Solutions

255.

Prove that there is no positive integer that, when written to base 10, is equal to its kth multiple when its initial digit (on the left) is transferred to the right (units end), where 2 £ k £ 9 and k ¹ 3.

If k = 4, the first digit of the number cannot exceed 2, and so must be even to achieve an even product. Thus, for some positive integers m and x £ 10^m - 1, we must have 4(2 ×10^m + x) = 10x + 2, whence

Solution 2. [A. Critch] Suppose that multiplication is positive for some k ¹ 3. Let the number be d ·10^m+ u for a positive digit d, a positive integer m and a nonnegative integer u < 10^m - 1. Then k(d ·10^m + u) = 10u + d, whence

Comment. When k = 3, the first digit must be 1, 2 or 3. It can be shown that 2 and 3 do not work, so that we must have 3(10^m + x) = 10x + 1 for x = (3 ×10^m - 1)/7. This actually gives a result when m º 5 (mod 6). Indeed, when m = 5, we obtain the example 142857.

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256.

Find the condition that must be satisfied by y₁, y₂, y₃, y₄ in order that the following set of six simultaneous equations in x₁, x₂, x₃, x₄ is solvable. Where possible, find the solution.

Solution. We have than y₁(y₂ - y₃) = x₂ - x₃ = y₄(y₂ - y₃), whence (y₁ - y₄)(y₂ - y₃) = 0. Similarly, (y₁ - y₂)(y₃ - y₄) = 0 = (y₁ - y₃)(y₂ - y₄). From this, we deduce that three of the four y_i must be equal. Suppose, wolog, that y₁ = y₂ = y₃ = u and y₄ = v. Then the system can be solved to obtain x₁ = x₂ = x₃ = u²/2 and x₄ = uv - (u²/2) = ¹/₂u(2v - u). (This includes the case u = v.)

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257.

Let n be a positive integer exceeding 1. Discuss the solution of the system of equations:

Solution 1. First, suppose that a = 1. Then all of the equations in the system become x₁ + x₂ + ¼+ x_n = 1, which has infinitely many solutions; any n-1 of the x_i's can be chosen arbitrarily and the remaining one solved for.

Henceforth, assume that a ¹ 1. Adding all of the equations leads to

When a ¹ 1 - n, then

Solution 2. As above, we dispose first of the case a = 1. Suppose that a ¹ 1.Taking the difference of adjacent equations leads to (a - 1)(x_i+1 - x_i) = aⁱ - a^i-1, so that x_i+1 = x_i + a^i-1 for 1 £ i £ n-1. Hence x_i = x₁ + (1 + a + ¼+ a^i-2) for 2 £ i £ n. From the first equation, we find that

Comment. Beware of the "easy" questions. Many solvers had only a superficial analysis which did not consider the possibility that a denominator might vanish, and almost nobody picked up the (n, a) = (2, -1) case. When you write up your solution, it is good to dispose of the singular cases first before you get into the general situation.

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258.

The infinite sequence { a_n ; n = 0, 1, 2, ¼} satisfies the recursion

an+1 = an2 + (an - 1)2

for n ³ 0. Find all rational numbers a₀ such that there are four distinct indices p, q, r, s for which a_p - a_q = a_r - a_s.

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259.

Let ABC be a given triangle and let A¢BC, AB¢C, ABC¢ be equilateral triangles erected outwards on the sides of triangle ABC. Let W be the circumcircle of A¢B¢C¢ and let A", B", C" be the respective intersections of W with the lines AA¢, BB¢, CC¢.

Solution. A rotation of 60^° about the vertex A takes triangle ACC¢ to the triangle AB¢B, and so BB¢ = CC¢. Similarly, it can be shown that each of these is equal to AA¢. Suppose that BB¢ and CC¢ intersect in F. From the rotation, ÐBFC¢ = 60^° = ÐBAC¢, so that AFBC¢ is concyclic.

hence ÐC¢FB = ÐC¢AB = 60^°. Also ÐAFC¢ = ÐABC¢ = 60^°, ÐAFB¢ = 60^° and so ÐBFC = ÐC¢FB¢ = 120^°. Since ÐBFC + ÐBA¢C = 180^°, the quadrilateral BFCA¢ is concyclic and ÐBFA¢ = ÐBCA¢ = 60^°. Hence ÐAFA¢ = ÐAFC¢+ÐC¢FB + ÐBFA¢ = 180^°, so that A, A¢ and F are collinear, and AA¢, BB¢ and CC¢ intersect at F.

From Ptolemy's Theorem, AB ·C¢F = AF ·BC¢+ FB ·AC¢ , whence C¢F = AF + BF. Similarly, A¢F = BF + CF and C¢F = AF + BF. Indeed, AA¢ = BB¢ = CC¢ = AF + A¢F = AF + BF + CF.

[J. Zhao] Let O be the circumcentre of triangle A¢B¢C¢ and let the respective midpoints of A¢A", B¢B", C¢C" be X, Y, Z. Since OX ^A¢A", OX ^FX. Similarly, OY ^FY and OZ ^FZ, so that X, Y, Z lie on the circle with diameter OF. Suppose, wolog, that F lies on the arc ZX. Then ÐXZY = ÐXFY = ÐA¢FB" = 60^° and ÐZXY = ÐZFY = 60^°, so that XYZ is an equilateral triangle and Ptolemy's theorem yields that FY = FX + FZ.

Hence

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260.

TABC is a tetrahedron with volume 1, G is the centroid of triangle ABC and O is the midpoint of TG. Reflect TABC in O to get T¢A¢B¢C¢. Find the volume of the intersection of TABC and T¢A¢B¢C¢.

Applying the same reasoning all around, we see that each side of one tetrahedron intersects a face of the other in its centroid one third of the way along its length. Thus GA¢ intersects TBC in A₁, GB¢ intersects TAC in B₁, GC¢ intersects TAB in C₁, TA intersects GB¢C¢ in A₂, TB intersects GA¢C¢ in B₂ and TC intersects GA¢B¢ in C₂. Note that the A_i¢ = A_j, B_i¢ = B_j, C_i¢ = C_j for i ¹ j.

The intersection of the two tetrahedra is a parallelepiped with vertices T, A₂, B₂, C₂, A₁, B₁, C₁, G and faces TA₂C₁B₂, TB₂A₁C₂, TC₂B₁A₂, GA₁C₂B₁, GB₁A₂C₁, GC₁B₂A₁ (to see that, say, TB₂A₁C₂ is a parallelogram, note that a dilation with centre T and factor 3/2 takes it to a parallelogram with diagonal TD). The volume of this parallelpiped is three times that of the skew pyramid TB₂A₁C₂A₂ with base TB₂A₁C₂ and altitude dropped from A₂, which in turn is twice that of tetrahedron TA₂B₂C₂. But the volume of tetrahedron TA₂B₂C₂ is 1/27 = (1/3)³ that of TABC since it can be obtained from TABC by a dilation with centre T and factor 1/3. Hence the volume of the parallelpiped common to both tetraheda TABC and GA¢B¢C¢ is 6 ×(1/27) = 2/9 is the volume of either of these tetrahedra.

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261.

Let x, y, z > 0. Prove that

x x + Ö (x+y)(x+z) +  y y + Ö (x+y)(y+z) +  z z + Ö (x+z)(y+z) £ 1 .