Find the last two digits of 9^{9^{9}}

Here is a cute example of the power of theory of congruences. Monster numbers can be tamed !!

Question :

Find the last two digits of 9^{9^{9}}.

Solution:

A famous mathematician, George Polya said that a good problem solving technique is to solve an analagous less difficult problem.

So, for example, if the problem posed was “find the last two digits of 2479”. How do we go about it? Find the remainder upon division by 100. Now, how does it relate to congruences ? Modulo 100 numbers !

So, the problem reduces to — find out 9^{9^{9}} \equiv 9 \pmod {10}.

Now, what is the stumbling block…the exponent 9^{9} makes the whole problem very ugly. But,

9^{9} \equiv 9 \pmod {10}, which means 9^{9}-9=10k, that is, 9^{9} = 9 + 10k,

also, use the fact 9^{9} \equiv 89 \pmod {100}

Hence, 9^{9^{9}}=9^{9+10k} = 9^{9}.9^{10k}

9^{9^{9}} \pmod {100}  = 9^{9}.9^{10k} \pmod {100} \equiv 89. 9^{10k} \pmod {100}

So, now we need to compute 9^{10k} \pmod {100} = (9^{10})^{k} \pmod {100} = (89.9)^{k} \pmod {100} = 89^{k}. 9^{k} \pmod {100}

Hence, 9^{9^{9}} \pmod {100} \equiv 89^{k+1}.9^{k} \pmod {100} \equiv (89.9)^{k}.89 \pmod {100} \equiv (1)^{k}.89 \pmod {100} \equiv 89 \pmod {100}.

-Nalin Pithwa.

A beautiful example of use of theory of congruences in engineering

The theory of congruences created by Gauss long ago is used in error control coding or error correction. The theory of congruences is frequently used to append an extra check digit to identification numbers, in order to recognize transmission errors or forgeries. Personal identification numbers of some kind appear in passports, credit cards, bank accounts, and a variety of other settings.

Some banks use (perhaps) an eight-digit identification number a_{1}a_{2}\ldots a_{8} together with a final check digit a_{0}. The check digit is usually obtained by multiplying the digits a_{i} for 1 \leq i \leq 8 by certain “weights” and calculating the sum of the weighted products modulo 10. For instance, the check digit might be chosen to satisfy:

a_{0} \equiv 7a_{1} + 3a_{2} + 9a_{3} + 7a_{4} + 3a_{5} + 9a_{6} + 7a_{7} + 3a_{8} {\pmod 10}

The identification number 815042169 would be printed on the cheque.

This weighting scheme for assigning cheque digits detects any single digit error in the identification number. For suppose that the digit a_{i} is replaced by a different a_{i}. By the manner in which the check digit is calculated, the difference between the correct a_{9} and the new a_9^{'} is

a_{9} - a_9^{'} \equiv k(a_{i} - a_{i}^{'}) \pmod {10}

where k is 7, 3, or 9 depending on the position of a_{i}^{'}. Because k(a_{i}-a_{i}^{'}) \not\equiv 0 \pmod {10}, it follows that a_{9} \neq a_{9}^{'} and the error is apparent. Thus, if the valid number 81504216 were incorrectly entered as 81504316 into a computer programmed to calculate check digits, an 8 would come up rather than the expected 9.

The modulo 10 approach is not entirely effective, for it does not always detect the common error of transposing distinct adjacent entries a and b within the string of digits. To illustrate, the identification numbers 81504216 and 81504261 have the same check digit 9 when our example weights are used. (The problem occurs when |a-b|=5). More sophisticated methods are available, with larger moduli and different weights, that would prevent this possible error.

-Nalin Pithwa.

Elementary problems in Ramsey number theory for RMO

Question 1:

Show that in any group of 6 people there will always be a subgroup of 3 people who are pairwise acquainted or a subgroup of 3 people who are pairwise strangers.

Solution 1:

Let \{ A, B, C, D, E, F\} be a group of 6 people. Suppose that the people known to A are seated in room Y and the people NOT known to A are seated in room Z; A is not in either room. Then, there are necessarily at least 3 people in either room Y or in room Z; (a) Suppose B, C, D to be in room Y. Either these 3 people are mutual strangers (and so the given theorem is true), or, at least two of them (say, B and C) know each other. In the latter case, A, B and C form a group of 3 mutual acquaintances — and again, the theorem is true. (b) In (a), replace room Y by Z and interchange the notion of ‘”acquaintances” and “strangers”‘.

Question 2:

Show that in any group of 10 people there is always (a) a subgroup of 3 mutual strangers or a subgroup of 4 mutual acquaintances, and (b) a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 2:

(a) Let A be one of the ten people; the remaining 9 people can be assigned to two rooms: those who are known to A are in room Y and those who are not known to A are in room Z. Either room Y has at least 6 people or room Z has at least 6 people. For, (i) suppose room Y has at least 6 people. Then, by previous problem number 1, there is either a subgroup of 3 mutual acquaintances or a subgroup of 3 mutual strangers (thus, the theorem is true) in this room. In the former case, A and these 3 people constitute 4 mutual acquaintances (ii) Suppose room Z has at least 4 people. Either these 4 people know one another or at least 2 of them, say B and C, do not know each other. In the former case, we have a subgroup of 4 mutual acquaintances. In the latter case A, B and C constitute 3 mutual strangers.

(b) In the previous scenario, let people who are strangers become acquaintances, and let people who are acquaintances pretend they are strangers. The situation is symmetric.

Question 3:

Show that in any subgroup of 20 people there will always be either a subgroup of 4 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 3:

Suppose A is one of these 20 people. People known to A are in room Y and people not known to A are room Z. Either room Y has at least 10 people or room Z has at least 10 people. (i) If Y has at least 10 people, then by part B of problem number 2 here, there is either a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers — as asserted — in this room. In the former case, A and these mutual acquaintances will form a subgroup of 4 mutual acquaintances. (ii) Switch ‘”acquaintances” and “strangers”‘ in (i).

Question 4:

Let p and q be 2 positive integers. A positive integer r is said to have the (p,q) – Ramsey property, if in any group of r people either there is a subgroup of p people known to one another or there is a subgroup of q people not known to one another. {By Ramsey’s theorem, all sufficiently large integers r have the (p,q)-Ramsey property.} The smallest r with the (p,q)-Ramsey property is called the Ramsey number R(p,q). Show that (a) R(p,q) = R (q,p). (b) R(p,1)=1, and (c) R(p,2)=p.

Solution 4:

(a) By parts (b) of the previous three questions, we have proved part a of the proof here.

(b) This is obvious.

(c) In any group of p people, if all of them are not known to one another, there will be at least 2 people who do not know each other.

Question 5:

Prove that R(3,3)=6.

Solution 5:

Question 1 and its proof in this blog article imply that R(3,3) \leq 6.

To prove that R(3,3)>5, it is sufficient to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation, there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.

Question 6:

Show that if m and n are integers both greater than 2, then

R(m,n) \leq R(m-1,n) + R(m,n-1).

(this recursive inequality gives a non-sharp upper bound for R(m,n)).

Solution 6:

Let p \equiv R(m-1,n), q=R(m,n-1) and r \equiv p + q. Consider a group \{ 1,2, 3, \ldots, r\} of r people. Let L be the set of people known to person 1 and M be the set of people NOT known to person 1. The two sets together have r-1 people, so either L has at least p people or M has at least q people. (a) If L has p \equiv R(m-1,n) people, then, by definition, it contains a subset of (m-1) people known to one another or it contains a subset of n people unknown to one another. In the former case, the (m-1) people and person 1 constitute m people known to one another.

Thus, in their case, a group of R(m-1,n) + R(m,n-1) people necessarily includes m mutual acquaintances or n mutual strangers. That is, R(m,n) \leq R(m-1,n) + R(m,n-1).

(b) By the usual symmetry argument, the same conclusion follows when M contains q people.

Question 7:

(Remark: A pretty property of Ramsey numbers related to combinatorics).

Show that if m and n are integers greater than 1, then R(m,n) \leq { {m+n-2} \choose {m-1}} — a non-recursive upper bound.

Solution 7:

When m=2, or n=2, (i) holds with equality (see problem 4 in this blog article). The proof is by induction on k=m+n. As we have just seen, the result is true when k=4. Assume the result true for k-1. Then,

R(m-1,n) \leq {{m+n-3} \choose {m-2}}  and R(m,n-1) \leq {{m+n-3} \choose {m-1}}

Now, Pascal’s identity gives:

{{m+n-3} \choose {m-2}} + {{m+n-3} \choose {m-1}} = {{m+n-2}} \choose {m-1} so that R(m-1,n) + R(m,n-1) \leq {{m+n-2}} \choose {m-1}

But, from the previous question and its solution, we get R(m,n) \leq R(m-1,n) + R(m, n-1)

PS: As Richard Feynman, used to say, you will have to “piddle” with smallish problems as particular cases of these questions in order to get a grip over theory or formal language of this introduction.

PS: Additionally, you can refer to any basic Combinatorics text like Brualdi, or Alan Tucker or even Schaum Series outline ( V K Balakrishnan).

How to find the number of proper divisors of an integer and other cute related questions

Question 1:

Find the number of proper divisors of 441000. (A proper divisor of a positive integer n is any divisor other than 1 and n):

Solution 1:

Any integer can be uniquely expressed as the product of powers of prime numbers (Fundamental theorem of arithmetic); thus, 441000 = (2^{3})(3^{2})(5^{3})(7^{2}). Any divisor, proper or improper, of the given number must be of the form (2^{a})(3^{b})(5^{c})(7^{d}) where 0 \leq a \leq 3, 0 \leq b \leq 2, 0 \leq c \leq 3, and 0 \leq d \leq 2. In this paradigm, the exponent a can be chosen in 4 ways, b in 3 ways, c in 4 ways, d in 3 ways. So, by the product rule, the total number of proper divisors will be (4)(3)(4)(3)-2=142.

Question 2:

Count the proper divisors of an integer N whose prime factorization is: N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}}\ldots p_{k}^{\alpha_{k}}

Solution 2:

By using the same reasoning as in previous question, the number of proper divisors of N is (\alpha_{1}+1)(\alpha_{2}+1)(\alpha_{3}+1)\ldots (\alpha_{k}+1)-2, where we deduct 2 because choosing all the factors means selecting the given number itself, and choosing none of the factors means selecting the trivial divisor 1.

Question 3:

Find the number of ways of factoring 441000 into 2 factors, m and n, such that m>1, n>1, and the GCD of m and n is 1.

Solution 3:

Consider the set A = {2^{3}, 3^{2}, 5^{3}, 7^{2}} associated with the prime factorization of 441000. It is clear that each element of A must appear in the prime factorization of m or in the prime factorization of n, but not in both. Moreover, the 2 prime factorizations must be composed exclusively of elements of A. It follows that the number of relatively prime pairs m, n is equal to the number of ways of partitioning A into 2 unordered nonempty, subsets (unordered as mn and nm mean the same factorization; recall the fundamental theorem of arithmetic).

The possible unordered partitions are the following:

A = \{ 2^{3}\} + \{ 3^{2}, 5^{3}, 7^{2}\} = \{3^{2}\}+\{ 2^{3}, 5^{3}, 7^{2}\} = \{ 5^{3}\} + \{ 2^{3}, 3^{2}, 7^{2}\} = \{ 7^{2}\}+\{ 2^{3}, 3^{2}, 5^{3}\},

and A = \{ 2^{3}, 3^{2}\} + \{ 5^{3}, 7^{2}\}=\{ 2^{3}, 5^{3}\} + \{3^{2}, 7^{2} \} = \{ 2^{3}, 7^{2}\} + \{ 3^{2}, 5^{3}\}

Hence, the required answer is 4+3=2^{4-1}+1=7.

Question 4:

Generalize the above problem by showing that any integer has 2^{k-1}-1 factorizations into relatively prime pairs m, n (m>1, n>1).

Solution 4:

Proof by mathematical induction on k:

For k=1, the result holds trivially.

For k=2, we must prove that a set of k distinct elements, Z = \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}, a_{k}\} has 2^{k-1}-1 sets. Now, one partition of Z is

Z = \{ a_{k}\} \bigcup \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}\} \equiv \{ a_{k}\} \bigcup W

All the remaining partitions may be obtained by first partitioning W into two parts — which, by the induction hypothesis, can be done in 2^{k-2}-1 ways — and then, including a_{k} in one part or other — which can be done in 2 ways. By the product rule, the number of partitions of Z is therefore

1 + (2^{k-2})(2)=2^{k-1}-1. QED.

Remarks: Question 1 can be done by simply enumerating or breaking it into cases. But, the last generalized problem is a bit difficult without the refined concepts of set theory, as illustrated; and of course, the judicious use of mathematical induction is required in the generalized case. 

Cheers,

Nalin Pithwa.

Some number theory (and miscellaneous) coaching for RMO and INMO: tutorial (problem set) III

Continuing this series of slightly vexing questions, we present below:

1. Prove the inequality \frac{A+a+B+b}{A+a+B+b+c+r} + \frac{B+b+C+c}{B+b+C+c+a+r} > \frac{C+c+A+a}{C+c+A+a++b+r}, where all the variables are positive numbers.

2. A sequence of numbers: Find a sequence of numbers x_{0}, x_{1}, x_{2}, \ldots whose elements are positive and such that a_{0}=1 and a_{n} - a_{n+1}=a_{n+2} for n=0, 1, 2, \ldots. Show that there is only one such sequence.

3. Points in a plane: Consider several points lying in a plane. We connect each point to the nearest point by a straight line. Since we assume all distances to be different, there is no doubt as to which point is the nearest one. Prove that the resulting figure does not containing any closed polygons or intersecting segments.

4. Examination of an angle: Let x_{1}, x_{2}, \ldots, x_{n} be positive numbers. We choose in a plane a ray OX, and we lay off it on a segment OP_{1}=x_{1}. Then, we draw a segment P_{1}P_{2}=x_{2} perpendicular to OP_{1} and next a segment P_{2}P_{3}=x_{3} perpendicular to OP_{2}. We continue in this way up to P_{n-1}P_{n}=x_{n}. The right angles are directed in such a way that their left arms pass through O. We can consider the ray OX to rotate around O from the initial point through points P_{1}, P_{2}, \ldots, P_{n} (the final position being P_{n}). In doing so, it sweeps out a certain angle. Prove that for given numbers x_{i}, this angle is smallest when the numbers x_{i}, that is, x_{1} \geq x_{2} \geq \ldots x_{n} decrease; and the angle is largest when these numbers increase.

5. Area of a triangle: Prove, without the help of trigonometry, that in a triangle with one angle A = 60 \deg the area S of the triangle is given by the formula S = \frac{\sqrt{3}}{4}(a^{2}-(b-c)^{2}) and if A = 120\deg, then S = \frac{\sqrt{3}}{12}(a^{2}-(b-c)^{2}).

More later, cheers, hope you all enjoy. Partial attempts also deserve some credit.

Nalin Pithwa.

Some number theory problems: tutorial set II: RMO and INMO

1. A simplified form of Fermat’s theorem: If x, y, z, n are natural numbers, and n \geq z, prove that the relation x^{n} + y^{n} = z^{n} does not hold.

2. Distribution of numbers: Find ten numbers x_{1}, x_{2}, \ldots, x_{10} such that (a) the number x_{1} is contained in the closed interval [0,1] (b) the numbers x_{1} and x_{2} lie in different halves of the closed interval [0,1] (c) the numbers x_{1}, x_{2}, x_{3} lie in different thirds of the closed interval [0,1] (d) the numbers x_{1}, x_{2}, x_{3} and x_{4} lie in different quarters of the closed interval [0,1],  etc., and finally, (e) the numbers x_{1}, x_{2}, x_{3}, \ldots, x_{10} lie in different tenths of the closed interval [0,1]

3. Is generalization of the above possible?

4. Proportions: Consider numbers A, B, C, p, q, r such that: A:B =p, B:C=q, C:A=r, write the proportion A:B:C = \Box : \Box : \Box in such a way that in the empty squares, there will appear expressions containing p, q, r only; these expressions being obtained by cyclic permutation of one another expressions.

5. Give an elementary proof of the fact that the positive root of x^{5} + x = 10 is irrational.

I will give you sufficient time to try these. Later, I will post the solutions.

Cheers,

Nalin Pithwa.