41st VTRMC, 2019, Solutions

1. Let M denote the minimal value of f (n). Clearly M≤2 + 7 + 7 + 1 = 17. We will show that M = 17, so assume by way of contradiction that M < 17. Choose n∈ℕ with f (n) = M, and write n in reverse order as 1a₁...a_d where a_d≠ 0 (so n is a (d + 1)-digit number). We have f (n)≡2771ⁿ≡(-1)ⁿmod 9. First assume that n is odd, so f (n)≡ -1 mod 9, so we must have a₁ + ... + a_d = 7. We also have 1 -a₁ + a₂ - ...≡2771ⁿ≡ -1 mod 11, so -a₁ + a₂ -a₃ + ... = -2. Adding these two equations, we obtain 2a₂ +2a₄ + ... = 5, a contradiction because the left hand side is an even integer and the right hand side is an odd integer. Now assume that n is an even integer. Then we have f (n)≡1 mod 9 and therefore a₁ + ... + a_d = 9. Also 1 -a₁ + a₂ - ...≡1 mod 11 and therefore -a₁ + a₂ -a₃ + ... = 0. Adding the last two equations, we obtain 2a₂ +2a₄ + ... = 9, again a contradiction and the result follows.

2. Since BX/XA = 9, we see that AX = AB/10 and we deduce that the area of AXC is 1/10 of the area of ABC, because they have the same height. Using the fact that the area of XYC is 9/100 of the area of ABC, we find that the area XYB is 81/100 of the area of ABC. Therefore the area of XBY is 9/10 of the area of XBC. Let H be the point on AB such that ∠AHC = 90^o. Since XBY and XBC have the same base, we see that MY = (9/10)CH. Now MBY and HBC are similar, consequently
   HB = (10/9)MB = (10/9)·(1/2)·XB = (10/9)·(1/2)·(9/10)AB = (1/2)AB.
   Therefore AC = BC and hence BC = 20.

3. Define g(x) = ∫₀^x(1 -t)f (t) dt. Then g(0) = 0 and

   g(1) = ∫₀¹f (x) dx - ∫₀¹xf (x) dx = ∑_d=0ⁿa_d/(d + 1) - ∑_d=0ⁿa_d/(d + 2) = ∑_d=0ⁿa_d/((d + 2)(d + 1)) = 0.

   
   
   By Rolle's theorem, there exists q∈(0, 1) such that g'(q) = 0, that is (1 -q)f (q) = 0. Since q≠1, we deduce that f (q) = 0 as required.

4. Let I = ∫₀¹x²/(x + √(1 -x²)) dx. We make the substitution x = sin t. Then dx = dt cos t and we see that I = ∫₀^π/2(sin²t cos t)/(sin t + cos t) dt. Also by making the substitution x = cos t, we see that I = ∫₀^π/2(cos²t sin t)/(sin t + cos t) dt and we deduce that
   2I = ∫₀^π/2(sin²t cos t + cos²t sin t)/(sin t + cos t) dt.
   Since sin²t cos t + cos²t sin t = sin t cos t(sin t + cos t), we find that 2I = ∫₀^π/2sin t cos t dt. Therefore 4I = ∫₀^π/2sin 2t dt and we conclude that I = 1/4.

5. We make the substitution t = 1/x. Let y' and y'' denote the first and second derivatives of y with respect to t, respectively.Then dy/dx = -t²y and d²y/dx² = 2t³y' + t⁴y'' and by substituting back into the original equation, we obtain y'' + (2t^-1 -2)y' + (1 - 2t^-1)y = 0. It is easy to see that y = e^t is a solution to this equation. We now use reduction of order to obtain a second solution, so let y = f (t)e^t be another solution, where f is to be determined. Then f'' + 2t^-1f' = 0, which has the solution f = t^-1. We deduce that e^1/x and xe^1/x are solutions to the original equation. Since these solutions are clearly linearly independent, it follows that the general solution to the original equation is y = C₁e^1/x + C₂xe^1/x, where C₁ and C₂ are arbitrary constants.

6. For each s∈S, there exist m, n, p, q∈ℕ and a, b∈{±1} such that s∈(am/n, bp/q) and S∩(am/n, bp/q) = {s}. Then we may define
   f (s) = 2^a+13^b+15^m7ⁿ11^p13^q.

7. For d∈ℕ, the number of d-digit integers in S is 9^d, because we have 9 choices for each digit, and all these integers are ≥10^d-1. Therefore the series is bounded by
   ∑_d∈ℕ9^d(10^d-1)^-99/100 = ∑_d∈ℕ9^d10^-99(d-1)/100.
   This is a geometric series with ratio between successive terms 9·10^-99/100; we show that this ratio is < 1. Rearranging, we find that we need to prove 10⁹⁹/9⁹⁹ > 9, equivalently (1 + 1/9)⁹⁹ > 9, which is true by the binomial series. It follows that the geometric series is convergent, and we conclude by the comparison test that the original series is convergent.