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MONTHLY PROBLEMS IN MATHEMATICS
October 2010 Problems and September Solutions
1. The sum of the digits (in base 10) is calculated for the number
, and then
the sum of the digits of the resulting number is calculated. This process
continues until a single digit is left. Which digit is this?
(Submitted by Bart Van Steirteghem)
2. Triangle ABC has side lengths equal to a, b, and c. Find a necessary and
sufficient condition for the angles of the triangle such that
can be
the side lengths of a triangle. (Submitted by Henry Ricardo)
3. If
, prove that
(Submitted by Homeira Pajoohesh)
4. Find two functions f and g such that f is differentiable at
and g is not
differentiable at 0. But the composite function
is differentiable at 0.
(Submitted by Jamshid Kholdi)
5. You have six weights. One pair is red, one pair white, one pair blue. In each
pair, one weight is a trifle heavier than the other but otherwise appears to be
exactly like its mate. The three heavier weights (one of each color) all weigh the
same. This is also true of the three lighter weights.
In two separate weighings on a balance scale, how can you identify which is the
heavier weight of each pair?
(Invented by Paul Curry)
Solutions are due October 28
Please send your solutions to:
Mahmoud Sayrafiezadeh
Department of Mathematics
Medgar Evers College/CUNY
If you would like your solution to appear in the Monthly Problems, please e-mail it as a MathType doc
file to [email protected]
Solutions to September Problems
Problem 1. Suppose that the product of k consecutive integers is equal to their
sum. Find all such integers. (Submitted by Philip Ording)
Solution 1 to Problem 1 by Raymond Thomas
The consecutive integers must comprise one of the following sets:
for
or
or
.
Proof: Let
- the product of the
with the integer and let
consecutive integers starting
- the sum of the same
consecutive integers. If 0 is a member of these integers then
and in order that
as well, we must have that the set of integers is of
the form
for
.■
We now examine the case where the integers are all positive .
We start with a lemma: Let
then
for
be distinct real numbers in
.
Proof of the lemma: We have that
and that
,however,
it is an easy induction to prove
for
.
(Alternatively, observe that
is the number of subsets of a set with
elements, and that a set with
elements has among its subsets,
singletons as well as the empty set – in other words, at least subsets and
so the total number of subsets
.)
■
So we have that
This lemma implies that if
equality between
or
, then
is if
- so the only hope for
. We now solve the equation
. It is an easy induction to prove that the left
hand side is strictly greater than the right hand side if
. A trivial check
shows that equality occurs only when
■
There remains the case where all the integers are negative . In this case, the
sum will be negative and in order for the product to be negative it must be
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that is odd. It is easy to show that in this event, the negatives of these
integers are corresponding solutions to the positive case for which there is
only one solution. Thus the only possibility is that the set of integers is
and this works by inspection. ■
It is only for odd values of >1 that we can find consecutive integers for
which
Solution 2 to Problem 1 by Silvius Klein
We have two main possibilities: 0 appears in our string of consecutive integers
or it doesn't.
If 0 does appear, then the product will be zero, and since the sum is zero too,
each positive integer in the string will have a negative counterpart. So in this
case we get the following solutions:
,
,
and
so on, or more generally:
for all
.
Assume now that 0 is not part of the string of consecutive integers, so they are
all positive or all negative. It is clearly enough to consider the case when they
are all positive (since if we had a solution with negative integers, their absolute
values will also form a solution).
Under this assumption, there are two possibilities: that the string of integers
contains (so it starts with) 1, or that it does not contain 1.
In the first case, when it starts with 1, we should have:
But
is at least
, so we should have that
which implies that
. This shows that we could have at most 3 numbers in the string.
does
not work, the sum is 3 and the product is 2, but
works, so that is a
solution (along with
).
Now we consider the case when the string does not contain 1, so all elements are
greater than or equal to 2. We will show that there are no solutions in this case.
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This is based on the following observation: given any two integers
we
have
. That is because
.
By induction we can then easily show that given any
integers (consecutive
or not) which are all greater than or equal to 2 and not all equal to each other,
their product will be strictly greater than their sum.
We conclude that our only solutions
for all
are:
,
and
Solution 3 to Problem 1 by Henry Ricardo
The sum of 1, 2, and 3 equals their product. We claim that there are no other
sets of positive consecutive integers satisfying our condition. In fact, we can
generalize this result.
Let
be a set of positive integers such
that
If k = 1, clearly any singleton
conditions of our problem trivially. Now suppose that
Then we have
satisfies the
We note that
So we must have
This implies that
If k = 2, then we have
It is easy to
see that the only solution is (2, 2), which doesn’t satisfy our hypotheses about
the integers
Now consider the case k = 3. If
then we have
and
must find solutions to
This contradiction forces
Then we
This equation can be rewritten as
Since
are integers, this gives us
If the solution is not restricted to positive integers, then
and sets of the form
the only sets satisfying the given condition.
are
Problem 1 was also solved by Mitch Joseph (Student), Jamshid Kholdi, Bart
Van Steirteghem and Cris Wellington.
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Problem 2. Show that if x and y are positive real numbers, then
(Submitted by Henry Ricardo)
Solution 1 to Problem 2 by Slivius Klein
[Note: The solution is five pages. MS]
We first note that the inequality is obvious if either
or
. We may then
assume that both and are less than 1.
Let us consider the function
. It is easy to verify that this function
is well defined and continuous for all pairs
where
and
except
for, and that is quite important, the origin
. (That is because
does not exist. )
We will have to show that the minimum of this function is greater than 1 (or
equal to 1, since we are including pairs
with
or
). There are two
main strategies for proving this type of inequality:
(S1) Reduce it to a one-dimensional problem, by considering lines (or curves)
that cover a two-dimensional domain, and then, along each such line, solving for
one variable (say ) in terms of the other variable (say ). The one-dimensional
problem can then be approached the usual way, by finding critical points, etc.
(S2) Find the critical points of this two-variable function, by setting the partial
derivatives equal to zero, and evaluate the function at those critical points. Then
find the minimum along the boundary of the domain. The overall minimum will
be the least of the values previously obtained. However, for such an approach to
work, the function would have to be continuous on a bounded (that is fine, since
we only consider pairs
in the unit square) and closed domain - there is a
big problem here, since we need to avoid the origin
.
We will have to use both strategies, each for a different region: (S1) will be used
in a small enough corner near the origin
, while (S2) will be used for the
rest of the unit square, as indicated in the picture below.
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(S1) First we show that if we choose a small enough corner (i.e. triangle) near
, along each line parallel to the hypotenuse the inequality holds. More
precisely, we show that if we choose
small enough, on each line
,
with
, we have
.
Indeed, on such a line, if we solve for , we get
, so our function of two
variables
becomes a function of one variable :
We will show that these functions
are (strictly) concave down, as we can
see in the picture below (which depicts their graphs for several values of the
parameter ). A (strictly) concave down function attains its minimum at one of
the endpoints of its domain (if the domain is a closed interval), and its values are
strictly larger than the minimum at all other points. Clearly
and
pairs
, so if
, then
, which would imply that
in that corner near the origin.
for all
To show that
is concave down, we will compute its second derivative and
show that it is negative. Computing its second derivative is not the easiest task,
since the function is quite complicated, but doing it carefully, or using a
computational tool we get:
We will show that if is small enough, then each of the quantities in the two
brackets above is negative, making the whole expression for
negative.
To simplify a little bit the notations, we will use again for
. Then, after
doing a little bit of simple algebra, the expression for
becomes:
Since
and
, it remains to show that if and are small enough,
. We will regard
as a (quadratic) polynomial in . We know
that a quadratic polynomial with positive leading term is negative when the
variable lies between the roots of the polynomial. We will show that it has
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two real roots: one,
that is negative and one,
that is greater than say,
(assuming is small enough). Choosing then between and will ensure that .
The discriminant of this polynomial is , which after some algebraic
manipulations becomes .
But as , it is easy to check that , so . We can then assume that provided was
chosen small enough, which in particular says that the quadratic equation has
two distinct solutions, (the smaller one) and , the larger one. By the quadratic
formula,
So choosing small enough, can be assumed larger than say, .
Now, the product of the roots is , which for small enough values of can be
shown to be negative. So while .
[The reason for small enough is the following: and if is small enough, then is
large enough, so because the logarithm increases slower than any power or
radical at infinity. Then .]
We have shown that for between 0 and , choosing small enough will ensure
that is between the roots of , so . Similarly, by symmetry, choosing small
enough will ensure that . This ends the proof for the small corner near the origin.
(S2) Now we will handle the remaining region depicted below. The region is
closed and bounded and the function is continuous there (since we are avoiding
the origin). Therefore, it has a minimum that is attained either on the boundary,
or at a critical point inside the domain.
On the boundary of this region, we have already shown that (we have done that
using the concavity argument for the line , on the lines and the function equals
1, and when or , is clearly greater than 1.)
We will now find the critical points and then evaluate at the critical points.
Finding the critical points will be quite tricky. We have:
We set the partial derivatives equal to zero and stare at the equations for a while.
Then the first thing we do is to solve for in the first equation, and for in the
second equation.:
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This shows that is the reciprocal of so
Moreover, we can solve for and get
We will eventually substitute this into the second equation (that we were staring
at earlier) to obtain an equation in the variable only. But before that, we will use
some algebraic manipulations, including taking the logarithm on both sides of
the equation.
Divide by both sides:
Apply the logarithm both sides and get:
. Now substitute for and for to get the following equation in the variable only:
We obtained that has to be a zero for the function
It turns out (see below), that this function is strictly decreasing on the interval .
Because of that, if it had a zero, it would only have one zero.
We do not hope to be able to find that zero directly. However, the crucial
observation is that because of the symmetry, the other variable, , would have to
be a zero for the same function . But this function has only one zero (or at most
one). So . Now we can use the fact that to solve for and and to conclude that
the only critical point of is (the other solution is , which is outside of our
domain).
The value of at this critical point is .
[ The reason the function is decreasing on , is that it is written as a sum of
decreasing functions on that interval:
and each of these two functions can be easily shown to be decreasing on .
which is the composition of the function and . But can be easily shown to be
increasing, while is decreasing, so the composition is decreasing. And finally,
is clearly decreasing. ]
We have shown that for all pairs in the first quadrant, except for the origin,. Can
this function be 1, for >0 and >0 ?
From our whole analysis above, the function could be equal to 1 only when the
pair is on the axis or on the axis, since otherwise 1 would be the minimum
value, so would have to be a critical point. The only critical point is and the
function is greater than 1 there.
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We conclude that whenever and .
Solution 2 to Problem 2 by Henry Ricardo
First we use Bernoulli’s inequality: for It follows that, for We
can write this last result as
Now, if x or y in our original problem is at least 1, we are done. Otherwise, let
In this case we apply the inequality derived in the last paragraph and conclude
that
Solution 3 to Problem 2 by Henry Ricardo
The inequality is trivially true if either x or y is Let and put Because of
the symmetry, we consider only Now
where But has a unique minimum at and is increasing for Since F(0) =
1,
F(1) =
Partial solution to Problem 2 by Raymond Thomas
If either x or y is greater than 1 the inequality is immediate.
There remains to consider the case where x and y are in ( 0 , 1). I have only a
partial solution: If and x, y are in (0,1) we have
and , hence
Problem 3. Prove that no finite field is algebraically closed.
(Note: A field F is algebraically closed if every non-constant polynomial with
coefficients in F has all its roots in F. For example, the field of complex
numbers is algebraically closed, but the field of rational numbers is not.)
(Submitted by Raymond Thomas)
Solution 1 to Problem 3 by Silvius Klein
Let be a finite field. We will define a polynomial over this field, that has no
roots in the field. The definition of is inspired by the way we prove that there
have to be infinitely many prime numbers:
This is a polynomial of degree , so it is not constant. However, if we evaluate it
at every element of the field, we clearly always get 1, so this polynomial has no
roots, showing that the field is not algebraically closed.
9
Solution 2 to Problem 3 by Raymond Thomas
Let have elements. Then , the multiplicative subgroup of has
elements and by Lagrange’s theorem, these satisfy the equation . Hence
every element of satisfies the polynomial equation so none
satisfy . Equivalently, the polynomial has no
roots in so is not algebraically closed . 
Problem 3 was also solved by Jamshid Kholdi, Henry Ricardo, Bart Van
Steirteghem and Cris Wellington. There was one wrong solution.
Problem 4. Let f be a real-valued, differentiable function on with for all .
Prove that
Solution 1 to Problem 4 by Henry Ricardo
Using the given conditions and the Fundamental Theorem of Calculus, we have
Solution 2 to Problem 4 by Silvius Klein
Using Leibniz-Newton's formula (i.e. the fundamental theorem of calculus), we have: .and
the inequality is proven.
Solution 3 to Problem 4 by Raymond Thomas
The stated conditions imply the following: 1)
2) Thus
=.
and
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Problem 5. Choose a point (any point) x on the unit circle. Consider an angle
(any angle) . Rotate the point x counterclockwise by the angle , and get another
point on the circle, say . Rotate by the angle and get the point , rotate by the
angle and get and so on, repeat this process infinitely many times. The
collection of points is called the orbit of x under the rotation by .
Question: does this collection of points fill up the whole circle?
What happens if is a rational multiple of ? (Submitted by Silvius Klein)
Solution 1 to Problem 5 by Raymond Thomas
The orbit of is countable and hence it cannot fill up the entire circle which is an
uncountable set. If is a rational multiple of , say where and have no common
factors, then the orbit is finite of order . On the other
hand, if where is irrational, then the orbit of is dense in the unit circle, i.e. the
closure of the orbit is the unit circle . We now prove:
The orbit of x is dense on the unit circle.
Proof: Denote the full orbit of by . Identifying the unit circle with the complex
numbers of modulus 1, we see that the full orbit of is for and this is an infinite
set (else is a root of unity and hence a rational multiple of contrary to our
assumption). Now assume that the full orbit is not dense, thus (the complement
of the closure of the full orbit ) is a non-empty open subset of .
As such, it is the union of finitely or countably infinitely many disjoint open
arcs. Let be one such arc of maximal length. Now consider the images of this
arc under rotation by : where I’ve used obvious notation.
We observe two things: 1) the sets are subsets of and 2) they are all disjoint –
for if any two coincide, is a root of unity contrary to assumption and if any two
overlap their union is an open interval in which is longer than which would
contradict the maximality of ’s length. But now we are faced with another
impossibility: infinitely many disjoint arcs of the same length on the unit circle –
another contradiction. Thus it must be that the full orbit of is dense in the circle
and hence so too is the set of points (also called the forward orbit ).
Solution 2 to Problem 5 (and comments) by Silvius Klein
Here's a picture describing the first few steps of the process:
11
When the rotation angle is a rational multiple of , after a number of steps (equal
to the denominator of that rational multiple) the points in the orbit of begin to
repeat, so the orbit has only a finite number of elements (see the picture below).
In particular, in this case, the orbit cannot fill up the whole circle.
When the rotation angle is not a rational multiple of , the orbit points are
topologically dense in the circle (between any two pints on the circle, an orbit
point can be found or, in other words, taking successive rotations of by the
angle , we can reach any arc of the circle, regardless how small). This can be
seen in the picture below, but it is not what the problem asks (it could be a
problem for a different month).
What we need to decide is if all these orbit points eventually cover the whole
circle. The answer is no. The set of all orbit points is infinite, but countable
(since it was obtained by performing successive rotations of a point). The circle,
however, has a greater cardinality, it is uncountable, since it is in a one to one
correspondence with the interval of real numbers . Therefore, the orbit cannot
fill up the whole circle.
Many other very interesting questions can be considered about these orbit
points, such as how they are distributed on the circle. The answers to these
questions involve more advanced math topics, such as Ergodic Theory and
Analytic Number Theory.
Problem 5 was also solved by Henry Ricardo and Bart Van Steirteghem
Problem 6. Find all solution of the equation for positive integers x, y and z. Or,
prove no solution exists.
Solution 1 to Problem 6 by Raymond Thomas
There are no positive integer solutions to the equation .
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Proof: First, observe that the function defined by is strictly increasing. Without
loss of generality we may assume , then
the increasing nature of ensures that . Since we also see that and the increasing
nature of implies that whence since all quantities are integers. This last
inequality yields the following: . Transitivity gives . Finally, combining the first
boxed inequality with this last one gives and equality between the first and last
term is impossible. ■
Solution 2 to Problem 6 by Silvius Klein
We show that there are no solutions. Since , and are positive integers, if , then
would have to be (strictly) greater than both and . Let's say that . Then , and we
have:
In this sequence of inequalities there is one which is strict, so we get
which says that the equality is not possible, so there are no solutions.
Solution 3 and 4 to Problem 6 by Henry Ricardo
There is no solution. Suppose there were positive integers x, y, and z such that
. We may assume that If x = y = 1, it is clear that there can be no positive
integer z satisfying the equation. If we have
since
Alternatively, consider the sequence This sequence increases very
rapidly. If then there certainly can’t be positive integers x and y with such that
Problem 6 was also solved by Jamshid Kholdi
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