Math IOQM

Q1. The age of a person in 2025 is a perfect square. His age was also a perfect square in 2012. His age will be a perfect cube (m) years after 2025. Determine the smallest value of (m). S1. Age in 2012 = p^2 p^2 + 13 = q^2 => 13 = q^2 - p^2 = (q-p)(q+p) => q - p = 1, q + p = 13 => q = 7, p = 6 => Age in 2012 = 36 => 49 + m = r^3 Smallest m  = 15 so that r^3 = 4^3 = 64 Answer = 15 Q2. The sum of two real numbers is a positive integer (n) and the sum of their squares is (n + 1012). Find the maximum possible value of (n). S2. a + b = n a^2 + b^2 = n + 1012 Root mean square inequality tells us that sqrt[(x1^2 + x2^2 ...xn^2)/n] >= (x1 + x2 ... xn)/n => sqrt[(a^2 + b^2)/2] >= (a+b)/2 => (n+1012)/2 >= n^2/4 => n^2 - 2n - 2024 <= 0 => n^2 - 2n - 1 <= 2025 => (n-1)^2 <= 2025 => |n-1| <= 45 => -45 <= n-1 <= 45 => -44 <= n <= 46 Max(n) = 46 Does it satisfy original equations? a + b = 46 (a + b)^2 - 2ab = 46 + 1012 = 46...

Q1. Let N=1×2×3×⋯⋯×n, where n is a positive integer. Find the largest possible value of n for which N is not divisible by some two-digit number. S1: First thought - for n>=5 10 will divide each factorial so what is the question even asking? But reading it carefully, we note that it's asking something else. Find the largest n! which is not divisible by at least one 2-digit number. 4! is not divisible by any(hence at least one) 2 digit number but it's not the largest. Largest such number could possibly be 99! . But it is divisible by all 2-digit numbers including 99. 98! is divisible by all except 99 but 11*9 = 99 so not correct. 97! is also divisible by 99 and 98(49*2). 96! is not divisible by 97 which is a prime. Hence 96!.   Q2. Real numbers a and b satisfy a+b=7 and ab=−3. What is the value of  a^3/b^2 + a +b^3/a^2 + b? S2: a^3/b^2 + a +b^3/a^2 + b = (a + b) + (a^5 + b^5)/(a^2b^2) We know the values of ab, a+b so we just need to focus on a^5 + b^5 which we can pro...

P1) If x1 and x2 are non-zero roots of ax^2 + bx + c = 0 and  -ax^2 + bx + c = 0 , then prove that a.x^2/2 + bx + c = 0  has a root between x1 and x2. S1) Plug x1 and x2 in first and second eqns. Then let P(x) = a.x^2/2 + bx + c = 0 P(x1) = a.x^2/2 P(x2) = -3/2.a.x^2 i.e. they have opposite signs => it has a root between x1 and x2. P2) Let P(x) = x^2 + ax + b be a quadratic polynomial in which a and b are integers. Show that there is an integer M such that P(n).P(n+1) = P(M) )  for any integer ( n ). S2) Let P(x) = (x - p)(x - q) where p,q are the roots and p + q = -a and pq = b. P(n) = (n - p).(n - q) P(n + 1) = (n + 1 - p).(n + 1 - q) Essentially we need to show that  P(n).P(n+1) = (M - p).(M - q) for some integer M. P(n).P(n + 1) = (n - p).(n - q). (n + 1 - p).(n + 1 - q) Now combine (n - p).(n + 1 - q).(n - q). (n + 1 - p) We want to combine like above so that we can use p + q = a and pq = b to get to P(M) form.  This will give us: M = n^2 + n(...

Q1. ABCD is a trapezium with side BC ∥ AD. If E is the midpoint of AB and the line through E parallel to DC meets AD and BC at X and Y respectively, prove that ABCD and XYCD have equal areas. Q2. ABCD is a parallelogram and O is any point. The parallelograms OAEB, OBFC, OCGD, ODHA are completed. Show that EFGH is a parallelogram. Q3. Let D be the midpoint of the side  BC of  triangle ABC. Prove that if AD > BD then angle A is acute; else if AD < BD, then angle A  is obtuse. S3: Since BD = CD, AD > BD => AD > CD Let angle BAD = x and angle DAC = y and hence angle A = x + y. In triangle ABD, AD > BD => Angle B > x In triangle ACD, AD > CD => Angle C > y B + C > x + y = A B + C > A A + B + C= 180 => A < 90 hence proved. Similarly the other case can be proved. S1: If I draw ABCD s.t. AD < BC then ABCD and XYCD have an overlapping pentagon between them which is EYCDA. What is extra is this: ABCD has triangle EBY XYCD has triangl...

Q1. "In a room of n people (where n >= 2), prove that there are always at least two people who have shaken hands with the exact same number of people." Q2. "In a tournament with n players (where n >=2 ), prove that at any given moment during the tournament, there are always at least two players who have completed the exact same number of games." Q3. Twenty pairwise distinct positive integers are all < 70. Prove that among their pairwise differences there are four equal numbers. Q4. Fifty-one small insects are placed inside a square of side 1. Prove that at any moment there are at least three insects which can be covered by a single disk of radius 1/7. Solution1: Possible number of handshakes: 0,1,2 ... n-1 But 0 and n-1 are not possible together, one of them has to be removed. So now there are n people and (n-1) possible values. By pigeonhole principle, at least 2 people need to have same number of handshakes. Solution 2: Same as above. Solution 3: 1,2...68...

Q1. A worker suffers a 20% cut in wages. He regains his original pay by obtaining a rise of: Q2. If m men can do a job in d days, then the number of days in which m + r men can do the job is: Q3. A boy walks from his home to school at 6 km per hour (kmph). He walks back at 2 kmph. His average speed, in kmph, is. Q4. A car travels from P to Q at 30 kilometres per hour (kmph) and returns from Q to P at 40 kmph by the same route. Its average speed, in kmph, is nearest to Q5. 5. A man invests Rs. 10,000 for a year. Of this Rs. 4,000 is invested at the interest rate of 5% per year, Rs. 3,500 at 4% per year and the rest at α% per year. His total interest for the year is Rs. 500. Then α equals Q6. Let (x_1, x_2, \ldots, x_{100}) be positive integers such that (x_i + x_{i+1} = k) for all (i), where (k) is a constant. If (x_{10} = 1), then the value of (x_1) is 7. If (a0 = 1), (a1 = 1) and (an = a_{n-1}a_{n-2} + 1) for (n > 1), then a465 and a466 are odd or even? 15.  A1. 25% A2. md/(m+r...