Indian Statistical Institute, ISI BStat & BMath 2023 UGB Solutions & Discussions

Problem 1.

Determine all integers such that every power of has an odd number of digits.

Solution:

We will prove that all powers of and integer have odd number of digits if and only if for some We note that the number of digits of an integer is exactly

Indeed, if then and has exactly digits. If then for any the number of digits of is odd:

Assume that all powers of have odd number of digits, but is not of the form If is a power of then is an odd power of hence has even number of digits. This is impossible.
So, is not a power of and is not an integer number. Let us write where is an integer and Since has odd number of digits, we deduce that

is odd. In particular, is even.

If for some integer then

is even (since is even). So, for any and for some integer But then has even number of digits:

The latter relation follows from the fact that

Problem 2.

Let and be defined inductively by

(a) Show that for .

and determine .

(b) Using (a) or otherwise, calculate

Solution:

(a) We note that If then

So, Since it follows that

(b)

since

Problem 3.

In a triangle , consider points and on and . respectively, and assume that they do not coincide with any of the vertices . If the segments and intersect at , consider the areas of the quadrilateral and the triangles , respectively.

(a) Prove that .

(b) Determine in terms of .

Solution:

(a) Consider altitudes of length respectively:

Denote length of by respectively, and let be the length of Then

The inequality is equivalent to

or to

After simplifications we obtain, equivalently,

Since the latter is equivalent to

or to

From triangles and we find that

From triangles and we find that

It remains to prove that

But it follows from simple inequalities

(b) Consider a segment and let be the area of be the area of

In particular, Triangles and share the same altitude on the line Hence, Triangles and share the same altitude on the line Hence, It follows that

Similarly, considering triangles and and and we deduce that

Hence,

Further,

Problem 4.

Let , be distinct natural numbers each of which has exactly 2023 positive integer factors. For instance, has exactly 2023 positive integer factors . Assume that no prime larger than 11 divides any of the ‘s. Show that there must be some perfect cube among the ‘s. You may use the fact that

Solution:

Each number has at most five distinct prime divisors: If where then the number of divisors of equals On the other hand,

It follows that Assume that there are no perfect cubes among If for some then the number of divisors of equals

Hence and is a perfect cube. If where are prime numbers among and then

If then and Hence is a perfect cube. The same holds when It follows that either or Equivalently, either or The number of possible combinations for is

If where are prime numbers among and then

Hence, values of are The number of possible combinations for is

There are at most natural numbers that are not perfect cubes, that have exactly 2023 divisors, and whose prime divisors are It follows that at least one of the numbers is a perfect cube.

Problem 5.

There is a rectangular plot of size . This has to be covered by three types of tiles – red, blue and black. The red tiles are of size , the blue tiles are of size and the black tiles are of size . Let denote the number of ways this can be done. For example, clearly because we can have either a red or a blue tile. Also, since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.

(a) Prove that for all .

(b) Prove that for all .
Here,

for integers .

Solution:

(a) At first we note that Indeed, if the first square is covered either by a red tile or by a blue tile, then there remains squares more to cover. If the first two squares are covered by a black tile, then there remains squares to be covered.

Consider a plot of size If the middle square is covered either by a red tile or by a blue tile, then there remains squares on the left and squares on the right of it to be covered. This gives possibilities. If the middle square is covered by a black tile, then either there remains squares on the left and squares on the right of this tile to be covered, or there remains squares on the left and squares on the right of this tile to be covered. This gives possibilities. Totally, we get

(b) Consider covering with exactly black tiles (in particular, ). Assume that these tiles occupy squares Remaining squares are partitioned by black tiles into groups (possibly empty):

Conversely, every such collection of groups define a configuration of black tiles. The number of such collections equals the number of non-negative integer solutions of the equation

which is known to be equal to When black tiles are placed, there remains squares to be covered by red or blue tiles. There are possibilities to do this. Totally, we get

Problem 6.

Let be a sequence of real numbers defined as and

Prove that for all .

Solution:

The function is strictly increasing on Since for all and it follows that for all We will verify the inequality by induction on If the inequality is true. If the inequality is also true, as

Assume the inequality was verified for Then

It is enough to prove that Or, equivalently,

Taking squares of both sides we obtain, equivalently,

or,

The inequality is true, since

Problem 7.

(a) Let be an integer. Prove that can be written as a polynomial with integer coefficients in the variables , and .

(b) Let . where are real numbers such that is an integral multiple of . Using (a) or otherwise, show that if , then for all positive integers .

Solution:

(a) We prove the result by induction on If then If then

If then

Assume that for every there exists a polynomial with integer coefficients in three variables, such that

Then

Hence,

Hence, for every can be expressed as a polynomial in with integer coefficients.

(b) According to part (a), for every there exists a polynomial in three variables with integer coefficients, such that

Moreover,

We note that

The first summand in the numerator is expressed as a polynomial applied to quantities

The second summand in the numerator is expressed as a polynomial applied to quantities

where is a conjugate of a complex number

Hence, we get a representation

It is enough to show that in order to deduce that all

Since and it follows that Since it follows that

Finally, since is an integral multiple of then

From relations it follows that for all

Problem 8.

Let be a continuous function which is differentiable on . Prove that either is a linear function or there exists such that .

Solution :

Assume that for all If then is constant. Assume that By the mean value theorem,

Assume that for some Consider the function Since then by the mean value theorem

Hence,

and

This is a contradiction. Hence, for all and

It follows that

If it is enough to apply the proved result to the fucntion