Elementary Number Theory, ISBN numbers and mathematics olympiads

Question 1:

The International Standard Book Number (ISBN) used in many libraries consists of nine digits a_{1} a_{2}\ldots a_{9} followed by a tenth check digit a_{10} (somewhat like Hamming codes), which satisfies

a_{10} = \sum_{k=1}^{9}k a_{k} \pmod {11}

Determine whether each of the ISBN’s below is correct.
(a) 0-07-232569-0 (USA)
(b) 91-7643-497-5 (Sweden)
(c) 1-56947-303-10 (UK)

Question 2:

When printing the ISBN a_{1}a_{2}\ldots a_{9}, two unequal digits were transposed. Show that the check digits detected this error.

Remark: Such codes are called error correcting codes and are fundamental to wireless communications including cell phone technologies.

More later,
Nalin Pithwa.

Any integer can be written as the sum of the cubes of 5 integers, not necessarily distinct

Question: Prove that any integer can be written as the sum of the cubes of five integers, not necessarily.

Solution:

We use the identity 6k = (k+1)^{3} + (k-1)^{3}- k^{3} - k^{3} for k=\frac{n^{3}-n}{6}=\frac{n(n-1)(n+1)}{6}, which is an integer for all n. We obtain

n^{3}-n = (\frac{n^{3}-n}{6}+1)^{3} + (\frac{n^{3}-n}{6}-1)^{3} - (\frac{n^{3}-n}{6})^{3} - (\frac{n^{3}-n}{6}).

Hence, n is equal to the sum

(-n)^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n-n^{3}}{6}-1)^{3}+ (\frac{n-n^{3}}{6}+1)^{3}.

More later,
Nalin Pithwa.

A random collection of number theory problems for RMO and CMI training

1) Find all prime numbers that divide 50!

2) If p and p^{2}+8 are both prime numbers, prove that p^{3}+4 is also prime.

3) (a) If p is a prime, and p \not|b, prove that in the AP a, a+b, a+2b, a+3b, \ldots, every pth term is divisible by p.

3) (b) From part a, conclude that if b is an odd integer, then every other term in the indicated progression is even.

4) Let p_{n} denote the nth prime. For n>5, show that p_{n}<p_{1}+p_{2}+ \ldots + p_{n-1}.

Hint: Use induction and Bertrand's conjecture.

5) Prove that for every n \geq 2, there exists a prime p with p \leq n < 2p.

More later,
Regards,
Nalin Pithwa

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.

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).