**POLYA033: (2002 Putnam A-Problems, 12/7/02). **

**[Putnam 2002:A1] ****Let k be a fixed positive integer. The n-th
derivative of 1/(x^k-1) has the form P_n(x)/(x^k-1)^{n+1} where
P_n(x) is a polynomial. Find P_n(1).**

[Putnam 2002:A2] **
Given any five points on a sphere, show that some four of them
must lie on a closed hemisphere.**

[Putnam 2002:A3] Let n>= 2 be an integer and T_n be the number
of nonempty subsets of S = {1, 2, 3, ..., n} with the property that
the average of the elements of S is an integer. Prove that T_n-n is always
even.

**[Putnam 2002:A4] **** In Determinant Tic-Tac-Toe, Player
1 enters a 1 in an empty 3 by 3 matrix. Player 0 counters with a 0 in
a vacant position, and play continues in turn until the 3 by 3 matrix
is completed with five 1's and four 0's. Player 0 wins if the determinant
is 0 and player 1 wins otherwise. Assuming both players pursue optimal
strategies, who will win and how?**

[Putnam 2002:A5] Define a sequence by a_0 = 1,together with
the rules a_{2n+1} = a_n and a_{2n+2} = a_n + a_{n+1} for each integer
n >= 0. Prove that every positive rational number appears in the set
**{a_{n-1}/a_n: n >=
1 } = {1/1, 1/2, 2/1, 1/3, 3/2, ... }.**

**[Putnam 2002:A6] ****Fix an integer b >= 2. Let f(1)
= 1, f(2) = 2, and for each n >= 3, define f(n) = n f(d), where d
is the number of base-b digits of n. For which values of b does
Sum_{n=1,... infty} 1/f(n) converge?
**

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**Discussion. [Problem A1] **** Let k be a fixed positive
integer. The n-th derivative of 1/(x^k-1) has the form P_n(x)/(x^k-1)^{n+1}
where P_n(x) is a polynomial. Find P_n(1).**

** *************************************************************************************************

**[Problem A2] **** Given any five points on a
sphere, show that some four of them must lie on a closed hemisphere.**

**[SCL, 12/8/02]: **Take the great circle C through any two points.
(If all five points are identical, we are obviously done.) C divides
the surface of the sphere, S, into two regions R and T. There are (at
most) three points in R union T, and thus, without loss of generality,
T has at most one point. Therefore, R union C is a closed hemisphere containing
at least four of the points.

**[JPE, 12/8/02]: **If P is a plane through the center of the sphere
and two of the given points, obviously, two at least of the three other
points lie on one of the closed hemispheres limited by P.

** *************************************************************************************************

**[Problem A3] Let n>= 2 be an integer and T_n
be the number of nonempty subsets of S = {1, 2, 3, ..., n} with the
property that the average of the elements of S is an integer. Prove that
T_n-n is always even.**

** [JPE, 12/8/02]:** Let E be the set of the subsets S of {1,2,..n}
such that

- S has at least 2 elements

- the average m(S) of the elements of S is an integer.

Let E1 (resp E2) be the set of the elements S of E such that m(S) is (resp is not) an element of S.

The mapping u| S -> S union {m(S)} is a bijection from E2 to E1 ( with reciprocal bijection S -> S-{m(S)}).

As E = E1 union E2 has T_n - n elements, the result follows.

[SCL, 12/7/02]: If Player 2 can get
3 zeroes in a row, horizontally or vertically, he will win. He will also
win if he can get some $2\times 2$ submatrix of zeroes, since then there
is either a row or all zeroes, or two identical rows. There is one more
way to get two identical rows (or columns), and that is if each has exactly
one zero and that is in the same column (row) for both rows (columns).

Since we are only interested in whether or not the determinant
the determinant is zero, we can rearrange rows and columns and take
transposes as needed without changing this property. Thus, without loss
of generality, Player 1 moves into cell (1,1). Player 2 has, up to rearrangements,
two possible responses: He can move to cell (1,2) or to cell (2,2).

Let us analyze the case in which Player 2 takes the cell (2,2)
(since this is sufficient). There are now, up to symmetries, four possibilities
for Player 1: cells (1,2),(1,3),(2,3),(3,3). We will assume that Player
1, whenever possible will move to block if Player 2 has a row with two
zeroes, and that Player 2 will move to win if he already has a row with
two zeroes. With these caveats, here are the responses for Player 2 in
each of the remaining four situations. As can now be seen, Player 2 can
force a win by creating a $2\times 2$ submatrix of zeroes.

Responses to (1,2): Player 2 now takes (2,3), (3,3), (3,2) in
this order.

Responses to (1,3): Player 2 now takes (2,3), (3,2), (3,3) in this
order.

Responses to (2,3): Player 2 now takes (3,2), then (3,1), then
(2,1).

Responses to (3,3): Player 2 now takes (1,2), then (2,3), then
(1,3).

**[ND, 12/10/02] (with minor corrections on 12/11/02):** We''ll
prove that the second player wins.

Interchanging rows and columns we can call the element the first player
selects A1. The rows now are Ai, Bi, Ci, i = 1, 2, 3 and the expansion of
the determinant is

D = A1B2C3-A1B3C2͠看A2B1C3Ͱ甋A3B2C1.

The second player selects C3 = 0 and again one of its neighbors say C2
= 0. Now the only terms in D that are not zero are

T1 = A2B3C1 and T2 = -A3B2C1 and the best case for the first player is
to be C1 = 1.

Hence whatever the first player selects, the second player can make zero
both terms T1, T2 in his third and fourth selection.

** *************************************************************************************************

**[Problem A5] **** Define a sequence by a_0 = 1,together
with the rules a_{2n+1} = a_n and a_{2n+2} = a_n + a_{n+1} for each
integer n >= 0. Prove that every positive rational number appears in
the set
**

[SCL, 12/8/02]:

In that problem, p(k) is the power of 2 in k: that is 2^s divides k and 2^{s+1} does not means p(k)=s. Then the sequence generated is x_0=0, x_n=1/(1+2p(n)-x_{n-1}).

The rationals appear to be generated in exactly the same sequence for both problems (with the exception of the term x_0). thus, a reasonable conjecture is that x(n)=a(n-1)/a(n) for n>=1. This may help in getting a proof.

a)If x>1 and x-1 = a[n-1]/a[n], we have l(x-1)=p<l(x) and x=a[2n]/a[2n+1]

b)If x<1 and x/(1-x) = a[n-1]/a[n], we have l(x/(1-x))=q<l(x) and x=a[2n-1]/a[2n].

The result follows by induction on l(x)

N. Calkin and H. S. Wilf

I suspect that D. Knuth was inspired by this paper and posed that problem 10906.

http://www.math.upenn.edu/~wilf/website/recounting.pdf

May be, this article inspired the author of the exercise but I don't think that we need this article for a solution.

In fact, the article explains the way the authers discovered the sequence giving every positive rationnal.

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**[Problem A6] **** Fix an integer b >= 2. Let f(1)
= 1, f(2) = 2, and for each n >= 3, define f(n) = n f(d), where
d is the number of base-b digits of n. For which values of b does Sum_{n=1,...
infty} 1/f(n) converge?**

As ln(n+1)-ln(n) < 1/n it follows that

sum(1/f(n),n=b^(d-1)..b^d-1) > ln(b)/f(d)

Hence, if sum(1/f(n),n=1..infi) converges to S, we have

S > S ln(b), and the given sum diverges for b>2.

Suppose now that b = 2.

As 1/n < ln(n)-ln(n-1) for n>1, we have

u[d] < ln(2^d - 1) - ln(2^(d-1)-1) .

The right hand side term decreases with d, hence,

u[d]<ln(7/3) for d>2.

We have 1/f(1)+1/f(2)+1/f(3)=1+1/2+1/6=5/3

Put M = 5/3/(1-ln(7/3)).

Suppose that there exists d such as

(1) sum(1/f(n),n=1..2^d-1)>M and consider the smallest value d0 of d for

which (1) is true. Clearly d0>2.

We have, by (1), M < 5/3+ln(7/3)sum(1/f(d),d=3..d0)

< 5/3+ln(7/3)M = M.

Contradiction. Hence, for b=2, the given sum converges.

We can summarize :

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