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## 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.
Solution 1. Note that the number of digits remains the same after multiplication. Thus, if k ³ 5, the left digit of the number must be 1 and so the multiple must end in 1. This is impossible for k = 5, 6, 8. If k = 7 or 9, then the number must have the form 10m + x where x £ 10n -1. Then k(10m + x) = 10x + 1, so that
 x = k ·10m - 1 10 - k ³ 7 ·10m - 1 3 > 2 ×10m ,
an impossibility.

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 £ 10m - 1, we must have 4(2 ×10m + x) = 10x + 2, whence

 x = 4 ×10m - 1 3 > 10m ,
again an impossibility. Finally, if k = 2, then d £ 4 and 2(d ·10m + x) = 10x + d, whence d(2 ·10m - 1) = 8x. Since 2 ·10m - 1 is odd, 8 must divide d, which is impossible. The desired result follows.

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

 (10m - 1)k < k ·10m - 1 £ d(k ×10m - 1) = (10 - k)u £ (10 - k)(10m - 1) ,
so that k < 10 - k and k is equal to 2 or 4. Since k is even, d must be even. Since
 10 - k = d æè k ×10m - 1 u öø > d k ×10m - k 10m - 1 = dk ,
d < (10/k) - 1. When k = 2, d must be 2, and we get 2(2 ×10m - 1) = 8u, or 2 ×10m - 1 = 4u, an impossibility. When k = 4, we get d < 1.5, which is also impossible. Hence the multiplication is not possible.

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(10m + x) = 10x + 1 for x = (3 ×10m - 1)/7. This actually gives a result when m º 5 (mod 6). Indeed, when m = 5, we obtain the example 142857.

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.

 x1 + x2 = y1 y2      x1 + x3 = y1 y3      x1 + x4 = y1 y4

 x2 + x3 = y2 y3      x2 + x4 = y2 y4      x3 + x4 = y3 y4 .

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 yi 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.)

257.
Let n be a positive integer exceeding 1. Discuss the solution of the system of equations:

 ax1 + x2 + ¼+ xn = 1

 x1 + ax2 + ¼+ xn = a

 ¼

 x1 + x2 + ¼+ axi + ¼+ xn = ai-1

 ¼

 x1 + x2 + ¼+ xi + ¼+ axn = an-1 .

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

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

 (n - 1 + a)(x1 + x2 + ¼+ xn) = 1 + a + a2 + ¼+ an-1 = 1 - an 1 - a .
If a = 1 - n, then the system is viable only if an = 1. This occurs, only if a = -1 and n is a positive integer i.e., when (n, a) = (2, -1). In this case, both equations in the system reduce to x2 - x1 = 1, and we have infinitely many solution. Otherwise, when a = 1 - n, there is no solution to the system.

When a ¹ 1 - n, then

 x1 + x2 + ¼+ xn = 1 - an (1 - a)(n - 1 + a) .
Taking the difference between this and the ith equation in the system leads to
 (a - 1)xi = ai-1 - æè 1 - an (1 - a)(n - 1 + a) öø
for each i and the system is solved.

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)(xi+1 - xi) = ai - ai-1, so that xi+1 = xi + ai-1 for 1 £ i £ n-1. Hence xi = x1 + (1 + a + ¼+ ai-2) for 2 £ i £ n. From the first equation, we find that

 (n - 1 + a)x1
 + 1 + (1 + a) + (1 + a + a2) + ¼+ ¼(1 + a + ¼+ an-2) = 1
 Þ (n - 1 + a)x1 + (1 - a2) + ¼+ (1 - an-1) 1 - a = 0
 Þ (n - 1 + a)x1 + n - 2 - a2(1 + a + ¼+ an-3) 1 - a = 0
 Þ (n - 1 + a)x1 + (n-2)(1-a) - a2(1 - an-2) (1 - a)2 = 0 .
Suppose that n = 1 - a. Then
 0 = (n-2)(1 - a) - a2(1 - an-2) = -(1 + a)(1 - a) -a2(1 - an-2) = an-2 - 1 ,
so that a must be -1 and n = 2, The system reduces to a single equation with an infinitude of solutions. If n ¹ 1 - a, then we can solve for x1 and then obtain the remaining values of the xi.

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.

258.
The infinite sequence { an ; 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 ap - aq = ar - as.
Solution. The recursion can be rewritten as
 an+1 = 2an2 - 2an + 1 Û2an+1 - 1 = (2an - 1)2 .
Let bn = 2an - 1, so that an = 1/2(bn + 1). Then ap - aq = ar - as is equivalent to bp - bq = br - bs. Since bn+1 = bn2 for each nonnegative integer n, we have that bn = b02n. If bp - bq = br - bs, then b0 must be the rational solution of a polynomial equation of the form,
 x2p - x2q - x2r + x2s = 0
where the left side consists of four distinct monomials. One possibility is b0 = 0. Suppose now that b0 ¹ 0. Dividing by the monomial with the smallest exponent, we obtain a polynomial equation for b0 whose leading coefficient and constant coefficients are each 1. So the numerator of b0 written in lowest terms, dividing the constant term, must be ±1 and the denominator, dividing the leading coefficient, must also be ±1. Hence, the only possibilities for b0 are -1, 0 and 1. These correspond to the possibilities 0, 1/2, 1 for a0, and each of these choices leads to a sequence for which an = a1 for n ³ 1 and for which there are two pairs of terms with the same difference (0).

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¢.
Prove that AA¢, BB¢, CC¢ are concurrent and that
 AA" + BB" + CC" = 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

 AA" + BB" + CC"
 = (A¢A" + B¢B" + C¢C") - (AA¢+ BB¢+ CC¢)
 = 2(A¢X + B¢Y + C¢Z) - (AA¢+ BB¢+ CC¢)
 = 2(A¢X ±FX + B¢Y -±FY + C¢Z ±FZ) - (AA¢+ BB¢+ CC¢)
 = 2(A¢F + B¢F + C¢F) - (AA¢+ BB¢+ CC¢)
 = 4(AF + BF + CF) - 3(AF + BF + CF)
 = AF + BF + CF = AA¢ = BB¢ = CC¢ .

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¢.
Solution. Denote by X' the reflection of a point X in O. In particular, T¢ = G. Let D be the midpoint of BC. Since TT¢ = TG and AA¢ intersect at O, the points A, G, D, T, A¢ are collinear. Let A1 be the intersection of DT and GA¢. Since the reflection in O takes any line to a parallel line, A¢G || AT, so that (from triangle DTA), DA1 : DT = DG : DA = 1:3 and A1 is the centroid of triangle TBC. Also
 GA1 : GA¢ = GA1 : AT = DA1 : DT = 1:3
so that GA1 = (1/3)GA¢.

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 Ai¢ = Aj, Bi¢ = Bj, Ci¢ = Cj 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.

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 .
Solution. Observe that
 (x + y)(x + z) - ( Ö xy + Ö xz )2 = x2 + yz - 2x Ö yz = (x - Ö yz )2 ³ 0
(with equality iff x2 = yz). Hence
x

 x + Ö (x + y)(x + z)
£  x

 x + Ö xy + Ö xz
=  Öx

Öx + Öy + Öz
,
with a similar inequality for the other two terms on the left side. Adding these inequalities together leads to the desired result.

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