Some random assorted (part A) problems in algebra for RMO and INMO training

You might want to take a serious shot at each of these. In the first stage of attack, apportion 15 minutes of time for each problem. Do whatever you can, but write down your steps in minute detail. In the last 5 minutes, check why the method or approach does not work. You can even ask — or observe, for example, that if surds are there in an equation, the equation becomes inherently tough. So, as a child we are tempted to think — how to get rid of the surds ?…and so on, thinking in math requires patience and introversion…

So, here are the exercises for your math gym today:

1) Prove that if x, y, z are non-zero real numbers with x+y+z=0, then

\frac{x^{2}+y^{2}}{x+y} + \frac{y^{2}+z^{2}}{y+z} + \frac{z^{2}+x^{2}}{x+z} = \frac{x^{3}}{yz} + \frac{y^{3}}{zx} + \frac{z^{3}}{xy}

2) Let a b, c, d be complex numbers with a+b+c+d=0. Prove that

a^{3}+b^{3}+c^{3}+d^{3}=3(abc+bcd+adb+acd)

3) Let a, b, c, d be integers. Prove that a+b+c+d divides

2(a^{4}+b^{4}+c^{4}+d^{4})-(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd

4) Solve in complex numbers the equation:

(x+1)(x+2)(x+3)^{2}(x+4)(x+5)=360

5) Solve in real numbers the equation:

\sqrt{x} + \sqrt{y} + 2\sqrt{z-2} + \sqrt{u} + \sqrt{v} = x+y+z+u+v

6) Find the real solutions to the equation:

(x+y)^{2}=(x+1)(y-1)

7) Solve the equation:

\sqrt{x + \sqrt{4x + \sqrt{16x + \sqrt{\ldots + \sqrt{4^{n}x+3}}}}} - \sqrt{x}=1

8) Prove that if x, y, z are real numbers such that x^{3}+y^{3}+z^{3} \neq 0, then the ratio \frac{2xyz - (x+y+z)}{x^{3}+y^{3}+z^{3}} equals 2/3 if and only if x+y+z=0.

9) Solve in real numbers the equation:

\sqrt{x_{1}-1} = 2\sqrt{x_{2}-4}+ \ldots + n\sqrt{x_{n}-n^{2}}=\frac{1}{2}(x_{1}+x_{2}+ \ldots + x_{n})

10) Find the real solutions to the system of equations:

\frac{1}{x} + \frac{1}{y} = 9

(\frac{1}{\sqrt[3]{x}} + \frac{1}{\sqrt[3]{y}})(1+\frac{1}{\sqrt[3]{x}})(1+\frac{1}{\sqrt[3]{y}})=18

More later,
Nalin Pithwa

PS: if you want hints, do let me know…but you need to let me know your approach/idea first…else it is spoon-feeding…

A Primer: Generating Functions: Part II: for RMO/INMO 2019

We shall now complicate the situation a little bit. Let us ask for the combinations of the symbols \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} with repetitions of each symbol allowed once more in the combinations. For example, let there be only two symbols \alpha_{1}, \alpha_{2}. Let us look for combinations of the form:

\alpha_{1}, \alpha_{2}, \alpha_{1}\alpha_{2}, \alpha_{1}\alpha_{1}, \alpha_{2}\alpha_{2}, \alpha_{1}\alpha_{1}\alpha_{2}, \alpha_{1}\alpha_{2}\alpha_{2}, \alpha_{1}\alpha_{1}\alpha_{2}\alpha_{2}

where, in each combination, each symbol may occur once, twice, or not at all. The OGF for this can be constructed by reasoning as follows: the choices for \alpha_{1} are not-\alpha_{1}, \alpha_{1} once, \alpha_{1} twice. This is represented by the factor (1+\alpha_{1}t+\alpha_{1}^{2}t^{2}). Similarly, the possible choices for \alpha_{2} correspond to the factor (1+\alpha_{2}t+\alpha_{2}^{2}t^{2}). So, the required OGF is (1+\alpha_{1}t+\alpha_{1}^{2}t)(1+\alpha_{2}t+\alpha_{2}^{2}t^{2})

On expansion, this gives : 1+(\alpha_{1}+\alpha_{2})t+(\alpha_{1}\alpha_{2}+\alpha_{1}^{2}+\alpha_{2}^{2})t^{2}+(\alpha_{1}^{2}\alpha_{2}+\alpha_{1}\alpha_{2}^{2})t^{3}+(\alpha_{1}^{2}\alpha_{2}^{2})t^{4}

Note that if we omit the term 1 (which corresponds to not choosing any \alpha), the other 8 terms correspond to the 8 different combinations listed in (*). Also, observe that the exponent r of the t^{r} tells us that the coefficient of t^{r} has the list or inventory of the r-combinations (under the required specification — in this case, with the restriction on repetitions of symbols) in it:

\bf{Illustration}

In the light of the foregoing discussion, let us now take up the following question again: in how many ways, can a total of 16 be obtained by rolling 4 dice once?; the contribution of each die to the total is either a “1” or a “2” or a “3” or a “4” or a “5” or a “6”. The contributions from each of the 4 dice have to be added to get the total — in this case, 16. So, if we write: t^{1}+t^{2}+t^{3}+t^{4}+t^{5}+t^{6}

as the factor corresponding to the first die, the factors corresponding to the other three dice are exactly the same. The product of these factors would be:

(*) (t+t^{2}+t^{3}+t^{4}+t^{5}+t^{6})^{4}

Each term in the expansion of this would be a power of t, and the exponent k of such a term t^{k} is nothing but the total of the four contributions which went into it. The number of times a term t^{k} can be obtained is exactly the number of times k can be obtained as a total on a throw of the four dice. So, if \alpha_{k} is the coefficient of t^{k} in the expansion, \alpha_{16} is the answer for the above question. Further, since (*) simplifies to (\frac{t(1-t^{6})}{1-t})^{4}, it follows that the answer for the above question tallies with the coefficient specified in the following next question: calculate the coefficient of t^{12} in (\frac{(1-t^{6})}{(1-t)})^{4}.6

Now, consider the following problem: Express the number N(n,p) of ways of obtaining a total of n by rolling p dice, as a certain coefficient in a suitable product of binomial expansions in powers of t. [ this in turn, is related to the observation that the number of ways a total of 16 can be obtained by rolling 4 dice once is the same as the coefficient of t^{12} in (\frac{1-t^{6}}{1-t})^{4}]:

So, we get that N(n,p)= coefficient of t^{n-p} in (\frac{1-t^{6}}{1-t})^{p}

Let us take an example from a graphical enumeration:

A \it {graph} G=G(V,F) is a set V of vertices a, b, c, …, together with a set E=V \times V of \it {edges} (a,b), (a,a), (b,a), (c,b), \ldots If (x,y) is considered the same as (y,x), we say the graph is \it{undirected}. Otherwise, the graph is said to be \it{directed}, and we say ‘(a,b) has a direction from a to b’. The edge (x,x) is called a loop. The graph is said to be of order |V|.

If the edge-set E is allowed to be a multiset, that is, if an edge (a,b) is allowed to occur more than once, (and, this may be called a ‘multiple edge’), we refer to the graph as a general graph.

If \phi_{5}(n) and \psi_{5}(n) denote the numbers of undirected (respectively, directed) loopless graphs of order 5, with n edges, none of them a multiple edge, find the series \sum \phi_{5}(n)t^{n} and \sum \psi_{5}(n)t^{n}.

Applying our recently developed techniques to the above question, a graph of 5 specified vertices is uniquely determined once you specify which pairs of vertices are ‘joined’. Suppose we are required to consider only graphs with 4 edges. This would need four pairs of vertices to be selected out of the total of 5 \choose 2 equal to 10 pairs that are available. So selection of pairs of vertices could be made in 10 \choose 4 ways. Each such selection corresponds to one unique graph, with the selected pairs being considered as edges. More informally, having selected a certain pairs of vertices, imagine that the vertices are represented by dots in a diagram and join the vertices of each selected pair by a running line. Then, the “graph” becomes a “visible” object. Note that the number of graphs is just the number of selections of pairs of vertices. Hence, \phi_{5}(4)=10 \choose 4.

Or, one could approach this problem in a different way. Imagine that you have a complete graph on 5 vertices — the “completeness” here means that every possible pair of vertices has been joined by an edge. From the complete graph which has 10 edges, one has to choose 4 edges — any four, for that matter — in order to get a graph as required by the problem.

On the same lines for a directed graph, one has a universe of 10 by 2, that is, 29 edges to choose from, for, each pair x,y gives rise to two possible edges (x,y) and (y,x). Hence,

\psi_{5}(4)=20 \choose 4.

Thus, the counting series for labelled graphs on 5 vertices is 1 + \sum_{p=1}^{10} {10 \choose p}t^{p}
and the counting series for directed labelled graphs on 5 vertices is
1+ \sum_{p=1}^{20}{20 \choose p}t^{p}.

Finally, the OGF for increasing words on an alphabet {a,b,c,d,e} with a<b<c<d<e is

(1+at+a^{2}t^{2}+\ldots)(1+bt+b^{2}t^{2}+\ldots)(1+ct+c^{2}t^{2}+\ldots)\times (1+dt+d^{2}t^{2}+\ldots)(1+et+e^{2}t^{2}+\ldots)

The corresponding OE is (1+t+t^{2}+t^{3}+\ldots)^{5} which is nothing but (1-t)^{-5} (this explains the following problem: Verify that the number of increasing words of length 10 out of the alphabet \{a,b,c,d,e \} with a<b<c<d<e is the coefficient of t^{10} in (1-t)^{-5} ).

We will continue this detailed discussion/exploration in the next article.

Until then aufwiedersehen,
Nalin Pithwa

Prof. Tim Gowers’ on recognising countable sets

https://gowers.wordpress.com/2008/07/30/recognising-countable-sets/

Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.

Prof. Tim Gowers’ on functions, domains, etc.

https://gowers.wordpress.com/2011/10/13/domains-codomains-ranges-images-preimages-inverse-images/

Thanks a lot Prof. Gowers! Math should be sans ambiguities as far as possible…!

I hope my students and readers can appreciate the details in this blog article of Prof. Gowers.

Regards,
Nalin Pithwa

More tough combinatorics tutorial for RMO: continued

Practice. Practice. Practice.

1) In how many ways, can a total of 16 be obtained by rolling 4 dice once?

2) Calculate the coefficient of t^{12} in (\frac{1-t^{6}}{1-t})^{4}.

3) Observe the identity of answers to Problem 1 and 2. If you think it is a mere coincidence, experiment with other numbers in place of 16 and 4. If you do not think it is a mere coincidence, explain it and go to Problem 4. If you cannot explain its identity, experiment with other numbers.

4) Express the number N(n,p) of ways of obtaining a total of n by rolling p dice, as a certain coefficient in a suitable product of binomial expansion in powers of t. If you cannot do this problem, go to problem 3.

5) Verify that the number of increasing words of length 10 out of the alphabet (a,b,c,d,e) with a<b<c<d<e is the coefficient of t^{10} in (1-t)^{-5}. Try to explain why this is so.

6) A child has a store of toy letters consisting of 4 A's, 3 B's, and 2 E's. (6a) How many different increasing four-letter words (given A<B<E) can it make? The child does not worry about the meanings of the words. (6b) Show that there exists a bijection between the set of increasing four-letter words of all possible lengths and the set of terms of the expansion (1+A+A^{2}+A^{3})(1+B+B^{2}+B^{3})(1+E+E^{2}). (6c) Can you obtain the answer to (6a) as certain numerical coefficient in a suitable expansion of products?

7) The Parliament of India, in which there are n parties represented, wants to form an all-party committee (meaning, a committee in which there is at least one member from each party) of r \geq n members. Assuming (a) that there exist sufficient members available for the committee in each party and (b) that the members of each party are indistinguishable among themselves(!), find the number of distinct ways of forming the committee. Show that this number may be expressed as a suitable coefficient in the binomial expansion of (1-t)^{n-r}

8) Find the number N of permutations of the multiset (a,a,b,b,b) taken three at a time. If you have to express N as the coefficient t^{3} or t^{3}/3! or 3!t^{3} (the choice is yours) in the expansion of one of
8a) (1+t)^{2}(1+t)^{3}
8b) (1+t)^{-2}(1+t)^{-3}
8c) (1+t+t^{2})(1+t+t+t^{2})
8d) (1+t+\frac{t^{2}}{2!})(1+t+\frac{t^{2}}{2!}+\frac{t^{3}}{3!})

which one would you choose? Can you justify your choice without using the value of N?

9) A family of 3, another family of 2, and two bachelors go for a joy ride in a giant wheel in which there are three swings A, B, C. In how many ways can they be seated in the swings (assuming there are sufficient number of seats in each swing) if the families are to be together? List all the ways.

10) n lines in a plane are in general position, that is, no two are parallel and no three are concurrent. What is the number of regions into which they divide the plane?

11) If (a_{n}) is a sequence of numbers satisfying a_{n}-na_{n-1}=-(a_{n-1}-(n-1)a_{n-2}), find a_{n}, given that a_{0}=1 and a_{1}=0.

12) If a_{n} is the number of ways in which we can place parentheses to multiply the n numbers x_{1}, x_{2}, \ldots, x_{n} on a calculator, find a_{n} in terms of the a_{k}‘s where k=1,2, \ldots, (n-1).

13) A graph G=G(V,E) is a set V of vertices a, b, c, \ldots, together with a set E=V \times V of edges (a,b), (a,a), (b,a), (c,b), \ldots. If (x,y) is considered the same as (y,x), we say this graph is undirected. Otherwise, the graph is said to be directed, and we say ‘(a,b) has a direction from a to b’. The edge (x,x) is called a loop. The graph is said to be of order |V|.

If the edge-set E is allowed to be a multiset, that is, if an edge (a,b) is allowed to occur more than once (and this may be called a ‘multiple edge’), we refer to the graph as a general graph.

\phi_{5}(n) and \psi_{5}(n) denote the numbers of undirected (respectively,directed) loopless graphs of order 5, with n edges, none of them a multiple edge, find the series \sum \phi_{5}(n) t^{n} and \sum \psi_{5}(n) t^{n}.

Cheers,
Nalin Pithwa.

Eight digit bank identification number and other problems of elementary number theory

Question 1:

Consider the eight-digit bank identification number a_{1}a_{2}\ldots a_{8}, which is followed by a ninth check digit a_{9} chosen to satisfy the congruence

a_{9} \equiv 7a_{1} + 3a_{2} + 9a_{3} + 7a_{4} + 3a_{5} + 9a_{6} + 7a_{7} + 3a_{8} {\pmod {10}}

(a) Obtain the check digits that should be appended to the two numbers 55382006 and 81372439.

(b) The bank identification number 237a_{4}18538 has an illegitimate fourth digit. Determine the value of the obscured digit.

Question 2:

(a) Find an integer having the remainders 1,2,5,5 when divided by 2, 3, 6, 12 respectively (Yih-hing, died 717)

(b) Find an integer having the remainders 2,3,4,5 when divided by 3,4,5,6 respectively (Bhaskara, born 1114)

(c) Find an integer having remainders 3,11,15 when divided by 10, 13, 17, respectively (Regiomontanus, 1436-1476)

Question 3:

Question 3:

Let t_{n} denote the nth triangular number. For which values of n does t_{n} divide t_{1}^{2} + t_{2}^{2} + \ldots + t_{n}^{2}

Hint: Because t_{1}^{2}+t_{2}^{2}+ \ldots + t_{n}^{2} = t_{n}(3n^{3}+12n^{2}+13n+2)/30, it suffices to determine those n satisfying 3n^{3}+12n^{2}+13n+2 \equiv 0 {\pmod {2.3.5}}

Question 4:

Find the solutions of the system of congruences:

3x + 4y \equiv 5 {\pmod {13}}
2x + 5y \equiv 7 {\pmod {13}}

Question 5:

Obtain the two incongruent solutions modulo 210 of the system

2x \equiv 3 {\pmod 5}
4x \equiv 2 {\pmod 6}
3x \equiv 2 {\pmod 7}

Question 6:

Use Fermat’s Little Theorem to verify that 17 divides 11^{104}+1

Question 7:

(a) If gcd(a,35)=1, show that a^{12} \equiv {\pmod {35}}. Hint: From Fermat’s Little Theorem, a^{6} \equiv 1 {\pmod 7} and a^{4} \equiv 1 {\pmod 5}

(b) If gcd(a,42) =1, show that 168=3.7.8 divides a^{6}-1
(c) If gcd(a,133)=gcd(b,133)=1, show that 133| a^{18} - b^{18}

Question 8:

Show that 561|2^{561}-1 and 561|3^{561}-3. Do there exist infinitely many composite numbers n with the property that n|2^{n}-2 and n|3^{n}-3?

Question 9:

Prove that any integer of the form n = (6k+1)(12k+1)(18k+1) is an absolute pseudoprime if all three factors are prime; hence, 1729=7.13.19 is an absolute pseudoprime.

Question 10:

Prove that the quadratic congruence x^{2}+1 \equiv 0 {\pmod p}, where p is an odd prime, has a solution if and only if p \equiv {pmod 4}.

Note: By quadratic congruence is meant a congruence of the form ax^{2}+bx+c \equiv 0 {\pmod n} with a \equiv 0 {\pmod n}. This is the content of the above proof.

More later,
Nalin Pithwa.