Please submit your solutions to
Valeria Pandelieva
641 Kirkwood Avenue
Ottawa, ON K1Z 5X5
no later than July 21, 2002.

151.

Prove that, for any natural number
$n$, the
equation
$x(x+1)(x+2)\dots (x+2n1)+(x+2n+1)(x+2n+2)\dots (x+4n)=0$
does not have real solutions.

152.

Andrew and Brenda are playing the following game.
Taking turns, they write in a sequence, from left to right,
the numbers 0 or 1 until each of them has written 2002 numbers
(to produce a 4004digit number). Brenda is the winner if the
sequence of zeros and ones, considered as a binary number
(i.e., written to base 2), can be written as the sum of
two integer squares. Otherwise, the winner is Andrew. Prove that
the second player, Brenda, can always win the game, and explain her
winning strategy (i.e., how she must play to ensure winning
every game).

153.

(a) Prove that, among any 39 consecutive natural
numbers, there is one the sum of whose digits (in base 10) is
divisible by 11.


(b) Present some generalizations of this problem.

154.

(a) Give as neat a proof as you can that, for any
natural number
$n$, the sum of the squares of the numbers
$1,2,\dots ,n$ is equal to
$n(n+1)(2n+1)/6$.


(b) Find the least natural number
$n$ exceeding 1 for which
$({1}^{2}+{2}^{2}+\dots +{n}^{2})/n$ is the square of a natural number.

155.

Find all real numbers
$x$ that satisfy the equation
${3}^{[(1/2)+\mathrm{log}{}_{3}(\mathrm{cos}x+\mathrm{sin}x)]}{2}^{\mathrm{log}{}_{2}(\mathrm{cos}x\mathrm{sin}x)}=\sqrt{2}\hspace{1em}.$
[The logarithms are taken to bases 3 and 2 respectively.]

156.

In the triangle
$\mathrm{ABC}$, the point
$M$ is from the
inside of the angle
$\mathrm{BAC}$ such that
$\angle \mathrm{MAB}=\angle \mathrm{MCA}$
and
$\angle \mathrm{MAC}=\angle \mathrm{MBA}$. Similarly, the point
$N$
is from the inside of the angle
$\mathrm{ABC}$ such that
$\angle \mathrm{NBA}=\angle \mathrm{NCB}$ and
$\angle \mathrm{NBC}=\angle \mathrm{NAB}$. Also, the point
$P$ is from the inside of the angle
$\mathrm{ACB}$ such that
$\angle \mathrm{PCA}=\angle \mathrm{PBC}$ and
$\angle \mathrm{PCB}=\angle \mathrm{PAC}$.
(The points
$M$,
$N$ and
$P$ each could be inside or outside of
the triangle.) Prove that the lines
$\mathrm{AM}$,
$\mathrm{BN}$ and
$\mathrm{CP}$ are concurrent
and that their intersection point belongs to the circumcircle of the
triangle
$\mathrm{MNP}$.