Indian Statistical Institute, ISI MMATH 2023 PMB Solutions & Discussions
ISI MMath PMB 2023 Subjective Questions, solutions and discussions
Problem 1.
Let . Prove that there exists
such that abc
for all
.
Topic: Set Theory, Inequalitites
Difficulty level: Easy
Solution:
Let By the AM-GM inequality
Hence, for all
Equality in the AM-GM inequality is achieved if and only if all numbers
are equal. Hence, to get the maximal possible product
among all
we need to look for
with identical coordinates. If
and
then
and
Let
Since
we have Further,
for all
Problem 2.
Prove that is a solution of the differential equation
on for some nonzero real polynomials
if and only if
.
Topic: Differential Equations
Difficulty Level: Medium
Solution:
In this solution is assumed to be an integer.} Since
must be twice differentiable on
we need to have
If
then
If for
then
for all
which is not the case.
Let If we take
and
then
So, is a solution to the equation of interest.
Assume that
is a solution of
Then
The function is a polynomial that is equal to zero on the interval
It follows that all coefficients of
are equal to zero. However, the coefficient of
near
equals
The obtained contradiction shows that the only function
which is a solution of the equation
is (i.e. when
).
Problem 3.
Let and
. Consider
and
as metric spaces via their inclusions in
and
respectively. Construct a bijection
such that both,
and
are contimuous.
Topic: Metric Space
Difficulty Level: Medium
Solution:
Consider the mapping defined by
This mapping is well defined, as
The mapping is continuous, as functions
and
are continuous. The mapping
defined by
is also continuous (as for
). In fact,
Indeed,
since
and
So, the mapping is the needed mapping.
Problem 4.
Let . A system consists of
components, each of which functions independently with probability
. The system is said to function if exactly
of the
components function. Compute the conditional probability that at least one of the first two components function given that the system functions.
Topic: Probability
Difficulty Level: Easy
Solution:
The system functions if and only if of the
components function, and other
components do not function. There are
ways to choose
components that function, hence the probability that the system functions equals
The system functions while first two components do not function if and only if there are exactly
components among components
that function. So, the probability that the system functions and first two components do not function equals
It follows that the probability that first two components do not function conditionally on the event that the system functions equals
Correspondigly, the probability that at least one of the first two components function given that the system functions equals
Problem 5.
Let be a linear transfarmation where
is a positive integer. Prove that
for all
.
Topic: Linear Algebra
Difficulty Level: Medium
Solution:
Denote by the identical operator. Since
we have that
Let denote the dimension of the subspace
Then
If all inequalities are strict, then
which is impossible. So, there exists such that
hence
We will verify by induction that for all
For
this is obvious. If
then
where the latter equality follows from the choice of So, for all
Problem 6.
Let be a square matrix with values in
such that every column has at most one +1 and at most one -1 . Show that
.
Topic: Linear Algebra
Difficulty Level: Medium
Solution:
Denote by the size of
i.e.
is
matrix. We will prove the result by induction on
If
then
and Assume the needed statement is true for
and let
By assumption, each column has at most one and at most one
Consider following cases.
Case 1. There exists a column with zero elements. In other words, either all elements of the column are equal to zero, or there exists one element which is equal to
or to
Then there are indices
such that
Expanding the determinant with respect to -th column we find that
where the matrix is obtained by deleting
-th row and
th column from the matrix
. Every column of the matrix
is contained in a column of the matrix
Hence, every column of the matrix
contains at most one
and at most one
By inductive assumption,
and
Case 2. Every column of the matrix contains exactly one
and exactly one
Consider the sum of all rows of the matrix
For each
Indeed, in the latter sum there is one one
and
zeros. Hence, rows of the matrix
are linearly dependent and
In any case,
Problem 7.
Let be a finite group having an odd number of elements. Suppose
is an element of
of order 3 such that the cyclic subgroup
generated by
is normal in
. Prove that
commutes with every element of
.
Topic: Group Theory
Difficulty Level: Easy
Solution:
Let Since
is a normal subgroup of
In particular,
If then
which is not the case.
Assume that The order of
is an odd number. Denote it by
Then
We will prove by induction that for
For
this was already shown. Assume the result is proved for
Then
Hence, the result is true for all Taking
we get
which is impossible. So,
Hence, and
commutes with all elements of
Problem 8.
Let be odd primes such that
divides
. If
is a positive integer coprime to
, prove that
divides
.
Topic: Number Theory
Difficulty Level: Easy
Solution:
Remark. We need to assume that If
then the statement is false. For example, we can take
Then
is not divisible by
}
By Fermat’s little theorem divides
So, it is enough to prove that
divides
Since
divides
it follows that
for some integer
Hence
It follows that divides
By Fermat’s little theorem
divides
Hence,
divides
and
divides