31st VTRMC, 2009, Solutions

1. Let x and y meters denote the distance between the walker and the jogger when the dog returns to the walker for the nth and n + 1st times respectively. Then starting at the walker for the nth time, it will take the dog x seconds to reach the jogger, whence the jogger will be distance 2x from the walker, and then it will take the dog a further 2x/4 seconds to return to the walker. Thus y = x + x + 2x/4 = 5x/2, and also the time taken for the dog to return to the walker will be x + 2x/4 = 3x/2 seconds. It follows that the total distance travelled by the dog to return to the walker for the nth time is
   3*(3/2)(1 + 5/2 + (5/2)² + ... + (5/2)^n-1) = 9((5/2)ⁿ - 1)/(2(5/2 - 1))
   meters. Therefore f (n) = 3((5/2)ⁿ - 1).

2. By examining the numbers 5,10,15,20,25,30,35,40 (at most 40 and divisible by 5), we see that the power of 5 dividing 40! is 9. Also it is easy to see that the power of 2 dividing 40! is at least 9. Therefore 10⁹ divides 40! and we deduce that p = q = r = s = t = u = v = w = x = 0.

   Next note that 999999 divides 40!. This is because 999999 = 9*111111 = 9*111*1001 = 9*3*37*11*91 = 27*7*11*13*37. Since 999999 = 10⁶ - 1, it follows that if we group the digits of 40! in sets of six starting from the units digit and working right to left, then the sum of these digits will be divisible by 999999. Therefore abcdef + 283247 + 897734 + 345611 + 269596 + 115894 + 272000 is divisible by 999999, that is

   abcdef + 2184082 = 999999y
   for some integer y. Clearly y = 3 and we conclude that abcdef = 815915.

3. By symmetry, f (x) is twice the integral over the triangle bounded by u = v, v = 0 and u = x, so f (x) = 2∫₀^x∫₀^ue^u2v2 dvdu and it follows that f'(x) = 2∫₀^xe^x2v2 dv. Set t = xv, so dv = dt/x and f'(x) = (2/x)∫₀^x2e^t2 dt. Therefore
   f''(x) = (-2/x²)∫₀^x2e^t2 dt + 2x(2/x)e^x4.
   We conclude that xf''(x) + f'(x) = 4xe^x4 and hence 2f''(2) + f'(2) = 8e¹⁶.

4. Let the common tangent at X meet AQ and PB at Y and Z respectively. Then we have ∠APX = ∠AXY (because YZ is tangent to α at X) = ∠ZXB (because opposite angles are equal) = ∠PQB (because YZ is tangent to β at X). It follows that AP is parallel to QB, as required.

5. Let a₃X³ + a₂X² + a₁X + a₀I denote the minimum polynomial of A, where a_i∈C and at least one of the a_i is 1. It is the unique monic polynomial f (X) of minimal degree such that f (A) = 0. Since AB = 0, we see that A is not invertible, hence 0 is an eigenvalue of A, consequently a₀ = 0. Set D = a₃A² + a₂A + a₁I. Then D≠ 0, because a₃X³ + a₂X² + a₁X is the minimum polynomial of A. Also AD = DA = a₃A³ + a₂A² + a₁A + a₀I = 0 and the result is proven.

6. Suppose n⁴ -7n² + 1 is a perfect square; we may assume that n is positive. We have n⁴ -7n² +1 = (n² -3n + 1)(n² + 3n + 1). Suppose p is a prime that divides both n² - 3n + 1 and n² + 3n + 1. Then p | 6n and we see that p divides 2,3 or n. By inspection, none of these are possible. Thus no prime can divide both n² - 3n + 1 and n² + 3n + 1, consequently both these numbers are perfect squares, in particular n² + 3n + 1 is a perfect square. By considering this mod 3, we see that n≡0 mod 3, so we may write n = 3m where m is a positive integer. Thus 9m² + 9m + 1 is a perfect square. However (3m + 1)² < 9m² +9m + 1 < (3m + 2)², a contradiction and the result follows.

7. It is easy to check that f (x) = ax + b satisfies the differential equation for arbitrary constants a, b; however it cannot satisfy the numerical condition. We need to find another solution to the differential equation. Since the equation is linear, it's worth trying a solution in the form f (x) = e^rx, where r is a complex constant. Plugging into the differential equation, we obtain re^rx = e^r(x+1) - e^rx, which simplifies to e^r = 1 + r. We need a solution with r≠ 0. Write r = a + ib, where a, b∈R. We want to solve e^a+ib = 1 + a + ib. By equating the real and imaginary parts, we obtain

   e^acos b = 1 + a

   e^asin b = b.

   
   
   Eliminating a, we find that g(b) : = -b cos b + (1 + log b - log(sin b))sin b = 0. Since lim_{x-> 0+}x log x = 0, we see that lim_{b-> 2π+}(log(sin b))sin b = 0 = lim_{b-> 3π-}(log(sin b))sin b. Therefore
   lim_{b-> 2π+}g(b) = -2π < 0 and lim_{b-> 3π-}g(b) = 3π > 0.
   Since g is continuous on (2π, 3π), we see that there exists q∈(2π, 3π) such that g(q) = 0. Set p = log q - log(sin q). Then e^p+iq + p + iq = 1. Since the differential equation is linear, the real and imaginary parts of f will also satisfy the differential equation, and in particular f (x) : = e^pxsin qx will satisfy the differential equation. A routine calculation shows that f''(0) = 2pq≠ 0, and so the answer to the question is ``yes".

   Remark It can be shown that p≈2.09 and q≈7.46.