Math IOQM

Prove that every n s.t. gcd(n,10) = 1 has a multiple which is made up of only 9's Proof: Let us pick one 'n'. When we divide 10^1, 10^2, 10^3 .. with 'n' there can be at max 'n' different remainders. So there must exist integers a,b with a>b s.t. 10^a and 10^b have same remainders when divided by 'n'. So 10^a - 10^b is divisible by 'n'. 10^b(10^(a-b) - 1) is div by n. gcd(n,10) = 1 => 10^(a-b) - 1 is div by 'n' and this is entirely made up of 9s. H.P.

 Q52 (Intro): Determine all positive integers n such that n has a multiple whose digits are nonzero. Q52 (Advanced)

1) ABCD is a parallelogram. Through C, a straight line RQ is drawn outside the figure and AP, BQ, DR are drawn ⟂ to RQ. Show that: DR + BQ = AP . S1) [RMO] Key idea: DR || BQ so they can form a trapezium. Since LHS is DR + BQ which also appears in Trapezium midline theorem we will try to use Trapezium midline theorem which says this: If AB || CD in trapezium ABCD, then line EF joining midpoints of AD and BC is || to AB,CD and also EF = (AB + CD)/2. Let O be intersection of diagonals AC and BD in the ||gram ABCD. O is their midpoint. Let OM be perpendicular to RQ. DR || BQ and DBQR is a trapezium. OM || DR || BQ and OM goes through midpoint of BD => M is the midpoint of RQ. OM = (BQ + DR)/2 We are almost there now. If we could show that OM = AP/2, we are done. OM || AP In Triangle APC, it's a strong hint to use Triangle midpoint theorem. O is midpoint of AC and OM || AP => M is midpoint of PC. => OM = AP/2 H.P. Another solution: Drop a perpendicular DO from D to AP. DRPO is...

1. Show that 10^(2n-1) + 1 is divisible by 11 for all natural number 'n'. Solution: 10 = -1 mod 11 10^(2n-1) = (-1)^(2n-1) mod 11 = -1 mod 11 => 10^(2n-1) + 1 = 0 mod 11 H.P. 2. Show that 11^(n+2) + 12^(2n+1) is divisible by 133 for all natural 'n'. Solution: 133 = 7*19 So LHS should be div by both 7 and 19 Try 7 first: 11 mod 7 = 4, 12 mod 7 = 5 LHS = 4^(n+2) + 5^(2n+1) 5 mod 7 = -2 => 5^2 mod 7 = 4 mod 7 LHS = 16.4^n + 5.4^n = 21.4^n mod 7 = 0 mod 7 Now do the same with 19. LHS = 121.11^n + 12.12^2n 12^2n = 144^n 144 = 11 mod 19 144^n = 11^n mod 19 121 = 7 mod 19 LHS = 7.11^n + 12.11^n = 19.11^n = 0 mod 19 H.P. 3. Find all integers a such that the quadratic expression (x+a)(x+1991) + 1 can be factored as (x+b)(x+c), where b and c are integers. Solution: Since -b is a root of RHS =>  (-b+a)(-b+1991) = -1 Case 1: (-b+a) = 1 (-b+1991) = -1 => b = 1992, a = 1993 Case 2: (-b+a) = -1 (-b+1991) = 1 => b = 1990, a = 1989 Answer: a = 1989, 1993.

Q1) Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides. Q2) Prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from all its vertices.

  Q1) Determine real x,y s.t. x^8 + y^8 = 8xy - 6 S1) LHS increases much faster than RHS, so only for some small values both sides will equal. Also when we have number of equations less than variables, here are few possible approaches: 1. make perfect squares 2. Some inequalities 3. Some useful substitutions. This problem does look like AM GM case since we have power of 8 and 8 terms: x^8,y^8,1....1(6 times). Also AM GM works only for non-negative numbers, which is true here. x^8 + y^8 + 1 ... 1)/8 >= (x^8.y^8.1 ... 1)^(1/8) = |xy| => x^8 + y^8 + 1 ... 1 >= 8|xy| >= 8xy But our question says they are equal which is only possible when all individual numbers are equal. So x^8 = y^8 = 1 => x,y = -+1 and upon checking they do satisfy. Answer. Q2. S2. x^2.y.z = 4^2 y^2.zx = 9^2 z^2.xy = 16^2 => x^4.y^4.z^4 = 4^2.9^2.16^2 = 2^4.3^4.4^4 => xyz = 2.3.4 = 24 => x = 16/24 = 2/3 y = 81/24 = 27/8 z = 256/24 = 32/3