Math IOQM

  Now, BMP ~ BAE. Why? Since Angle B is same in both. BM/BA = BP/BE = 1/2. So SAS similarity criteria. So: MP/AE = 1/2 => x + z = 2MP => x = MP. But MP = AB/2 = (x+y)/2 => 2x = x + y => x = y. So ABE is equilateral and hence all angles are 60. So DAE = 60 as well. Q4. Solution: After applying the transformation 4 times you will get: 81M + 80 = 16N + 405 => 81(M-5) = 16(N-5) Minimum values of M,N when LHS  = RHS, M = 5, N = 5 and both sides become 0. Next equal values will be when M-5 = 16 and N-5 = 81 since there are no common factors in 81 and 16. So M + N = 10 Q5. Let a , b ∈ Z > 0 a,b\in\mathbb Z_{>0} with a 3 − b 3 − a b = 25 > 0    ⇒    a > b . a^3-b^3-ab=25>0\;\Rightarrow\; a>b. Put d = a − b ≥ 1 d=a-b\ge1 . Then a 3 − b 3 − a b = ( a − b ) ( a 2 + a b + b 2 ) − a b = d ( 3 b 2 + 3 b d + d 2 ) − b ( b + d ) = d 3 + ( 3 d − 1 ) a b . a^3-b^3-ab=(a-b)(a^2+ab+b^2)-ab = d(3b^2+3bd+d^2)-b(b+d) = d^3+(3d-1)ab. So d 3 + ( 3 d − 1 ) a b = 2...

Q1. Let n be a positive integer such that 1 ≤ n ≤ 1000 . Let M n be the number of integers in the set X n = { 4 n + 1 , 4 n + 2 , … , 4 n + 1000 } . Let a = max ⁡ { M n : 1 ≤ n ≤ 1000 } , and b = min ⁡ { M n : 1 ≤ n ≤ 1000 } . a = \max\{M_n : 1 \leq n \leq 1000\}, \quad \text{and} \quad b = \min\{M_n : 1 \leq n \leq 1000\}. Find a − b a - b . Solution: Quick tip: If n^2 = k then (n+1)^2 = k + (2n + 1) We will use this here. Also as the numbers grow larger the gap between 2 perfect squares becomes less and less. For. e.g. there are 10 perfect squares between 1 and 100 but only 4 perfect squares between 101 and 200. So X1 = {sqrt(5) ... sqrt(1004)} will have the most number of perfect squares. While X1000 = {sqrt(4001)...sqrt(5000)} will have the least. In X1 the first integer square root is 3 and we know that 1024 is the square of 32 so the last integer will be 31. Total: 31 - 3 + 1 = 29 integers in X1. In X1000, We know that 64^2 = 4096 so 64 is the first integer. 70^2 = 4900 and ...

 Q1. Define n a ! n_a! for n n and a a positive to be n a ! = n ( n − a ) ( n − 2 a ) ( n − 3 a ) … ( n − k a ) n_a! = n(n - a)(n - 2a)(n - 3a) \ldots (n - ka) where k k is the greatest integer for which n > k a n > ka . If the quotient 72 8 ! / 18 2 ! 72_8! / 18_2! is equal to 2 x 2^x , find x x . Solution: Given number equals: 72.(72-8).(72-16)...(72-64)/18.(18-2).(18-4)...(18-16) = 4^9 = 2^18 Q2.  If  p , q p, q are two real numbers satisfying the relationships 2 p 2 − 3 p − 1 = 0 2p^2 - 3p - 1 = 0 and q 2 + 3 q − 2 = 0 q^2 + 3q - 2 = 0 and p q ≠ 1 pq \ne 1 , find the value of p q + p + 1 q \frac{pq + p + 1}{q} . Solution: Clearly p, 1/q are roots of 2x^2 - 3x - 1 = 0. p + 1/q = 3/2 p/q = -1/2 (pq + p + 1)/q = p + p/q + 1/q = 1 = Answer Q3. Convex quadrilateral ABCD has A B = 9 AB = 9 and C D = 12 CD = 12 . Diagonals AC and BD intersect at E, A C = 14 AC = 14 , and △ A E D \triangle AED and △ B E C \triangle BEC have equal areas. What is AE? Solutio...