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 10
^{m} + x where x
£ 10
^{n} 1. Then
k(10
^{m} + x) = 10x + 1, so that
x = 
k ·10^{m}  1
10  k

³ 
7 ·10^{m}  1
3

> 2 ×10^{m} , 

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 £ 10^{m}  1, we must have
4(2 ×10^{m} + x) = 10x + 2, whence
x = 
4 ×10^{m}  1
3

> 10^{m} , 

again an impossibility. Finally, if k = 2, then d
£ 4
and 2(d ·10
^{m} + x) = 10x + d, whence d(2 ·10
^{m}  1) = 8x. Since 2 ·10
^{m}  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 ·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
(10^{m}  1)k < k ·10^{m}  1 £ d(k ×10^{m}  1) = (10  k)u £ (10  k)(10^{m}  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 
æ è


k ×10^{m}  1
u


ö ø

> d 
k ×10^{m}  k
10^{m}  1

= dk , 

d < (10/k)
 1. When k = 2, d must be 2, and
we get 2(2 ×10
^{m}  1) = 8u, or 2 ×10
^{m}  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(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.

256.

Find the condition that must be satisfied by
y_{1}, y_{2}, y_{3}, y_{4} in order that the following set
of six simultaneous equations in x_{1}, x_{2}, x_{3}, x_{4} is solvable.
Where possible, find the solution.
x_{1} + x_{2} = y_{1} y_{2} x_{1} + x_{3} = y_{1} y_{3} x_{1} + x_{4} = y_{1} y_{4} 

x_{2} + x_{3} = y_{2} y_{3} x_{2} + x_{4} = y_{2} y_{4} x_{3} + x_{4} = y_{3} y_{4} . 

Solution. We have than y_{1}(y_{2}  y_{3}) = x_{2}  x_{3} = y_{4}(y_{2}  y_{3}), whence (y_{1}  y_{4})(y_{2}  y_{3}) = 0.
Similarly, (y_{1}  y_{2})(y_{3}  y_{4}) = 0 = (y_{1}  y_{3})(y_{2}  y_{4}).
From this, we deduce that three of the four y_{i} must be equal.
Suppose, wolog, that y_{1} = y_{2} = y_{3} = u and y_{4} = v.
Then the system can be solved to obtain x_{1} = x_{2} = x_{3} = u^{2}/2 and x_{4} = uv  (u^{2}/2) = ^{1}/_{2}u(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:
ax_{1} + x_{2} + ¼+ x_{n} = 1 

x_{1} + ax_{2} + ¼+ x_{n} = a 

x_{1} + x_{2} + ¼+ ax_{i} + ¼+ x_{n} = a^{i1} 

x_{1} + x_{2} + ¼+ x_{i} + ¼+ ax_{n} = a^{n1} . 

Solution 1. First, suppose that a = 1. Then all of the
equations in the system become x_{1} + x_{2} + ¼+ x_{n} = 1,
which has infinitely many solutions; any n1 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
(n  1 + a)(x_{1} + x_{2} + ¼+ x_{n}) = 1 + a + a^{2} + ¼+ a_{n1} = 
1  a^{n}
1  a

. 

If a = 1
 n, then the system is viable only if a
^{n} = 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 x
_{2}  x
_{1} = 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
x_{1} + x_{2} + ¼+ x_{n} = 
1  a^{n}
(1  a)(n  1 + a)

. 

Taking the difference between this and the ith equation in
the system leads to
(a  1)x_{i} = a^{i1}  
æ è


1  a^{n}
(1  a)(n  1 + a)


ö ø



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)(x_{i+1}  x_{i}) = a^{i}  a^{i1}, so that
x_{i+1} = x_{i} + a^{i1} for 1 £ i £ n1.
Hence x_{i} = x_{1} + (1 + a + ¼+ a^{i2}) for 2 £ i £ n.
From the first equation, we find that


+ 1 + (1 + a) + (1 + a + a^{2}) + ¼+ ¼(1 + a + ¼+ a^{n2}) = 1 
 
 
Þ (n  1 + a)x_{1} + 
(1  a^{2}) + ¼+ (1  a^{n1})
1  a

= 0 
 
 
Þ (n  1 + a)x_{1} + 
n  2  a^{2}(1 + a + ¼+ a^{n3})
1  a

= 0 
 
 
Þ (n  1 + a)x_{1} + 
(n2)(1a)  a^{2}(1  a^{n2})
(1  a)^{2}

= 0 . 

Suppose that n = 1
 a. Then
0 = (n2)(1  a)  a^{2}(1  a^{n2}) = (1 + a)(1  a) a^{2}(1  a^{n2}) = a^{n2}  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 x
_{1} and then obtain the remaining values of
the x
_{i}.
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 { a_{n} ; n = 0, 1, 2, ¼} satisfies the recursion
a_{n+1} = a_{n}^{2} + (a_{n}  1)^{2} 

for n ³ 0. Find all rational numbers a_{0} such that there are
four distinct indices p, q, r, s for which a_{p}  a_{q} = a_{r}  a_{s}.
Solution. The recursion can be rewritten as
a_{n+1} = 2a_{n}^{2}  2a_{n} + 1 Û2a_{n+1}  1 = (2a_{n}  1)^{2} . 

Let b
_{n} = 2a
_{n}  1, so that a
_{n} =
^{1}/
_{2}(b
_{n} + 1).
Then a
_{p}  a
_{q} = a
_{r}  a
_{s} is equivalent to
b
_{p}  b
_{q} = b
_{r}  b
_{s}. Since b
_{n+1} = b
_{n}^{2} for each
nonnegative integer n, we have that b
^{n} = b
_{0}^{2n}.
If b
_{p}  b
_{q} = b
_{r}  b
_{s}, then b
_{0} must be the rational
solution of a polynomial equation of the form,
x^{2p}  x^{2q}  x^{2r} + x^{2s} = 0 

where the left side consists of four distinct monomials.
One possibility is b
_{0} = 0. Suppose now that b
_{0} ¹ 0.
Dividing by the monomial with the smallest exponent, we obtain
a polynomial equation for b
_{0} whose leading coefficient and
constant coefficients are each 1. So the numerator of b
_{0}
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 b
_{0} are
1, 0 and 1.
These correspond to the possibilities 0,
^{1}/
_{2}, 1 for
a
_{0}, and each of these choices leads to a sequence for which
a
_{n} = a
_{1} 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


= (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 A
_{1} 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), DA
_{1} : DT = DG : DA = 1:3 and A
_{1} is the centroid
of triangle TBC. Also
GA_{1} : GA¢ = GA_{1} : AT = DA_{1} : DT = 1:3 

so that
GA
_{1} = (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 A_{1}, GB¢ intersects TAC in B_{1}, GC¢ intersects
TAB in C_{1}, TA intersects GB¢C¢ in A_{2}, TB intersects
GA¢C¢ in B_{2} and TC intersects GA¢B¢ in C_{2}. 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, A_{2}, B_{2}, C_{2}, A_{1}, B_{1}, C_{1}, G and faces
TA_{2}C_{1}B_{2}, TB_{2}A_{1}C_{2}, TC_{2}B_{1}A_{2}, GA_{1}C_{2}B_{1},
GB_{1}A_{2}C_{1}, GC_{1}B_{2}A_{1} (to see that, say, TB_{2}A_{1}C_{2}
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 TB_{2}A_{1}C_{2}A_{2} with base TB_{2}A_{1}C_{2}
and altitude dropped from A_{2}, which in turn is twice that
of tetrahedron TA_{2}B_{2}C_{2}. But the volume of tetrahedron
TA_{2}B_{2}C_{2} 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
Solution. Observe that
(x + y)(x + z)  (  Ö

xy

+  Ö

xz

)^{2} = x^{2} + yz  2x  Ö

yz

= (x   Ö

yz

)^{2} ³ 0 

(with
equality iff x
^{2} = yz). Hence

x

£ 
x

= 
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.