Math IOQM

 V_p(n) = k => highest power of prime p in number 'n' is k. This is known as p-adic valuation. For e.g. V2(12) = 2 => p-adic valuation of 2 wrt 12 is 2. Also it's additive, ie V_p(m.n) = V_p(m) + V_p(n) V2(36) = V2(18.2) = V2(18) + V2(2) = 1 + 1 = 2 Legendre's formula:

 for any integer k > 1, there will always be a prime p s.t. k < p < 2k.

Q1. Find all positive integers n for which 2^n + 7^n is a perfect square. S1. for n = 1, it's 9. Ok. for n even  2^n = (-1)^n mod 3 = 1 mod 3 7^n = 1^n mod 3 = 1 mod 3 2^n + 7^n = 2 mod 3 for all even 'n'. But any positive number square = 0,1 mod 3. So no even n is possible. for n odd and n>=3 2^n = 0 mod 4 3^n = (-1)^n = -1 mod 4 = 3 mod 4 2^n + 7^n = 3 mod 4 Any positive square  = 0,1 mod 4 So only answer is n = 1 Q2. Given five distinct integers, consider the ten differences formed by pairs of these numbers. (Note that some of these differences may be equal.) Determine the largest integer that is certain to divide the product of these ten differences, regardless of which five integers were originally given. S2. 1,2,3,4,5 => 288 = 2^5.3^2 Now show 288 will always divide any such product. 3k,3k+1,3k+2 3 buckets, 5 objects. No matter how you distribute, you will get 2 factors of 3. Because at least 1 bucket will have more than 1 object. Case 1: 1 bucket with 3 objects...

Proof: Compute ma and mb using appollonius and equate. 2[ma^2 + a^2/4] = b^2 + c^2 __[1] 2[mb^2 + b^2/4] = a^2 + c^2 __[2] ma^2 - mb^2 = (b^2 - a^2)/2 - (a^2 - b^2)/4 = 3/4[b^2 - a^2] H.P.

Yes. Why? In first triangle: a/SinA = 2R Next triangle: p/SinP = 2R Angles are same => sides are same H.P.

Q1. In a convex quadrilateral ABCD , angle ABD = 30 , angle BCA = 75 , angle ACD = 25 , and CD = CB . Extend CB  to meet the circumcircle of triangle DAC  at E . Prove that CE = BD . S1. Q2. Let ABC be a triangle with circumcircle Gamma, and let G be the centroid of triangle ABC. Extend AG, BG, and CG to meet the circle Gamma again at A1, B1, and C1, respectively. Suppose that triangle ABC ~ A1B1C1, specifically with: angle BAC = angle A1B1C1,  angle ABC = angle A1C1B1,  and angle ACB= B1A1C1. Prove that ABC and A1B1C1 are equilateral triangles. S2. Step 1. Do angle chasing. Same chord same angle. Step 2. Use given angle equalities. A1 = C => 5 + 6 = 4 + 6 => 4 = 5 B1 = A => 1 + 4 = 1 + 2 => 4 = 2 C1 = B => 3 + 5 = 3 + 2 => 2 = 5 Now draw the modified figure: And all the medians intersect at Centroid G. We know that medians divide the triangle into 3 equal areas. [GAB] = [GBC] = [GCA] Each of these triangles as angle 2. [GAB] = GB.AB.sin(2)/2 [GBC] ...

Q1. Let ABCDE be a paper pentagon with AB = AE, angle A = angle B = angle E = 90, BC = 3, CD = 5, and DE = 2. Construct a perpendicular from  A to CD using only a ruler and drawing not more than six lines. All lines have to be drawn inside the pentagon. S1. Make A = (0,0) assign co-ordinates and you can get radius r = 6. Also the circle centered at A with radius AB = r will be touched at B, the tangent from C to B is of length 3. Similarly the tangent DE = 2. So the other tangents from D,C to the same circle will have lengths 2,3 respectively. Using the right triangles ABC, AED, we can show that AC = 3.sqrt(5) and AD = 2.sqrt(10) Now let K be the point on CD s.t. CK = 3 and KD = 2 (such a point will exist since CD = 5). In triangle ACD, apply stewart's theorem to get AK = 6. So, point K will also lie on the circle. And CD is the tangent to it. Triangle ABC and AKC are congruent. So angle ACK = ACB, hence CA is angle bisector of angle BCD. Similarly AED,AKD are congruent and DA is a...