Competitive Math Made Easy

 The sequence of numbers that starts as $1,1,$ and continues with each new number calculated as the sum of the previous two numbers, i.e., $1,1,2,3,5,8,13,21,34,55,89,144....$ is called the sequence of Fibonacci numbers. It was first described in 1202 by Leonardo of Pisa, better known as Fibonacci. It appeared in his book Liber Abaci as the solution to a problem involving the growing number of rabbits over time. Since then, Fibonacci numbers have seen many interesting applications and connections to a variety of topics, both within mathematics and in other disciplines, such as computer science, physics, chemistry, biology, and architecture.  This pattern of Fibonacci numbers is given by $u_1 = 1, u_2 = 1$ and the recursive formula $u_n = u_{n-1} + u_{n-2}, n > 2$. Simple Properties of Fibonacci Numbers $1.$ Sum of the Fibonacci Numbers The sum of the first n Fibonacci numbers can be expressed as $u_1 + u_2 + ... + u_{n-1} + u_n = u_{n+2} − 1.$ $2.$ Sum of Odd Terms The sum ...

 Let's take a look at some interesting sequences and series problems. 1. (USAMTS) Evaluate the value of  $$S = \sqrt{1+\cfrac{1}{1^2}+\cfrac{1}{2^2}} + \sqrt{1+\cfrac{1}{2^2}+\cfrac{1}{3^2}} + ... + \sqrt{1+\cfrac{1}{1999^2}+\cfrac{1}{2000^2}}$$ Solution Notice that $$1+\cfrac{1}{n^2}+\cfrac{1}{(n+1)^2} = \cfrac{n^4+2n^3+3n^2+2n+1}{n^2(n+1)^2} = \cfrac{n^2+n+1)^2}{n^2(n+1)^2} = \Bigg(1+\cfrac{1}{n(n+1)} \Bigg)^2$$ Thus, $$S= \Bigg(1+\cfrac{1}{1 \cdot 2} \Bigg) + \Bigg(1+\cfrac{1}{2 \cdot 3} \Bigg) + \Bigg(1+\cfrac{1}{3 \cdot 4} \Bigg) + ... + \Bigg(1+\cfrac{1}{1999 \cdot 2000} \Bigg)$$ $$= 1999 + \Bigg(1 - \cfrac{1}{2} \Bigg) + \Bigg(\cfrac{1}{2} - \cfrac{1}{3} \Bigg) + ... + \Bigg(\cfrac{1}{1999} - \cfrac{1}{2000} \Bigg) = 1999 + 1 - \cfrac{1}{2000} = \boxed{\cfrac{3,999,999}{2000}}$$ 2. (HMMT) Find the value of  $$S = \cfrac{1}{3^2+1} + \cfrac{1}{4^2+2} + \cfrac{1}{5^2+3} + ...$$ Solution For this problem, we can use partial fraction decomposition and try to find an alt...

Two triangles are called similar , if we can get two congruent triangles after enlarging or compressing the sides of one of them according to an equal ratio. That is, two triangles are similar means they have a same shape but may have different sizes. Criteria for Similarity of Two Triangles  (I) Each pair of corresponding angles are equal (A.A.A.);  (II) All corresponding sides are proportional (S.S.S.);  (III) Two pairs of corresponding sides are proportional, and the included corresponding angles are equal (S.A.S.);  (IV) For two right triangles, a pair of two corresponding acute angles are equal (A.A.);  (V) Among the three pairs of corresponding sides two pairs are proportional (S.S.). Basic Properties of Two Similar Triangles (I) For two similar triangles, their corresponding sides, corresponding heights, corresponding medians, corresponding angle bisectors, corresponding perimeter are all proportional with the same ratio.  (II) Consider the similarit...

When solving problems relating to clocks, keeping the following two points in mind will be helpful. The Minute hand covers $360^{\circ}$ in 60 minutes, or $6^{\circ}$ per minute. The Hour hand covers $360^{\circ}$ in 12 hours, or   $30^{\circ}$ per hour. Using these facts, the we can easily derive the results below: In a period of 12 hours, the hour and minute hand coincide 11 times. In a period of 12 hours, the hour and minute hand form a $180^{\circ}$ 11 times. In a period of 12 hours, any other angle between the minute and hour hand is formed 22 times. The time gap between any two coincidences is $\cfrac{12}{11}$ hours or $65\cfrac{5}{11}$ minutes. One common problem involving clocks is to find the angle between the minute and hour hand at a given time. To do so, one can use the formula  $$\theta = |5.5m - 30h |$$ Where $\theta =$ desired angle, and the time can be written as $h:m$. Let's use what we have learned to solve a couple problems. 1. What is the angle between the ...

Let $ABCD$ be a convex quadrilateral with $AB=BC=AC$. Suppose that a point P lies in the interior of the quadrilateral such that $AD=AP=DP$ and $\angle PCD=30^{\circ}$. Given that $CP=2$ and $CD=3$, find that length of $AC$.  The first step in solving this question is to draw the segment $PB$. Doing this allows us to see that $\triangle APB \cong \triangle ADC$ because $AP=AD$, $AB=AC$, and $\angle DAC = \angle PAB$ because $\angle DAP = \angle CAB = 60^{\circ}$. Thus, $PB=3$.  Next, let's set $\angle ACD = x$. We see that $\angle PCA = 30-x$, and $\angle BCP = 30 +x$. Due to the triangle congruency we established earlier, $\angle PBA = x$ as well. This means that we can write $\angle PBC = 60 - x$. Now,  $$\angle CPB = 180^{\circ} - \angle BCP - \angle PBC = 180^{\circ} - (30+x) - (60-x) = 90^{\circ}$$.  Thus, $\triangle BPC$ is a right triangle, and we know that $CP=2$ and $PB=3$, so we can easily find $CB$ to be  $$CB = \sqrt{2^2+3^2} = \boxed{\sqrt{13}}$$

The Pigeon Hole principle follows a simple observation that if you have 10 pigeons and 9 pigeonholes and you distribute these pigeons randomly, its quite obvious that there will be one pigeonhole with more than 1 pigeon. In other words, there are n boxes (the Pigeonholes), into which more than n objects (the Pigeons), have been placed, then at least one of the boxes must have received more than one of the objects. This principle is frequently applied in problems where we need to determine a minimal number of objects to ensure that some integral number property is satisfied. Consider the following puzzle:  A box contains red, black and white balls. The objective is to pick balls satisfying some constraints. How many balls must be taken to ensure that there is a pair of same color? The key to solve this puzzle is pigeonhole principle.  The principle says:  If $n$ pigeons are nesting in $m$ pigeonholes, where $n > m$, then at least one pigeonhole has more than one pig...

Percent is another way of saying 'for every hundred'.     Any division where the divisor is $100$ is a percentage.         For example: $ \cfrac {20}{100} = 20 \%$  or                             $ \cfrac {45}{100} = 45 \%$ In essence saying that,                              $ \cfrac {x}{100} = x \%$ Since any ratio can also be expressed as a division, it can also be represented as a percentage.  For ex., a ratio of $\cfrac{1}{2}$ or $1:2$ can be converted to a percent.  $ \cfrac {1}{2}$ $=$ $ \cfrac {1 * 50}{2 * 50} = \cfrac {50}{100} = 50$ Per Cent $=  50\%$  Expressing $x \%$ as a Fraction Any percentage can be expressed as a decimal fraction by dividing the percentage by $100$.  As $x \% = x$ out of $100 = \cfrac {x}{100}$ $75 \% = 75$...

Let us observe the following pattern of numbers.     $(i)$   $5, 11, 17, 23, ............$    $(ii)$   $6, 12, 24, 48, ............$   $(iii)$   $4, 2, 0, -2,-4, ............$   $(iv)$   $\cfrac {2}{3}, \cfrac{4}{9}, \cfrac{8}{27}, \cfrac {16}{81},......... $ In example, $(i)$, every number (except 5) is formed by adding $6$ to the previous numbers. Hence a specific pattern is followed in the arrangement of these numbers. Similarly, in example $(ii)$, every number is obtained by multiplying the previous number by $2$. Similar cases are followed in examples $(iii)$ and $(iv)$.  SEQUENCE A systematic arrangement of numbers according to a given rule is called a sequence.     The numbers in a sequence are called its terms. We refer the first term of a sequence as $T_{1}$, the second term as  $T_{2}$ and so on. The $n$th terms of a sequence is denoted by  $T_{n}$, which may also be...

Let us look at some key terms that pertain to Polynomials before we deep dive into the study of Polynomials.  Constant: A number having a fixed numerical value Example: $3, \cfrac{4}{5}, 4.2, 6.\overline{3}$ Variable:  A number which can take various numerical values Example:   $ x,  y,  z$ Algebraic Expression: A combination of constants and variables connected by arithmetic operators Example: $2x^2 + 7, 5x^3 + 4xy + 2xy^2 + 7,$ etc   Terms: Several parts of an algebraic expression separated  by $+$ or $-$ signs are called the terms of the expression.  Example: In the expression $9x + 7y + 5$, $9x$, $7y$, and $5$ are terms.   Coefficient of a Term:  In the term $8x^2$, $8$ is the numerical coefficient of $x^2$ and $x^2$ is said to be the literal coefficient of 8.  Like Terms: Terms having the same literal coefficients are called Like Terms.  Example: $8xy$, $9xy$, and $10xy$ are Like Terms ...

A number which cannot be written in the form $ \cfrac {p}{q}$  where $p$ and $q$ are integers and $ q $  $ \neq 0$. Example: $ \sqrt {2}$, $ \sqrt {3}$, $ \sqrt {6}$, $ \sqrt {7}$, $ \sqrt {8}$, $ \sqrt {10}$ Theorem: If $p$ divides $x^3$, then $p$ divides $x$, where $x$ is a positive integer and $p$ is a prime number.  Proof :  Let $x = p_{1} p_{2}...p_{n}$ where $p_{1}, p_{2}, p_{3}.....p_{n}$  are primes, not necessarily distinct.  $ \rightarrow x^3 = p_{1}^3 p_{2}^3.....p_{n}^3$ Given that $p$ divides $x^3$ By fundamental theorem, $p$ is one of the primes of $x^3$.  By the uniqueness of fundamental theorem, the distinct primes of $x^3$ are same as the distinct primes of $x$.  $ \rightarrow p$ divides $x$ Similarly if $p$ divides $x^2$, then $p$ divides $x$, where $p$ is a prime number and $x$ is a positive integer. Example: Prove that  $ \sqrt {2}$ is irrational.  Solution : Let us assume that  $ \sqrt ...

In $\bigtriangleup ABC$, bisect the two sides $BC$ and $CA$ in $D$ and $E$ respectively and draw $DO$ and $EO$ perpendicular to $BC$ and $CA$.  Thus, $O$ is the center of the circumcircle. Join $OB$ and $OC$.  The two triangles $BOD$ and $COD$ are equal in all respect, so  $ \angle BOD = \angle COD$ Therefore, $ \angle BOD = \cfrac{1}{2} \angle BOC =  \angle BAC = \angle A$ Also, $BD = BO \sin \angle BOD$ Therefore, $\cfrac{a}{2} = R \sin \angle A$ or $R =  \cfrac{a}{2 \sin \angle A}$ Now, we know that sin $A = \cfrac{2}{bc}$ $\sqrt{s(s-a)(s-b)(s-c)} = \cfrac {2S}{bc}$ where $S$ is the area of the triangle.   Therefore, substituting the above values of sin $\angle A$ in eq $(1)$, we get $R = \cfrac {abc}{4S}$ giving the radius of the circumcircle in terms of the sides. 

Circular Permutations are type of arrangements where the objects are to be arranged around a circle or in a circular order. Observe that in circular permutations the order around the circle (or the relative positions ) alone need to be taken into consideration and not the actual positions. For example, suppose $5$ different things are arranged around a circle. Consider the 5 positions around a circle. $A$, $B$, $C$, $D$, $E$ can be arranged in 5 different positions in 5! ways. Consider one such arrangement say $ABCDE$ in that order around the circle. This arrangement and the $4$ new arrangements $BCDEA$, $CDEAB$, $DEABC$, and  $EABCD$ are not only really different arrangements because the same relative positions around the circle are maintained by the $5$ letters. Therefore, the number of different ways of arranging the  $5$ letters around a circle is $ \cfrac {5!}{5} = 4!$ Also, in the above we are considering the clockwise and anti-clockwise arrangements on the circle ...

Let us consider the following example - In how many different ways can the letters of the word 'POSSESSIVE' be arranged? Observe that there are $10$ letters in the word and all these letters are not distinct. There are $4$ $S's$ and $2E's$ and other letters are distinct. If all the letters were distinct, the number of different arrangements is obviously $^{10}P_{10}$ or $10!$. In this case, since some of the letters are alike the number of arrangements cannot be $10!$  Let us assume that the number of arrangements of the letters of the word 'POSSESSIVE' is $x$. Consider any of these permutations (or arrangements) A typical arrangement can be thought as PSOESVSIES. Keep all except the three $S's$ fixed in their places.  Suppose we replace the four $S's$ in the above by four distinct objects, say $S_{1}$,  $S_{2}$,  $S_{3}$,  $S_{4}.$ If we permute these objects among themselves in all possible ways, it will give rise to $^{4}P_{4}$    or $4!$ arra...

Let's look at some interesting applications of the basic arithmetic formulas in the problems below. Problem 1 - Evaluate $123456789 \cdot 999999999$.         Solution - $123456789 \cdot 999999999 = 123456789 \cdot (1000000000-1)$                         $=123456789000000000-123456789=\boxed{123456788876543211}$ Problem 2 - Find the value of $\frac{13579}{(-13579)^2+(-13578)(13580)}$.         Solution - Using $(a-b)(a+b) = a^2 - b^2$, we have $$\frac{13579}{(-13579)^2+(-13578)(13580)} = \frac{13579}{(13579)^2-(13579^2-1)} = \boxed{13579}$$ Problem 3 - Evaluate $\frac{83^3+17^3}{83 \cdot 66+17^2}$.         Solution - By the use of the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ $$\frac{83^3+17^3}{83 \cdot 66+17^2} = \frac{(83+17)(83^2-83 \cdot 17 + 17^2)}{83 \cdot 66 + 17^2} = \frac{100 \cdot (83 \cdot 66 + 17^2)}{83 \cdot 66 + 17^2} = \boxed{100}$$ Problem 4 ...

1. Let $n$ be a fixed positive number. Two integers $a$ and $b$ are said to be congruent modulo $n$, symbolized by  $a\equiv b \pmod{n}$  if $n$ divides the difference $a-b$., i.e., provided that $a-b = kn$ for some integer $k$. For ex.   $23\equiv 3 \pmod{5}$,  $19\equiv 3 \pmod{4}$ ,   $12\equiv 5 \pmod{7}$ ,   $5\equiv 5 \pmod{3}$ , 2. For arbitrary integers $a$ and $b$,  $a\equiv b \pmod{n}$  iff $a$ and $b$ leave the same non negative remainder when divided by $n$ 3. Let $n>1$ be fixed and $a, b, c, d$ be arbitrary integers. Then the following properties hold: $a\equiv a \pmod{n}$          Ex.,  $7\equiv 7 \pmod{3}$          $7-7 = 0$ is a multiple of $3$. Hence $3$ divides $0$ If  $a\equiv b \pmod{n}$ , then  $b\equiv a \pmod{n}$          Ex.,  $17\equiv 3 \pmod{7}$          This congruen...

An equation, in which the highest power of the variable is $2$ is called a quadratic equation. The standard form of a quadratic equation is $ax^2+bx+c=0$, where $a, b,$ and $c$ are constants and $a \neq -$ Solution of Quadratic Equations: There are two main methods to find the solutions, or roots, of a quadratic equation. Factorization $\rightarrow$ Let $ax^2+bx+c = a(x - \alpha)(x - \beta) = 0$. Here, the roots of the equation are $x = \alpha$ and $x = \beta$. Hencec, factorizing the equation and equating each factor to zero is one method to find the roots. The second method is known as the Quadratic Equation, which is:  $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where the equation of the quadratic is $ax^2+bx+c = 0$.  Nature of the Roots:  In a quadratic equation in standard form, the term $b^2-4ac$ is known as the discriminant of the equation, which plays an important role in finding the nature of the roots. It is often denoted by $D$. If $a, b, c \in \ma...

In $\triangle ABC$, let the name the incenter as point $I$ and the feet of radii as $D, E,$ and $F$ as shown below. Since $I$ is the incenter, $ID = IE = IF = r$ Now, using the $A = \frac{bh}{2}$ formula on $\triangle ICB, \triangle IAC, \text{and } \triangle IBA$, we get   $[ICB] = \frac{1}{2}ID \cdot BC = \frac{ra}{2}$ $[IAC] = \frac{1}{2}IE \cdot AC = \frac{rb}{2}$ $[IBA] =  \frac{1}{2}IE \cdot AB = \frac{rc}{2}$ Adding all three of our equations, we have  $$\frac{ra}{2} + \frac{rb}{2} + \frac{rc}{2} = [ABC]$$ Letting $s = \text{semiperimeter} = \frac{a+b+c}{2}$, we get $$[ABC] = r \cdot s$$

Factorization is the process of writing a number or expression as a product of several factors, usually smaller or simpler. For example $8$ can be factorized as $2 \cdot 4$, and $(x-3)(x+3)$ is the factorization of the polynomial $x^2-9$. Here are some common factorizations worth memorizing: $$(a-b)(a+b)=a^2-b^2$$ $$(a \pm b)^2 = a^2 \pm 2ab + b^2$$ $$(a \pm b)(a^2 \mp ab + b^2) = a^3 \pm b^3$$ $\text{Proof }$ $$(a+b)(a^2 - ab + b^2) = a^3 - a^2b + ab^2 + a^2b - ab^2 +b^3 = a^3 + b^3$$ $$\text{Use ($-b$) to replace $b$ in the above, we obtain,}$$ $$(a-b)(a^2 + ab + b^2) = a^3 - b^3$$ $$(a \pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3$$ $\text{Proof }$ $$(a+b)^3 = (a + b) \cdot (a + b)^2 = (a+b)(a^2+2ab+b^2)$$ $$(a+b)^3 = a^3 + 2a^2b+ab^2+a^2b+2ab^2+b^3$$ $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ $$\text{Use ($-b$) to replace $b$ in the above, we obtain,}$$ $$(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$ $$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$$ $\te...

Let $A, B, C$ be three letters , then we can combine any two of them in the following ways: $$ AB, BC, AC$$ Similarly, if $A, B, C, D$ are four letters, then we can combine any two of them in the following manner: $$ AB, AC, AD, BC, BD, CD$$ Similarly, we can combine any $3$ of $A, B, C, D$ as : $$ ABC, ABD, ACD, BCD$$ We can generalize by saying that the number of all combinations of $n$ distinct things taken $r$ at a time $(r \le n)$ is ${n}\choose{r}$  $= \frac{n!}{(n-r)!r!}$ Note In combinations, the order of the letters (or things) is not considered. Here, AB and BA are the same, so they are only counted once, unlike permutations. The term "combination" is generally used for selection of things and "permutations" are used for rearrangements. Combinations with Restrictions Number of combinations of $n$ things taken $r$ at a time in which $x$ particular things always occur is $${n-x}\choose{r-x}$$ Number of combinations of $n$ things t...

In circular permutations, things are to be arranged in the form of a ring or a circle, e.g. arrangements of people around a circular table. In circular permutation there are no end points, i.e. there are no beginning or ending positions. So, the number of circular permutations of n objects are $$\frac{n!}{n}=(n-1)!$$ Thus in a circular permutation, one thing is kept fixed and the remaining  $(n-1)$ things are arranged in $(n-1)!$ ways. If the clockwise and counter clockwise orders are not distinguishable, then the number of ways = $\frac{1}{2}(n-1)!$ Let us look at a few examples.  Ex:   In how many ways can $6$ boys be seated at a circular table? Sol:   We keep one boy in a fixed position and to find the number of permutations to arrange the remaining 5 boys, we simply take $5!$, which gets us to our answer $\boxed {120} $.  Ex:  In how many ways can $6$ boys be arranged at a round table so that $2$ particular boys can be seat...