12th Vietnamese Mathematical Olympiad 1974 Problems

12th Vietnamese Mathematical Olympiad 1974 Problems

A1.  Find all positive integers n and b with 0 < b < 10 such that if a_n is the positive integer with n digits, all of them 1, then a_2n - b a_n is a square.  

A2.  (1) How many positive integers n are such that n is divisible by 8 and n+1 is divisible by 25?
(2) How many positive integers n are such that n is divisible by 21 and n+1 is divisible by 165?
(3) Find all integers n such that n is divisible by 9, n+1 is divisible by 25 and n+2 is divisible by 4.

B1.  ABC is a triangle. AH is the altitude. P, Q are the feet of the perpendiculars from P to AB, AC respectively. M is a variable point on PQ. The line through M perpendicular to MH meets the lines AB, AC at R, S respectively. Show that ARHS is cyclic. If M' is another position of M with corresponding points R', S', show that the ratio RR'/SS' is constant. Find the conditions on ABC such that if M moves at constant speed along PQ, then the speeds of R along AB and S along AC are the same. The point K on the line HM is on the other side of M to H and satisfies KM = HM. The line through K perpendicular to PQ meets the line RS at D. Show that if ∠A = 90^o, then ∠BHR = ∠DHR.  

B2.  C is a cube side 1. The 12 lines containing the sides of the cube meet at plane p in 12 points. What can you say about the 12 points?

Solution

12th VMO 1974

Problem A2

Find all positive integers n and b with 0 < b < 10 such that if a_n is the positive integer with n digits, all of them 1, then a_2n - b a_n is a square.

Answer

b = 2 works for any n
b = 7 works for n = 1

Solution

a_n = (10ⁿ-1)/9, so we need 10²ⁿ - b10ⁿ + (b-1) to be a square. For b = 2, this is true for all n. Note that (10ⁿ - c)² = 10²ⁿ - 2c10ⁿ + c², so if b = 2c, we need c² = b-1 = 2c - 1 and hence c = 1. So b = 4, 6, 8 do not work. Equally 10²ⁿ > 10²ⁿ - 10ⁿ > (10ⁿ - 1)² = 10²ⁿ - 2·10ⁿ + 1, so it cannot be a square for b = 1. Similarly, (10ⁿ - 1)² = 10²ⁿ - 2·10ⁿ + 1 > 10²ⁿ - 3·10ⁿ + 2 > 10²ⁿ - 4·10ⁿ + 4 = (10ⁿ - 2)², so b = 3 does not work. Similarly, (10ⁿ - 2)² = 10²ⁿ - 4·10ⁿ + 4 > 10²ⁿ - 5·10ⁿ + 4 > 10²ⁿ - 6·10ⁿ + 9 = (10ⁿ - 3)², so b = 5 does not work. Similarly, (10ⁿ - 3)² = 10²ⁿ - 6·10ⁿ + 9 > 10²ⁿ - 7·10ⁿ + 6 > 10²ⁿ - 8·10ⁿ + 16 = (10ⁿ - 4)² for n > 1, so b = 7 does not work, except possibly for n = 1. Since 11 - 7 = 2², b = 1 does work for n = 1. Finally, (10ⁿ - 4)² = 10²ⁿ - 8·10ⁿ + 16 > 10²ⁿ - 9·10ⁿ + 8 > 10²ⁿ - 10·10ⁿ + 25 = (10ⁿ - 4)² for n > 1, so b = 9 does not work, except possibly for n = 1. It is easy to check it does not work for n = 1.