^n - 1) = D Then a^m - 1 mod D = 0 => a^m = 1 mod D and a^n = 1 mod D => (a^m)^x = 1 mod D and (a^n)^y = 1 mod D => a^g = 1 mod D using [1] => D divides (a^g -1) So D divides (a^g -1) which is a common divisor of (a^m - 1) and (a^n - 1) while D is their gcd. GCD can ...

Q. Let S be the sum of the base 10 logarithms of all the proper divisors of 1000000. What is the integer nearest to S? Solution: Before that prove that product of divisors of an integer N is P = N^(d(N)/2) where d(N) = number of divisors of N. Let there be k divisors of N which are 1 = d1, d2, ... dk = N. So k = d(N). If we multiply them pairwise d1.dk = N = d2.d(k-1) = dk.d1 Multiply all together. (d1.d2..dk)^2 = N^k = P^2 => P = N^(k/2) H.P. Now: S = log(d1) + log(d2)... log(dk) = log(d1.d2...dk) = log(N^(k/2)) = (k/2).log(N) = (1/2).k.6 = 3.k k = number of divisors = 7*7 So S = 49*3 = 147 But from this we need to subtract log of improper divisor(number itself). Answer = 147 - 6 = 141 Q. For k≥2, show each of the following: (a)  n=2^(k−1) satisfies the equation σ(n)=2n−1. (b) If 2^k −1 is prime, then  n=2^(k−1)(2^k−1) satisfies the equation σ(n)=2n. (c) If 2^k −3 is prime, then  n=2^(k−1)(2^k −3) satisfies σ(n)=2n+2. Solution: (a) Do the sum of the GP 1 + 2 + ... 2^(...

  Q. If point O is inside triangle ABC, then prove that AB + BC > AO + OC Solution: Extend AO to meet BC at D. Consider triangle ABD. AB + BD > AD = AO + OD In triangle ODC: OD + CD > OC Add both: AB + BD + OD + CD > AO + OD + OC Cancel OD AB + BD + DC > AO + OC =>  AB + BC > AO + OC H.P. Q. Prove that the length of median AM in triangle ABC is not greater than half the sum of sides AB and AC. Prove also that the sum of the lengths of the three medians is not greater than the triangle's perimeter. Solution: Extend the median AM to AD s.t. AM = MD ABDC is a ||gram. Why? Its diagonals BC and AD bisect each other. So AB = CD and AC = BD In triangle ABD, AB + BD > AD, similarly AC + CD > AD Add both and substitute to get AB + AC > 2AM. Do it for all the medians to get 2*perimeter > 2*(sum of medians) H.P. Q. A fly sits on one vertex of a wooden cube. What is the shortest path it can follow to the opposite vertex? Answer: sqrt(5) units. Solution: Fi...