Math IOQM

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. Show that there are no positive integers n satisfying σ(n)=10. Hint: Note that for n>1 σ(n)>n. Solution: Extending the hint, sigma(n) >= n + 1 (since 1,n are divisors of n). Now we can check for n = 1 to 9 and show that none of them equals 10. Another algebraic way is this: sigma(n) = (1+p+p^2...p^a)(1+q+q^2..q^b)... If n = p^a.q^b... where p,q... are prime. For it to be 10 each of the polynomials should divide 10 which has 1,2,5,10 as divisors. Since each of them is more than 1 and smallest prime is 2 => each factor is >= 3 So we can't build 10 with 2 as one factor and 5 as another. There will have to be a single polynomial adding upto 10. And n = p^a, i.e. only one prime factor. Let's try with a = 1 n = 1 + p = 10 => p = 9.No. a = 2 1 + p + p^2 = 10 => p(p+1) = 9. No. should be even. a = 3 1 + p + p^2 + p3 = 10 not possible since even with smallest prime 2, it exceeds 10. Q. Show that sigma(a.b) = sigma(a).sigma(b) where sigma is sum of divisors and a,...

  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...

Q1. The two sides of a triangle are 2 cm and 7 cm. Find the number of possible lengths of the third side. (The length of three sides of the triangle is an integer value) Solution: a + 2 > 7 a + 7 > 2 7 + 2 > a => a < 9 a > -5 a > 5 => 5 < a < 9 => a = 6,7,8 Answer: 3 Q2. Only using triangle inequality The three sides of a triangle are 9 cm, 7 cm and 12 cm. Which of the following can be a median of the triangle? 12,14,15,16 Solution: If the median is falling upon side length 12,then the triangles are: 9,6,x(median length) 7,6,x Only value of x satisfying both triangle inequalities here is: 12(from the given options). Similarly, If the median is falling upon side length 9,then the triangles are: 12,4.5,y(median length) 7,4.5,y From the given options, none satisfies both. Similarly, If the median is falling upon side length 7,then the triangles are: 12,3.5,z(median length) 9,3.5,z From the given options, 12 satisfies both. So answer is 12. Though it's ...