31st VTRMC, 2009, Solutions
- 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))
f (n) = 3((5/2)n - 1).
- 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.
- By symmetry, f (x) is twice the integral over the triangle bounded
by u = v, v = 0 and u = x, so
f (x) = 2∫0x∫0ueu2v2 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.
- 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
= ∠PQB (because YZ is tangent to
β at X).
It follows that AP is parallel to QB, as required.
a3X3 + a2X2 + a1X + a0I denote the minimum polynomial of
ai∈C 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.
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.
n4 -7n2 + 1 is a perfect square; we may assume that n is
n4 -7n2 +1 = (n2 -3n + 1)(n2 + 3n + 1).
Suppose p is a prime that divides both n2 - 3n + 1 and n2 + 3n + 1.
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.
- 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, b∈R. We want to solve
ea+ib = 1 + a + ib. By equating
the real and imaginary parts, we obtain
||= 1 + a
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.
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