Solutions for October Problems

Comment on problems 339 and 342. In both these problems, a condition was left out and made each of them trivial. Accordingly, problem 339 is marked out of 4 and problem 342 out of 3, for the basic solution. However, additional marks were provided for students who recognized that the problems might have been misstated and provided work that led to the intended solutions. While I try to make sure that the problems are correct, and certainly on contests, the problems are generally gone over very carefully, mistakes do occur. If on a contest, you feel that a mistake has been made in formulating a problem, then you should state clearly a non-trivial version of the problem and solve that. In the solutions below, the corrected version of the problem is given.

339.

Let a, b, c be integers with abc ¹ 0, and u, v, w be integers, not all zero, for which

au2 + bv2 + cw2 = 0 .

Let r be any rational number. Prove that the equation

ax2 + by2 + cz2 = r

is solvable for rational values of x, y, z.

Solution 2. [R. Peng] Suppose that r = p/q. Then the equation with r on the right is satisfied by

Comment. If not all a, b, c are zero, then it is trivial to prove that the equation with r on the right has a solution; the only rôle that the equation with 0 on the right plays is to ensure that a, b, c do not all have the same sign. However, some made quite heavy weather of this. The hypothesis that integers are involved should alert you to the fact that some special character of the solution is needed. It is unreasonable to ask that the solution be in integers, but one could seek out rational solutions.

340.

The lock on a safe consists of three wheels, each of which may be set in eight different positions. Because of a defect in the safe mechanism, the door will open if any two of the three wheels is in the correct position. What is the smallest number of combinations which must be tried by someone not knowing the correct combination to guarantee opening the safe?

We now show that at least 32 combinations are needed. Suppose, if possible, that a set S of combinations has three members whose first entry is 0: (0, a, b), (0, c, d), (0, e, f). There will be twenty-five combinations of the form (0, y, z) with y ¹ a, c, e, z ¹ b, d, f, that will not match in two entries any of these three. To cover such combinations, we will need at least 25 distinct combinations of the form (x, y, z) with 1 £ x £ 7. None of the 28 combinations identified so far match the 7 ×6 = 42 combinations of the form (u, v, w), where v ÃŽ { a, c, e }, w ÃŽ {b, d, f }, (v, w) ¹ (a, b), (c, d), (e, f). Any combination of the form (u, v, t) or (u, t, w) can cover at most three of these and of the form (t, v, w) at most 7. Thus, S will need at least 37 = 3 + 25 + (42/7) members to cover all the combinations. A similar argument obtains if there are only three members in S with any other given entry. If there are only one or two members in S with a given entry, say first entry 0, then at least 36 combinations would be needed to cover all the entries with first entry 0 and the other entries differing from the entries of these two elements of S.

Thus, a set of combinations will work only if there are at least four combinations with a specific digit in each entry, in particular at least four whose first entry is k for each of k = 0, ¼, 7. Thus, at least 32 entries are needed.

Comment. Some solvers formulated the problem in terms of the minimum number of rooks (castles) required to occupy or threaten every cell of a solid 8 ×8 ×8 chessboard.

341.

Let s, r, R respectively specify the semiperimeter, inradius and circumradius of a triangle ABC.

(a) Determine a necessary and sufficient condition on s, r, R that the sides a, b, c of the triangle are in arithmetic progression.

(b) Determine a necessary and sufficient condition on s, r, R that the sides a, b, c of the triangle are in geometric progression.

Solution 1. (a) a, b, c are in arithmetic progression if and only if

(b) First, note that

Solution 2. The three sides of the triangle are the three real roots of the cubic equation

(a) The condition is that 2s/3 satisfies the cubic equation:

(b) The condition is that ³ Ö{4Rrs} satisfies the cubic equation: 2s(4Rrs)^1/3 = s² + 4Rr + r² or 32Rrs³ = (s² + 4Rr + r²)³.

Solution 3. [B.H. Deng] Assume that b lies between a and c, inclusive. (a) The three sides are in arithmetic progression if and only if b = ²/₃s or a + c = 2b. Since 4Rrs = abc, this is equivalent to 6Rr = ac, which in turn is equivalent to

(b) The three sides are in geometric progression if and only if b³ = abc = 4Rrs and ac = b². This holds if and only if

342.

Prove that there are infinitely many solutions in positive integers, whose greatest common divisor is equal to 1, of the system

	a + b + c	= x + y
	a3 + b3 + c3	= x3 + y3 .

Let p = 3b - 8a, q = 2a. We can get solutions by solving p² - 10q² = 9. Three solutions are (p, q) = (3, 0),(7, 2), (13, 4). The fundamental solution of u² - 10v² = 1 is (u, v) = (19, 6). So from any solution (p, q) = (r, s) of p² - 10q² = 9, we get another (p, q) = (19r + 60s, 6r + 19s). For these to yield solutions (a, b, c; x, y) of the original system, we require q to be even and p + 4q to be divisible by 3. Since 19r + 60s º r (mod 3) and 6r + 19s º s (mod 2), if (p, q) = (r, s) has these properties, then so also does (p, q) = (19r + 60s,6r + 19s). Starting with (r, s), we can define integers p and q, and then solve the equations x + y = 3b, y - x = 1. Since p and so b are odd, these equations have integer solutions. Here are some examples:

Solution 2. [D. Dziabenko] Let a = 3d. c = 2b - 3d, so that x + y = 3b and a, b, c are in arithmetic progression. Then

Solution 3. [F. Barekat] Let m = a + b + c = x + y and n = a³ + b³ + c³ = x³ + y³. Then

Let c = u(a + b). Then

For example, v = 0 yields the equation (a - 4)(b - 4) = 15 from which we get the possibilities

Solution 4. [D. Rhee] An infinite set of solutions is given by the formula

Comment. M. Fatehi gave the solution

343.

A sequence { a_n } of integers is defined by

a0 = 0 ,   a1 = 1 ,   an = 2an-1 + an-2

for n > 1. Prove that, for each nonnegative integer k, 2^k divides a_n if and only if 2^k divides n.

It is straightforward to show by induction from the recursion that a_n is odd whenever n is odd and even whenever n is even. Suppose now that n is even. Then a_n+1 = 2a_n + a_n-1 º a_n-1 º a₁ = 1 (mod 4), so that a_n-1 + a_n+1 = 2b_n for some odd number b_n. Hence a_2n = 2a_nb_n. For k = 0, we have that 2^k |a_n if and only if 2^k |n. Suppose that this has been established for k = r.

Suppose that n = 2^r+1m for some integer m. Then n/2 is divisible by 2^r, and therefore so is a_n/2. Hence a_n = 2a_n/2b_n/2 is divisible by 2^r+1. On the other hand, suppose that n is not divisible by 2^r+1. If n is not divisible by 2^r, then a_n is not so divisible by the induction hypothesis, and so not divisible by 2^r+1. On the other hand, if n = 2^rc, with c odd, then a_n is divisible by 2^r. But n/2 = 2^r-1c, so a_n/2 is not divisible by 2^r. Hence a_n = 2a_n/2b_n/2 is not divisible by 2^r+1. The result follows.

Solution 2. For convenience, imagine that the sequence is continued backwards using the recursion a_n-2 = a_n - 2a_n-1 for all integer values of the index n. We have for every integer n, a_n+1 = 2a_n + a_n-1 Þ 2a_n+1 = 4a_n + 2a_n-1Þ a_n+2 - a_n = 4a_n + a_n - a_n-2 Þa_n+2 = 6a_n - a_n-2. Suppose, for some positive integer r, we have established that, for every integer n,

Observe that, since a_n+1 º a_n-1 (mod 2) and a₀ = 0, a₁ = 1, a_n is even if and only if n is even. When n is even, then a_n+2 º a_n-2 (mod 4), so that a_n is divisible by 4 if and only if n is.

Let m ³ 2 be a positive integer. Suppose that it has been established for 1 £ s £ m, that 2^s divides a_n if and only if 2^s divides n. Then 2^s+1 will divide a_n only if n = 2^s p for some integer p. Now

Suppose that it has been established for 1 £ i £ p that when n = 2^s i, 2^s+1 |n if and only if p is even. We have that

Solution 3. The characteristic equation for the recursion is t² - 2t - 1 = 0, with roots t = 1 ±Ö2. Solving the recursion, we find that

Comment. Y. Zhao obtained by induction that

344.

A function f defined on the positive integers is given by

f(1) = 1 ,   f(3) = 3 ,   f(2n) = f(n) ,

	f(4n + 1)	= 2f(2n + 1) - f(n)
	f(4n + 3)	= 3f(2n + 1) - 2f(n) ,

for each positive integer n. Determine, with proof, the number of positive integers no exceeding 2004 for which f(n) = n.

Since 4n + 1 = a_r2^r+2 + a_r-12^r+1 + ¼+ a₁2³ + a₀2+ 0·2 + 1,

Since 4n + 3 = a_r2^r+2 + a_r-12^r+1 + ¼+ a₁2³ + a₀2+ 1·2 + 1,

We show by induction that f(n) = g(n) for every positive integer n. This is true for 1 £ n £ 4. Suppose it holds for 1 £ n £ 4m. Then

For f(n) = n, it is necessary and sufficient that n is a palindrome when written to base 2. We need to find the number of palindromes between 1 and 2004 = (11111010100)₂ inclusive. The number of (2r-1)- and 2r-digit palindromes is each 2^r-1 as the first and last digits must be 1 and there are r-1 other matching pairs of digits or central digits that can be set to either 0 or 1. The number of palindromes up to 2¹¹ - 1 = 2047 is 2(1 + 2 + 4 + 8 + 16) + 32 = 94. The only palindromes between 2004 and 2048 are (11111011111)₂ and (11111111111)₂, and these should not be counted. Therefore, there are exactly 92 palindromes, and therefor 92 solutions of f(n) = n between 1 and 2004, inclusive.

345.

Let C be a cube with edges of length 2. Construct a solid figure with fourteen faces by cutting off all eight corners of C, keeping the new faces perpendicular to the diagonals of the cuhe and keeping the newly formed faces identical. If the faces so formed all have the same area, determine the common area of the faces.

The corner faces of the final solid are hexagons formed by large equilateral triangles with smaller equilateral triangles clipped off each vertex; the other faces are squares (diamonds) in the middle of the faces of the cube. Let the square faces have side length x. The vertices of this face are distant 1 - (x/Ö2) from the edge of the cube, so that smaller equilateral triangles of side Ö2(1 - (x/Ö2)) = Ö2 - x are clipped off from a larger equilateral triangle of side 2(Ö2 - x) + x = 2Ö2 - x. The areas of the hexagonal faces of the solid figure are each

Solution 2. Let the cut be made distant u from a vartex. As in Solution 1, we argue that 1 < u < 2. Then the edge of the square face of the final solid is distant u/Ö2 from the vertex of the cube and Ö2(1 - (u/2)) from the centre of the face. Thus, the square face has side length Ö2(2 - u) and area 8 - 8u + 2u².

The hexagonal face of the solid consists of an equilateral triangle of side Ö2u with three equilateral triangles of side Ö2(u - 1) clipped off. Its area is (Ö3/2)[-2u² + 6u - 3. For equality of area of all the faces, we require that