Try the Free Math Solver or Scroll down to Tutorials!

 Depdendent Variable

 Number of equations to solve: 23456789
 Equ. #1:
 Equ. #2:

 Equ. #3:

 Equ. #4:

 Equ. #5:

 Equ. #6:

 Equ. #7:

 Equ. #8:

 Equ. #9:

 Solve for:

 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
 Ineq. #2:

 Ineq. #3:

 Ineq. #4:

 Ineq. #5:

 Ineq. #6:

 Ineq. #7:

 Ineq. #8:

 Ineq. #9:

 Solve for:

 Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:

# THE SIXTY-NINTH ANNUAL WILLIAM LOWELL PUTNAM EXAMINATION

Problem A5

Let n ≥ 3 be an integer. Let f(x) and g(x) be polynomials with real coefficients such that the
points in R2 are the vertices of a regular n-gon in
counterclockwise order. Prove that at least one of f(x) and g(x) has degree greater than or equal
to n - 1.

Solution: Consider the polynomial p(x) = f(x) + ig(x), whose degree is the maximum of the
degrees of f and g and which maps { 1, 2, ... , n} onto the vertices of this polygon. With a suitable
choice of a complex constant (equal to the center of the polygon), and a nonzero complex number
, we can replace this polynomial with the polynomial , whose degree is the
same as that of p(x), that is, equal to the maximum of the degrees of f(x) and g(x), and which
maps to , in that order, where . We need to show that
the degree of q(x) is at least n - 1.

Suppose not, that is, suppose . The coefficients
satisfy the following linear system of n - 1 equations:

Now this system has a Vandermonde matrix, and can therefore be solved for the coefficients
, which are uniquely determined. (The determinant of the Vandermonde
matrix is   The point of this analysis is that there is one and only one
polynomial of degree at most n - 2 that maps onto in that order.
But that polynomial is obviously

What we need to show is that q(1) cannot be 1. But that is obvious, since

Problem A6

Prove that there exists a constant c > 0 such that in every nontrivial finite group G there exists
a sequence of length at most with the property that each element of G equals the product
of some subsequence. (The elements of G in the sequence are not required to be distinct. A
subsequence of a sequence is obtained by selecting some of the terms, not necessarily consecutive,
without reordering them, for example, 4, 4, 2 is a subsequence of 2, 4, 6, 4, 2, but 2, 2, 4 is not.)

Solution: If we make the convention that the identity element e can be represented as an empty
product, the constant c can be taken equal to 6. With this convention, the restriction to non-trivial
groups can be removed also, since a string of 0 elements will represent the only element in the
trivial group feg, and we'll certainly have 0 ≤ 6 ln(1) = 0.

Let G be any finite group, and let a any element of G different from the identity. Since G is
finite, a has finite order, say k ≥ 2, that is, ak = e. Let r be the unique positive integer such that
. Then each of the elements a j , 0 ≤ j < k (where a0 = e) has a unique representation
as a product of the elements , since each integer j with 1 ≤ j < k has a unique
binary representation as a sum of powers 2i with 0 ≤ i ≤ r - 1. Thus we can now represent the
k distinct elements a j , 0 ≤ j < k, as products of the r elements , 0 ≤ j ≤ r - 1. Notice that
. Thus, at this stage, we can represent k elements of the
group using a string of at most elements, taken from the sequence

If G consists only of the cyclic group H generated by a, we are done, and indeed c = 3 > 2/ ln(2)
will suffice in this case. If H ≠ G, let b be an element not in H, and let l ≥ 2 be the smallest
positive integer such that . Just as above, we know that any power b j , 0 ≤ j < l, can be
represented as a product of the s powers , where . We now form the
sequence

which contains elements. The set K of products of subsequences of
T consists of the elements for which 0 ≤ h < l, 0 ≤ i < k, and 0 ≤ j < l. At least kl
of these are distinct, namely the elements . For if with , then
for some v with 0 ≤ v < k. Since , this means and
hence

If G = K, we are now done, since that means ≥ kl, and we have represented every element
of the group by a product taken from the sequence T, which consists of at most
) elements.

If G ≠ K, let d ∈ G\K, and let m ≥ 2 be the smallest positive integer such that . We
then form the sequence

where , so that .
The products of subsequences of the sequence U form a set L of elements containing all products
of the form

where 0 ≤ f < m, 0 ≤ g < l, 0 ≤ h < k, 0 ≤ i < l, and 0 ≤ j < m. At least klm of these are
distinct, namely the products . For if , with , then
for some w with 0 ≤ w < k. Since , it follows
that , and then, by what was proved about the set K above, that and .

If L = G, we are now done, having represented every element of G as a product taken from
a sequence containing fewer than
6 ln() elements.

It is now clear that this process can be continued until the group G is entirely exhausted.