By Valentin Boju, Louis Funar

This quantity deals a set of non-trivial, unconventional difficulties that require deep perception and mind's eye to resolve. They disguise many themes, together with quantity thought, algebra, combinatorics, geometry and research. the issues commence as basic workouts and turn into tougher because the reader progresses throughout the booklet to turn into tough sufficient even for the skilled challenge solver. The introductory difficulties specialise in the elemental tools and instruments whereas the complicated difficulties target to strengthen challenge fixing concepts and instinct in addition to advertise additional learn within the region. options are incorporated for every problem.

Show description

Read or Download The Math Problems Notebook PDF

Similar combinatorics books

Proofs from THE BOOK

This revised and enlarged 5th version gains 4 new chapters, which comprise hugely unique and pleasant proofs for classics akin to the spectral theorem from linear algebra, a few more moderen jewels just like the non-existence of the Borromean jewelry and different surprises. From the Reviews". .. inside of PFTB (Proofs from The e-book) is certainly a glimpse of mathematical heaven, the place shrewdpermanent insights and lovely principles mix in excellent and wonderful methods.

Combinatorial Algebraic Geometry: Levico Terme, Italy 2013, Editors: Sandra Di Rocco, Bernd Sturmfels

Combinatorics and Algebraic Geometry have loved a fruitful interaction because the 19th century. Classical interactions contain invariant conception, theta features and enumerative geometry. the purpose of this quantity is to introduce fresh advancements in combinatorial algebraic geometry and to strategy algebraic geometry with a view in the direction of purposes, akin to tensor calculus and algebraic facts.

Finite Geometry and Combinatorial Applications

The projective and polar geometries that come up from a vector area over a finite box are relatively important within the building of combinatorial items, akin to latin squares, designs, codes and graphs. This ebook offers an creation to those geometries and their many functions to different components of combinatorics.

Extra info for The Math Problems Notebook

Example text

Monthly) Denote by P the set of points of the plane. Let : P × P → P be the following binary operation: A B = C, where C is the unique point in the plane such that ABC is an oriented equilateral triangle whose orientation is counterclockwise. Show that is a nonassociative and noncommutative operation satisfying the following “medial property”: 22 3 Geometry (A B) (C D) = (A C) (B D). 7. Consider two distinct circles C1 and C2 with nonempty intersection and let A be a point of intersection. Let P , R ∈ C1 and Q, S ∈ C2 be such that P Q and RS are the two common tangents.

If p is prime, then by Wilson’s theorem, (p − 2)! ≡ −(p − 1)! ≡ 1 (mod p), which implies that (p − 2)! (p − 2)! − 1 = . , and therefore sin π (p − 2)! p 2 p = sin π [(p − 2)! − 1] = −1. 2 Consequently, the nth term of the sum is zero if at least one among n, n + 2 is composite, and 1 if both n and n + 2 are prime. Finally, one adds 2 units in order to count the pairs (3, 5), (5, 7). 35. 1. Find all solutions of the equation 3x+1 + 100 = 7x−1 . 2 2 2. Find two solutions of the equation 3x + 3x = 2x + 4x , and prove that there are no others.

See also: 44 5 Number Theory Solutions • T. Andreescu and D. , 2006. 14. If N = 2 + 2 28n2 + 1 ∈ Z for a natural number n, then N is a perfect square. 14. If 28n2 + 1 = k, then N = 2 + 2k. If N ∈ Z, then k ∈ 21 Z. But k is the square root of an integer, and whenever such a square root is a rational number, then it is actually an integer. This proves that k ∈ Z. Now, 28n2 = (k − 1)(k + 1), and therefore k is odd, k = 2t + 1. Thus, 7n2 = t (t + 1). We have two cases: 1. If 7 divides t + 1, then, t = 7s − 1 and so (7s − 1)s = n2 , for s ∈ Z.

Download PDF sample

Rated 4.93 of 5 – based on 24 votes