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∫0x0ueu2v2 dvdu and it follows that f'(x) = 2∫0xex2v2 dv. Set t = xv, so dv = dt/x and f'(x) = (2/x)∫0x2et2 dt. Therefore

f''(x) = (-2/x2)∫0x2et2 dt + 2x(2/x)ex4.

We conclude that xf''(x) + f'(x) = 4xex4 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 aiC and at least one of the ai 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) = erx, where r is a complex constant. Plugging into the differential equation, we obtain rerx = er(x+1) - erx, which simplifies to er = 1 + r. We need a solution with r≠ 0. Write r = a + ib, where a, bR. We want to solve ea+ib = 1 + a + ib. By equating the real and imaginary parts, we obtain

 eacos b = 1 + a easin b = b.

Eliminating a, we find that g(b) : = -b cos b + (1 + log b - log(sin b))sin b = 0. Since limx-> 0+x log x = 0, we see that limb-> 2π+(log(sin b))sin b = 0 = limb-> 3π-(log(sin b))sin b. Therefore

limb-> 2π+g(b) = -2π < 0 and limb-> 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 ep+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) : = epxsin 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.

Peter Linnell 2009-11-01