# 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.

# Mathematics Olympiads: A curious calculation and its cute proof !!

Explain why the following calculations hold:

$1.9 + 2 =11$
$12.9 + 3 = 111$
$123.9 + 4 = 1111$
$1234.9 + 5 = 11111$
$12345.9 + 6 = 111111$
$123456.9 + 7 = 1111111$
$1234567.9 + 8 = 11111111$
$12345678.9 + 9 = 111111111$
$123456789.9 + 10 = 11111 11111$

Hint:

Show that $(10^{n-1}+2.10^{n-2}+3.10^{n-3}+ \ldots + n)(10-1) + (n+1)=\frac{10^{n+1}-1}{9}$

More later,
Nalin Pithwa

# A good way to start mathematical studies …

I would strongly suggest to read the book “Men of Mathematics” by E. T. Bell.

It helps if you start at a young age. It doesn’t matter if you start later because time is relative!! 🙂

Well, I would recommend you start tinkering with mathematics by playing with nuggets of number theory, and later delving into number theory. An accessible way for anyone is “A Friendly Introduction to Number Theory” by Joseph H. Silverman. It includes some programming exercises also, which is sheer fun.

One of the other ways I motivate myself is to find out biographical or autobiographical sketches of mathematicians, including number theorists, of course. In this, the internet is an extremely useful information tool for anyone willing to learn…

Below is a list of some famous number theorists, and then there is a list of perhaps, not so famous number theorists — go ahead, use the internet and find out more about number theory, history of number theory, the tools and techniques of number theory, the personalities of number theorists, etc. Become a self-learner, self-propeller…if you develop a sharp focus, you can perhaps even learn from MIT OpenCourseWare, Department of Mathematics.

Famous Number Theorists (just my opinion);

1) Pythagoras
2) Euclid
3) Diophantus
4) Eratosthenes
5) P. L. Tchebycheff (also written as Chebychev or Chebyshev).
6) Leonhard Euler
7) Christian Goldbach
8) Lejeune Dirichlet
9) Pierre de Fermat
10) Carl Friedrich Gauss
11) R. D. Carmichael
12) Edward Waring
13) John Wilson
14) Joseph Louis Lagrange
15) Legendre
16) J. J. Sylvester
11) Leonoardo of Pisa aka Fibonacci.
15) Srinivasa Ramanujan
16) Godfrey H. Hardy
17) Leonard E. Dickson
18) Paul Erdos
19) Sir Andrew Wiles
20) George Polya
21) Sophie Germain
24) Niels Henrik Abel
25) Richard Dedekind
26) David Hilbert
27) Carl Jacobi
28) Leopold Kronecker
29) Marin Mersenne
30) Hermann Minkowski
31) Bernhard Riemann

Perhaps, not-so-famous number theorists (just my opinion):
1) Joseph Bertrand
2) Regiomontanus
3) K. Bogart
4) Richard Brualdi
5) V. Chvatal
6) J. Conway
7) R. P. Dilworth
8) Martin Gardner
9) R. Graham
10) M. Hall
12) F. Harary
13) P. Hilton
14) A. J. Hoffman
15) V. Klee
16) D. Kleiman
17) Donald Knuth
18) E. Lawler
19) A. Ralston
20) F. Roberts
21) Gian Carlo-Rota
22) Bruce Berndt
23) Richard Stanley
24) Alan Tucker
25) Enrico Bombieri

Happy discoveries lie on this journey…
-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}.

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