28th VTRMC, 2006, Solutions
- If we write such an integer n in base 3, then it must end in
200...0, because n contains no 1's. But then n2 will end
100...0 and we conclude that there are no positive
integers n for which neither n nor n2 contain a 1 when written
out in base 3.
- The format of such a sequence must either consist entirely of A's and
B's, or must be a block of A's, followed by a single B, followed by a
block of C's, followed by a string of A's and B's. In the former
case, there are 2n such sequences. In the latter case, the number
of such sequences which have k A's and m C's (where
2n-m-k-1. Therefore the number of such sequences with k A's
∑m=1n-k-12n-m-k-1 = 2n-k-1 - 1.
We deduce that the total number of such sequences is
∑k=0n-2(2n-k-1 -1) + 2n = 2n -2 - (n - 1) + 2n = 2n+1 - (n + 1).
We conclude that
S(10) = 211 - 11 = 2037.
- From the recurrence relation
F(n) = F(n - 1) + F(n - 2), we obtain
|F(n + 5)
||= F(n + 4) + F(n + 3) = 2F(n + 3) + F(n + 2)
||= 3F(n + 2) + 2F(n + 1) = 5F(n + 1) + 3F(n).
F(n + 20) = 34F(n) = F(n)mod 5 and we deduce that
F(2006) = F(6)mod 20. Since
F(6) = 5F(2) + 3F(1) = 8.
it follows that F(2006) has remainder 3 after being divided by 5.
F(n + 5) = 2F(n + 3) + F(n + 2) tells us that
F(n + 5) = F(n + 2)mod 2
F(2006) = F(2) = 1 mod 2. We conclude that F(2006) is
an odd number which has remainder 3 after being divided by 5,
consequently the last digit of F(2006) is 3.
cn = (- b3n-2)n - (- b3n-1)n + (- b3n)n. Then the
can be written as the sum of the three series
∑n=1∞(- 1)nb3n-2, ∑n=1∞ - (- 1)nb3n-1, ∑n=1∞(- 1)nb3n.
Since each of these three series is alternating in sign with the
absolute value of the terms monotonically decreasing with limit 0,
the alternating series test tells us that each of the series is
convergent. Therefore the sequence
sk : = ∑n=13k(- 1)nbn is convergent, with limit S say.
limn- > ∞bn = 0, it follows that
∑n=1∞(- 1)nbn is also convergent with sum S.
- We will model the solution on the method reduction of order;
let us try a solution of the form
y = u sin t where u is a function of t. Then
y' = u'sin t + u cos t and
y'' = u''sin t + 2u'cos t - u sin t. Plugging
y'' + py' + qy = 0, we obtain
u''sin t + u'(2 cos t + p sin t) + u(p cos t + q sin t - sin t) = 0. We set
u''sin t + u'(2 cos t + p sin t) = 0 and p cos t + q sin t - sin t = 0.
There are many possibilities. We want u = t2 to satisfy
u'' + u'(2 cos t + p sin t)/sin t = 0. Since u = t2 satisfies
u'' - u'/t = 0, we set
2 cot t + p = - 1/t, and then
||= - 1/t - 2 cot t,
||= 1 - p cot t = 1 + (cot t)/t + 2 cot2t,
||= t2sin t.
Then p and q are continuous on
(0,π) (because 1/t and cot t are continuous on
y = sin t and y = f (t) satisfy
u'' + pu' + qu = 0. Also f is infinitely
differentiable on the whole real line
(- ∞,∞) and
f (0) = f'(0) = f''(0) = 0.
β = ∠QBP and
γ = ∠QCP. Then the sine
rule for the triangle ABC followed by the double angle formula for
sines, and then the addition rules for sines and cosines yields
|(AB + AC)/BC
||= (sin 2β + sin 2γ)/(sin(2β +2γ)) = (2 sin(β + γ)cos(β - γ))/(2 sin(β + γ)cos(β + γ))
||= (cosβcosγ + sinβsinγ)/(cosβcosγ - sinβsinγ) = (1 + tanβtanγ)/(1 - tanβtanγ).
tanβtanγ = (PQ/BQ)(PQ/QC) = 1/2, we see that
(AB + AC)/BC = 3 and the result is proven.
- We will call the three spheres A, B, D and let their centers be
P, Q, R respectively. Then PQR is an equilateral triangle with
sides of length 1. So we will let
O = (0, 0, 0),
P = (0,√3/2, 0),
Q = (- 1/2, 0, 0),
R = (1/2, 0, 0),
X = (0, 1/(2√3), 0). Then M can be
described as the cylinder C with cross-section PQR which
is bounded above and below by the spheres A, B, D. Let V denote
the space above ORX. We now have the following diagram.
By symmetry, the mass of M is
12zdV. Also above QRX, the mass M is bounded above by
the A, which has equation
z = √1-x2-(y-√3/2)2,
and the equation of the line XR in the xy-plane is
x + √3y = 1/2. Therefore the mass of M is
||= 6∫01/(2√3)∫01/2-√3y(1 - x2 - (y - √3/2)2) dxdy
||= 2∫01/(2√3)[3x - x3 -3x(y - √3/2)2]01/2-√3ydy
||= ∫01/(2√3)(1 - 2√3y)(1/2 + 4√3y - 6y2) dy
||= ∫01/(2√3)(12√3y3 -30y2 +3√3y + 1/2) dy
||= [3√3y4 -10y3 +3√3y2/2 + y/2]0√3/6
||= √3(1/48 - 5/36 + 1/8 + 1/12) = 13/(48√3).