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To start with Angle Bisector Theorem A classic example from one of the simcat's which makes use of both external and internal angle bisector theorems. Find DE : DB, if BC = 3 units, EB = 1 unit and DE is an angle bisector of CDB and DB is an angle bisector of ADE.

We are asked to find DE/DB, From the above figure we see that DE is internal angle bisector of Triangle(DCB) whereas DB is the External angle bisector of Triangle(DCE). So, From an external angle bisector in Triangle(DCE) we have, DC/DE = BC/BE..............i From an internal angle bisector in Triangle(DCB) we have, DC/DB = EC/EB .............ii From i & ii we get, DE/DB = 2/3. In addition to the above Angle Bisector theorem, I have attached a file with respect to angle bisectors that will help you in comprehending the below concept. AP / PM = (b + c) / a where, P is the incentre and a, b, c are the respective sides of the triangle. This concept will be really handy in solving the problems. Thanks in advance. (attached thumbnail pg 7 post 68)

For any set of n positive numbers , Arithmetic Mean >= Geometric Mean >=Harmonic mean The equality occurs only when all the numbers are equal . This can be used if the sum(or product ) of some numbers are given and the maximum(or minimum) value of the product(or sum ) is asked ... Ex : xy=27 minimum value of 3x+4y ? taking 3x and 4y to be two numbers and applying our rule A.M : (3x+4y)/2 G.M : (12xy)^1/2=(4*3*27)^1/2 = 18 A.M >= GM

3x+4y >= 2*18 Hence the minimum value of (3x+4y) is 36 how to put 9 identical rings in 4 fingers. concept= n+r-1Cr-1 this formula is used to distribute n identical things among r people. if the 9 rings are named, a1,a2,a3....a7,a8,a9 we want to distribute them among 4 fingers, means we want to make four groups out of this 9 rings. so if we arrange the 9 rings side by side. _a1_a2_a3 _a4_ a5_ a6_ a7_a8_a9_ we need three separator to divide them in 4 groups. and we can put the separator at any of the black space above. suppose I put first separator after a1, second after a4, 3rd after a6 so groups are a1 a2,a3,a4 a5,a6 a7,a8,a9 that means I can say I have total 12 items ( 9+3) to arrange them selves. that is 12! but 9 rings are identical and 3 separators are also identical so final answer shd be = 12!/9!*3 if we replace 9 with n,3 with r we get n+r-1Cr-1 2. how to distribute 9 different rings among 4 fingers. just a single change rings are different, so we dont have to divide by 9!, so answer = 12!/3! general formula = n+r-1Pr-1 both of the above Q are of arrangement and distribution. examples where this concept can be used. 1. 2. 3. 4. distribute 1o chocolates aming 6 children such that no children is empty handed. find whole number solutions for X+Y+Z = 22 find natural number solutions for X+Y+Z= 22 total number of terms in (a+b+c+d)^15

concept 3. total number of squares which can be made from size in N*N size square. = 1^2 + 2^2 + 3^3 .....N^2 like for 2*2 square we can have total 5 squares, 4 square of 1*1, and the 2*2 square itself. concept 4. total number of rectangles which can be made from N*N square. = 1^3+2^3+3^3.....n^3 concept 5. A plane( restricted) is to be divided in N distinct parts, find the minimum number of lines to do so. formula = > sigma X = N-1 X is the minimum number of lines. suppose we want to divide plane in 16 distinct parts sigma5 = 15 so answer is 5 lines.

Sigma denotes the sum of all the term eg1+2+3+...n = sigma (n) 1+3+6+10+.....(1+2+..n)=Sigma[Sigma(n)]

Suppose we have to find the number of ways in which an ordered pair (a, b), where a and b are natural numbers, can be choosen such that LCM of a and b is (p^x)*(q^y)*(r^z)*... Let a = (p^x1)*(q^y2)*(r^z3)*... and b = (p^x2)*(q^y2)*(r^z2)*... Since LCM of a and b is (p^x)*(q^y)*(r^z)*..., we can say that max(x1, x2) = x => one of x1 and x2 has to be x, this can be done in (x + 1)^2 - x^2 = (2x + 1) ways Similarly, max(y1, y2) = y => (2y + 1) ways & max(z1, z2) = z => (2z + 1) ways So, total number of such ordered pairs = (2x + 1)(2y + 1)(2z + 1).. Had the question been that in how many ways two numbers can be choosen such that their LCM is (p^x)*(q^y)*(r^z)*..., then the answer would be {(2x + 1)(2y + 1)(2z + 1).... + 1}/2 Number Systems (Concept 1)

Lets brush up some painted cube funda We assume the cube is divided into n^3 small cubes. no. of small cubes with ONLY 3 sides painted : 8( all the corner cubes ) no. of small cubes with ONLY 2 sides painted : A cube is painted on 2 sides means, it is on the edge of the bigger cube ,and we have 12 edges, each having n cubes. but since the corner cubes are painted on 3 sides, we need to neglect them. so in effect, for each side we will have (n-2) small cubes with only 2 sides painted. thus, then number is, 12 * (n-2) no of small cubes with ONLY 1 side painted : for each face of the cube ( 6 faces ) we have (n-2)^2 small cubes with only one side painted. and we have 6 faces in total. so th number is, 6*(n-2)^2 no of small cubes with NO sides painted : if we remove the top layer of small cubes from the big cube we will end up a chunk of small cubes with no sides painted. this number will be equal to, (n-2)^3. Also, remember for Cuboids with all different sizes, the following are the results:

a x b x c (All lengths different) Three faces - 8 (all the corner small cubes of the cuboid) Two faces - There are two (a-2) units of small cubes on one face of the cuboid and there is a pair of such faces. Hence, number of such small cubes corresponding dimension a of the cuboid = 4(a-2). Similarly, for others. So, total with two faces painted = 4(a - 2) + 4(b - 2) + 4(c - 2) One face - Since each face of the cuboid is a combination two different dimensions, hence for the face which is a combination of a and b dimensions, the number of small cubes is 2* (a-2)(b-2) [multipliesd by 2 because there are 2 such faces for the combination] Similarly, for others. So, total with one face painted = 2(a - 2)(b - 2) + 2(a - 2)(c - 2) + 2(b - 2)(c - 2) Zero faces - The entire volume of small cubes except for two cubes in each of the rows and columns will not be painted at all. hence this is the simplest ... (a - 2)(b - 2)(c - 2) You can put different integer values for number of small cubes producing different edge lengths of cuboid to get varied results. To verify for a cube, put a=b=c=L, you get Three faces - 8 Two faces - 12(L - 2) One face - 6(L - 2)^2 Zero faces - (L - 2)^3 Problems on Intersection of Straight lines , Circles, Formation of Points and Formation of Triangles, Quadrilaterals Etc Basic Concept Fundas 1. If there are n number of straight lines , They intersect each other in nc2 ways 2. If there are m number of circles , They intersect each other in 2*(mc2) ways = m (m-1)= 2p2 ways 3. When n straight lines and m circles intersect each other , they intersect in at most 2 * m * n = 2* ( no. of circles ) * ( no.of straight lines) 4. When n parallel lines intersect m straight lines , Then no. of parallelograms possible = nc2 * mc2= mn (m-1) (n-1)/4 5. There is one case when collinear and non- collinear points are given , and asked how many triangles it can formed=> The funda for this - ( Triangles that can be formed with all points ) ( Triangles formed with collinear points )

And the same funda is applied whenever such variations in condition occurs 6. For quadrilaterals standard approach is followed

1. 4 points out of 8 points are collinear . Number of different quadrilaterals that can beformed by joining this is 1. 56 2. 53 3. 76 4. 60 => total points = 8 => Collinear points = 4 , Non collinear points = 4 => Standard approach = => 4c0 * 4c4 + 4c1* 4c3 + 4c2 * 4c2 => 1+ 16+36 => 53

No f quads with 0 points : 1 No of quads with 1 point 4 * 4C3 = 16 No of quads with 2 points 4C2 * 4C2 = 36 so in all 53. 2. The number of points of intersection of 8 different circles is 1. 16 2. 24 3. 28 4. 56 => Guess easy now ; Standard Formula = 2* nc2= n*(n-1) = np2 => 8 * (8-1)= 8*7 => 56 3. The max . number of points of intersections of 8 straight lines 1. 8 2.16 3. 28 4. 56 => Standard formula = nc2 => 8c2 = 8*7/ 2= 28 4. The max. no of points into which 4 circles and 4 straight lines intersect is , 1.26 2. 50 3. 56 4.72 => Since max. no of points is asked => ( no. of points possible due to intersection of 4 circles with each other ) +

( no. of points possible due to intersection of 4 straight lines ) + ( no. of points possible due to intersection of 4 circles with 4 straight lines ) => 4p2+ 4c2+ 2*4*4 => 4*3+2*3+32 => 12+6+32 => 50 5. If 5 straight lines are intersected by 4 straight lines , The number of parallelograms possible => Standard formula => 5c2 * 4c2 =>60 6. Now problems on Formation of triangles If there are 7 points out of 12 lie on the same straight line , then number of triangles thus formed is. => Here total points= 12 => These 12 points can formed 12c3 triangles => and 7 collinear points can also form 7c3 triangles => Thus total triangles possible = 12c3-7c3 => 185 7. The sides BC,CA,AB of triangle ABC have 3,4,5 interior points respectively on them. How many triangles can be formed using these points as vertices. a.200 b.205 c.400 d.410 => Here total number of points will be A+B+C+3+4+5 => 3+3+4+5= 15 Points => Total number of triangles possible = 15c3 => 3 points are collinear , 4 points are collinear , 5 points are also collinear => So triangles formed with these points is omitted , thus answer is

=> 15c3 3c3-4c3-5c3 => 205

However, I am afraid that there is a mistake made in deriving the above solution. The total vertices to be considered should be 3+4+5 = 12 instead of 15. We should not include the existing vertices (A, B, C) since the question was how many triangles can be formed from using the new points as vertices. Now, the solution becomes 12C3 - 3C3 - 4C3 - 5C3 = 220 1 4 10 which yields 205 as the answer. There are 6 straight lines in a plane, no 2 of which are parallel and no 3 of which pass through the same point. If their points of intersection are joined, then the number of additional lines thus introduced is (a) 45 (b) 78 (c) 105 (d) none of the foregoing => solution for this => Total points formed from the intersection : (6c2)c2 = 105 Number of existing lines from these points : 6*5c2 = 60 So, 105-60 = 45 9.which one is the list containing the number of points at which a circle can intersect a triangle ??? 2, 4 2, 4, 6 1, 2, 3 1, 2, 3, 4 1, 2, 3, 4, 5, 6 => 1,2,3,4,5,6 10.if there are 7 pts. on a circle and 10 non collinear pts outside circle in same plane how many circles can be made? => Total Points = 7+10 = 17 => For a circle formation 3 points are required - 17c3+ 1( This one is added cause 7 points can give one circle ) => 7 points are collinear , so circles are omitted - 7c3 => Thus the answer is 17c3 + 1 -7c3 3 points are required for a circle. And there are 17 points. hence 17C3 circles can be formed. But 7 points are on the same circle. hence 17C3 - 7C3 +1 (+1 because 7 points are on the same circle). 11.if there are collinear 7 pts and 8 collinear pts. on other line parallel to it and 3 non collinear pts outside two lines and all are in same plane , find no. of circ can be made

=> Total no. of points = 7+8+3= 18 Points => Number of circles that can be made with these points = 18c3 => And since 7 points and 8 points are collinear , the circles made with it are omitted 7c3 and 8c3 => Thus answer is 18c3-7c3-8c3 A cuboid with dimensions l, b and h is painted on surface and then cut into cubes of 1cm3 sizes. Now how many cubes have none of the faces painted, how many cubes have one face painted, how many cubes have two faces painted and how many cubes have three faces painted. Ans: No of cubes with no face(side) painted is (l-2)(b-2)(h-2) No of cubes with one face(side) painted is 2(l-2)(b-2) + 2(b-2)(h-2) + 2(l-2)(h-2) No of cubes with two faces(sides) painted is 4(l-2) + 4(b-2) + 4(h-2) No of cubes with three faces(sides) painted is 8 (always constant) No of cubes with four or more faces (sides) painted is zero. If problem statement says its cube instead of cubiod with k cm sides. Then the answers will be (k-2)3, 6(k-2)2, 12(k-2), 8 and zero respectively. (Multiplication Principle) If there are n choices for the first step of a two step process and m choices for the second step, the number of ways of doing the two step process is nm. The number of arrangements of n objects is n! The number of arrangements of r out of n objects is nPr = n!/(n-r)! The number of arrangements of n objects in a circle is (n-1)! The number of arrangements of n objects on a key ring is (n-1)!/2 The number of arrangements of n objects with r1 of type 1, r2 of type 2, ..., ri of type i is n!/(r1! r2!...ri!) The number of ways of choosing n out of r objects is nCr = n!/((n-r)! r!) The number of distributions of n distinct objects in k distinct boxes is kn. The number of ways of distributing n identical objects in k distinct boxes is (n+k-1)Cn. 1.if f(x) = ax^2 + bx + c, how to find maximum, minimum . USE first derivation test. f'(x)= 2aX+b. find value of X for which , 2ax+b= 0. when a>0, at this value of X, f(x) is minimum. when a0, so at x=-2, f(x) will attain minimum value. f(-2) = 4-8+3 = -1. if f(x) = -X^2 + 4X + 3 f"(x) = -2x + 4 X= 2. f(2) = -4+8+3 = 7 is the maximum value of f(x). generalizing for f(x) = ax^2 + bx + c, x=-b/2a (2ax+c=o)gives maximum or minimum value of f(x) depending upon a>0 or a 3x/2 + 3x/2 + 5y/3 + 5y/3 + 5y/3 = 15.--------------1 as I said, when sum of any quantities is constant, there product is maximum when they are equal. here sum is constant. so when 3x/2 = 5y/3. we get maximum value of x^2*y^3. taking 3x/2 = 5y/3 putting it in 1, => 5(3x/2) = 15. =>x=2. and y = 9/5. answer is 2^2*(9/5)^3. generalizing it, how to find maximum value of x^m*y^n where ax+by=P. a,b,x,y>0 x^m*y^n is maximum when ax/m = by/ n = p/m+n 4. when the product of any quantity is constant, sum of the all the quantity is minimum, when they are equal. xy^3 = 64. find minimum value of x+12y. we need to adjust x+12y, accordingly. x+12y = x+ (12y/3)*3 now, x*(12y/3)^3= 64 *64 ( coz xy^3 = 64)-----------1 the product is constant. so the sum of the quantities will be minimum when quantities are equal. take x= 12y/3 putting it in 1, we get x= 8 =>12y/3 = 8, y = 2. minimum value of x+12y = 8+24 = 32. generalizing it, how to find minimum value of ax+by where x^m*y^n=P a,b,x,y>0 ax+by is minimum when ax/m = by/n concept of CRT : before getting to chinese remainder theorem , let me explain whats the need for it? problem: find the smallest number when divided by 5 leaves 3 and when divided by 7 leaves 4

common approach divisor of 5 + 3 - 3,8,13,18,23,28,33 divisor of 7 + 4 - 4,11,18,25,32 18 is common to both series .. so we have the answer... same is the case with chinese remainder theorem find the remainder of 3^1001 divided by 1001 ... 1001 - 7*11*13 so find the remainder when 3^1001 divided by 1001 3^1001 / 7 ----> 3^5/7 , remainder - 5 3^1001/11 ----> 3/7 ,remainder - 3 3^1001/13 ----> 3^5/13 , remainder - 9 so we get 7a + 5 = 11b + 3 = 13c+9 now what is word interpretation of the above statement .. find the smallest number which when divided by 7 gives remainder 5 , when divided by 11 leaves remainder 3 and when divided by 13 leaves remainder 9? first take any two condition , i always prefer big numbers 11b + 3 = 13c + 9 divisor of 13 + 9 = 9,22,35,48,61,74,87,100,113,126,139 divisor of 11 + 3 - 3,14,25,36,47,58,69,80,91,102,113 so smallest number is 113 whats the next number then ? its of form LCM(11,13) + 113 = 143k + 113 so we have combined two conditions so now our job is to compare this with third one 143k + 113 = 7a + 5 143k + 108 = 7a

140+ 3k + 105 + 3 = 7a so 3k + 3 should give 0 remainder when divided by 7 so k = 6 final remainder is hence 143(6) + 113 = 971 this is all about chinese remainder theorem to sum up , use this theorem only when denominator is factorisable to prime factors.

Bases

1. A number is base N is divisible by N-1, when the sum of digits in base N is divisible by N-1 2. When digits of a number N1 in base N are rearranged to form a number N2, then N2-N1 is always divisible by N-1. 3. If a number in base N has even number of digits and that number is a palindrome, then the number is divisible by N+1Problems

1. A number 2342a121 is in base 8 and it is divisible by 7. Find the value of a. 2. A palindromic number in base 16 will always be divisible by which number? 3. A five digit number is in base 19. It is rearranged to form another 5 digit number. The difference of these numbers will be divisible by ??Zeller's Rule : With this technique named after its founder Zeller, you can solve any 'Dates and Calendars' problems. Zellers rule can be used to find the day on any particular date in the calendar in the history. All you have to know is the formula given below and how to use it. Zeller's Rule Formula: F = K + [(13xM - 1)/5] + D + [D/4] + [C/4] - 2C K = Date => for 25/3/2009, we take 25 In Zellers rule months start from march. M = Month no. => Starts from March. March = 1, April = 2, May = 3 Nov. = 9, Dec = 10, Jan = 11 Feb. = 12 D = Last two digits of the year => for 2009 = 09 C = The first two digits of century => for 2009 = 20

Example: 25/03/2009 F = 25 + [{(13 x1)- 1}/5] + 09 + 09/4 + 20/4 - (2 x 20) = 25 + 12/5 + 09 + 09/4 + 20/4 - 2x20 =25+2+09+2+5-40 [ We will just consider the integral value and ignore the value after decimal] = 43 - 40 = Replace the number with the day using the information given below. 1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday 7 = Sunday So it's Wednesday on 25th march, 2009. If the number is more than 7, divide the no. by 7. The remainder will give you the day. I hope you will find the above method very useful

SHORTCUT METHOD TO FIND RANK OF A GIVEN WORDThis shortcut method is used when the lettors of the given word are not repeated. Given word is MASTER The letters of the are M,A,S,T,E,R. Write the alphabetical order of the letters of the given word ' MASTER ' as A,E,M,R,S,T Now strike off the first letter M. A,E,M,R,S,T. Then count the no.of letters before M, and it is equal to 2,which is the coefficient of 5!. Again strike off the first letter A. A,E,M,R,S,T Then count the no.of letters before A and it is equal to 0 which is coefficient of 4! Again strike off the first letter S. A,E,M,R,S,T Then count the no.of letters before S and it is equal to 2 which is coeffcient of 3! Again strike off the first letter T. A,E,M,R, S, T Then count the no.of letters before T and it is equal to 2 which is coeffcient of 2! Again strike off the first letter E. A,E, M,R, S,T Then count the no.of letters before E and it is equal to 0 which is coeffcient of 1! Finally add 1 to the above values to get the rank of the word MASTER as follows: 2(5!) + 0(4!) +2(3!)+2(2!)+0(1!)+1=257 The concept of rank goes like this... Say for the number of letters before S..we have A,E,M & R before it...but we have already counted A & M..so we will go with E & R...i.e 2 letters remain... Similarly in case of T..we have A,E,M,R,S before it..but we have already counted A,M,S...so only E,R are left..i.e 2 letters.. Also in case of E we have A,M,R,S,T..but we have already taken care of all of them..no letters are pending..so 0 letters.. Hope this clears your confusion..Please PM me if further clarification is required..

Basic Formulae for Sequences and Series Some more formulae1. Greatest possible sum of the A.P - It is possible only when all the terms of A.P are non- negative Tn >or equal 0 2.Least possible sum of the A.P - it is possible only when all the terms of A.P are non- positive Tn < equal 0 3. Tn = Sn-S(n-1)

1.The perimeter of a right angle triangle=(2r+2R) And the area =r^2+2rR I think there is a small error with this formula for perimeter...The value should be P=2(2R+r).... Consider, 3,4,5 to be the sides of a right triangle...So area is 6 and the perimeter is 12... r=1 and R=10/4 2r+2R makes the perimeter 2+5=7... 2(2R+r)=10+2=12 which is the standard formula... Where r=inradius R=circumradius 2.For an right angle triangle R>=[root(2)+1]r

let me start from divisibility of 2 if last digit is divisible by 2 , then it is divisible by 2 ... why ? last digit is nothing but divide by 10 ... 10 is perfectly divisible by 2.. now we can extend it for 4 .. lets take last digit --> divide by 10 10 /4 --> not perfectly divisible now move to next digit--> divide by 100 100 is perfectly divisible by 4 .. now this goes on ... jus try for various numbers the above logic works for factor of 10^n .. other numbers cant be treated this way

------------------------------------------------------------------lets take an example of other number .. take 3 10/3 --> not divisible 100/3 --> not divisible 1000/3 --> not divisible and it goes on .. here comes next case , remainder 1 case if remainder is 1 , add the digits ---------------------------------------------------now here comes one more set of numbers .. like 7

10/7 --> not divisible but remainder is not 0 or 1 or -1 100/7 --> not divisible but remainder is not 0 or 1 or -1 1000/7 --> not divisible but remainder is -1 so every three digit we will have -1 remainder .. if 10^3 has -1 remainder , 10^6 has 1 as remainder... so its alternative say number is 123130 to checked divisibility by 7 .,.. so according to this concept we have , 130 - 123= 7 which is divisible by 7 hardly 2 sec answer ..

u might think wats so special about it .. divisor can be tricky like 143 235378/143 .. is it divisible ? by this method we can say 378 - 235 = 143.. thats all u r done applying this formula , we can test divisibility of 2,3,4,5,7,8,9,11,13,37,27,77,143, etc .... jus work on it .. u can easily find divisibility This is a very useful method to determine the last non zero digit of a factorial... Nomenclatures:

Z(x)= The last non zero digit of x! L(x)=The last non zero digit of x but don't you think that for higher values it will become recursive. like for x=79 u will have to calculate z(79/5) = z(15.XX)

Z(x)=L(4^y)*z(x/5)*z(x%10) Please don get frightened by the above formula...Ill tell you how it works... z(79)=l(4^7)*z(15)*z(9) z(15)=L(4^1)*z(3)*z(5)=4*6*2=8 last digit of 4^x, when x is odd is odd is 4 when x is even is 6 Hence,z(79)=4*8*8=6

As i told u it will sound cumbersome...You get down and solve its way easier

Suppose we want to find the non zero digit of 15! then, y= the higest multiple of 10 in the target number... 15=10*1+5,hence y=1 22=10*2+2,hence y=2 and so on So, Z(15)=L(4^1)*z(15/5)*z(15%10)=4*z(3)*z(5)=4*6*2=8..... How r u getting z(3) and z(5) ? by similar approach or there any table? It is good to remember the values for Z(1) to Z(9)... Z(1)=1 Z(2)=2 Z(3)=6 Z(4)=4 Z(5)=2 Z(6)=2 Z(7)=4 Z(8)=2 Z(9)=8

NO. OF SQUARES AND RECTANGLES IN A CHESSBOARD in a chessboard,there are 8*8 squares. in a 2*2 chessboard, there are 5 squares (4 small aquares, 1 big square). similarly in an n*n chessboard, there are 1^2 +2^2+....+n^2 squares. so in a 8*8 chessboard, n=8 => no.of squares = 1^2 +2^2+3^2+.....+8^2 = [n (n+1) (2n+1)]/6 (summation formula) = 204 rectangles :

in a 2*2 chessboard, there are 9 rectangles (4 1*1s,1 2*2,2 2*1s, 2 1*2s) for an n*n chessboard, there are 1^3 +2^3+3^3+.....+n^3 rectangles. so for an 8*8 chessboard, there are, rectangles = 1^3+2^3+....+8^3 => no. of rectangles = [{n^2}{(n+1)^2}/4 n=8, we get no. of rectangles = 1296 **No. of rectangles that are not squares in an 8*8 chessboard => 1296- 204 = 1092

CIRCULAR MOTIONConsider a circle with circumference 200 sq metres and two people A and B moving in clockwise direction with speeds 5m/s and 9m/s a) when do they meet at the starting point for the first time ? b) after how much time will they meet for the first time ? c) at how many distinct points they meet ? d) if they move in opposite direction then in how many distinct points do they meet ? ans a) time taken by A to reach the starting point for the first time is 200/5 = 40 sec, for the second time 80 seconds for the third time 120 seconds and so on ....... similarly time taken by B to reach the starting place is 200/9 for the second time 2 * 200/9 for the third time 3 * 200/9 so they meet for first time at the starting point at the LCM of their time periods which are 40 and 200/9 in this case and hence they will meet at t= 200 for the first time . ans b )this should be solved using the relative speed concept time taken for them to meet for the first time will be relative distance/ relative speed here relative distance is 200 and relative speed is 95=4. so ans is 200/4= 50 ans c)when 2 bodies are moving in circular motion in same direction the number of distinct points where they meet is the difference of the speeds . here it is 9-5 = 4 distinct points. the lcm of the speeds must be 1 ans d ) when 2 bodies move in circular motion in opposite direction then the number of distict points they meet is the sum of the speeds here it will be 14 . the lcm of the speeds must be 1 e) find the number of distict points at which 2 bodies with speed 4 and 8 meet when they move in clockwise direcion and when they move in anticlock wise direction ? here we first need to divide the HCF so we get 1:2 now using the formula stated above the number of distict point they meet when moving in same

direction will be 2-1 = the number of distict point they meet when moving in opposite direction is 2+1 = 3 f)consider three bodies a,b,c with speeds 5,9,13 respectively moving in clock-wise direction now number of distinct points at which all three meet ? a-b = 4 b-c = 4 a-c = 8 so number of distinct points will be the hcf of the differences which is 4 ans. g)consider three bodies a,b with speeds 5,9 respectively moving in clock-wise direction and c with speed 13 in anti-clockwise direction now number of distinct points at which all three meet ? a-b=4 a+c=18 b+c=22 hcf is 2 so they all will meet at 2 distinct points

last 2 digits of a number

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divide the number by 100 then it will give the last 2 digits if division is cumbersome u may use one of the following methods: points to remember : if last 2 digits are 25 then (abcdef25)^z where z is a natural number will always give the last 2 digit as 25. if last 2 digits are 76 then (abcdef76)^z where z is a natural number will always give the last 2 digit as 76. if the last digit is 1 say the number is (abcdefg1)^thgfds then the units digit will be 1 and the tens digit will be g*s e.g (1231)^4563 last 2 digits will be 91. (sdfdsf24)^ odd = last 2 digits 24 (dfdsfd24)^even = last 2 digits 76 (dbfh26)^odd = last 2 digits 26 (dfdsaf26)^even = last 2 digits 76 few examples : last 2 digits for (71)^45 = 51 (2)^32 = (2^10)^3 * 2^2 = (1024)^3 *2^2 = asjahsj76 *4 = sdnsj04 (6)^76 = (3*2)^76 = (81)^19 * (1024)^7 * (2)^6 = aks21 * 24 * 64 = ss56

last 2 digits of a number

]

divide the number by 100 then it will give the last 2 digits

if division is cumbersome u may use one of the following methods: points to remember : if last 2 digits are 25 then (abcdef25)^z where z is a natural number will always give the last 2 digit as 25. if last 2 digits are 76 then (abcdef76)^z where z is a natural number will always give the last 2 digit as 76. if the last digit is 1 say the number is (abcdefg1)^thgfds then the units digit will be 1 and the tens digit will be g*s e.g (1231)^4563 last 2 digits will be 91. (sdfdsf24)^ odd = last 2 digits 24 (dfdsfd24)^even = last 2 digits 76 (dbfh26)^odd = last 2 digits 26 (dfdsaf26)^even = last 2 digits 76 few examples : last 2 digits for (71)^45 = 51 (2)^32 = (2^10)^3 * 2^2 = (1024)^3 *2^2 = asjahsj76 *4 = sdnsj04 (6)^76 = (3*2)^76 = (81)^19 * (1024)^7 * (2)^6 = aks21 * 24 * 64 = ss56

p.s. correct me if i any mistakes thanks for your method...but you made a small mistake.i have marked in bold.

HOW to find the last non zero digit in x! I appreciate Siddharth's approach. Following is another approach: nice method...originally i used this method : finding the last non-zero digit of N! The last non-zero digit of a number N is (4^n)*(2n)! ..... (for N = 10,20,30....) where n=N/10 For nos other than multiples of 10, the result can be easily back tracked.. for eg.. if u are asked for 36!, then find for 40! and then back tract to 36!.. But finding 30! and moving to 36! might be tricky...since it involves a 35 and will introduce an unnecessary 0.. here N =40! and n=40/10 = 4 so the last digit is 4^n*(2n)! = 4^4*8! = 6*2 = 2 2 is the last non-zero digit of 40!, now if u want to calculate for 36!, then 36!*7*8*9*4 = 2 (i am writing all the last digit) 36!*6 = 2 so last non-zero digit of 36! = 2 Last non-zero digit of 10! = 8

For 20! = 8*8 = 4 [6 is dropped] This can continue for any number of 10s. For example, Last digit of 70! will be given by 8^7 For last digit of 8^7 = 2^21 =( 2^10)^2 * 2 = 76*2 =2 Therefore, last digit of 70! is 2. If you are asked for 73!, then just multiply the above 2 with 71*72*73 or simply 2 with 1*2*3 Therefore, last non zero digit for 73! become 2*6 = 2 [ignoring 10s digit] always got confused with power of powers. Following is what i found out: a^b^c = a^(b^c) Above is not equal to a^(b*c) In fact, (a^b)^c=a^(b*c) One correction. If Last non-zero digit of 10! = 8 then the last non-zero digit of 20! = 8x8x2 (this 2 comes from 20)=8 Last non zero digit of 70!= last non zero digit of (8^7) x2x3x4x5x6x7 That is incorrect. The power of 2 remains the same. Note that between 1 and 10 we have 4 and 8, which are pure multiples (I am sorry to coin this word) of 2. Just write down the numbers between 11 and 20 and you will see the gap. Let us consider a triangle ABC of any dimension. Now there is a point E on the line BC, such that it divides BC into the ratio, say, 3:5 and there is another point F on the line AC such that it divides AC into the ratio 2:7 Now the question is in what ratio does the line AE divides the line BF??? Ok since I have always been weak in geometry, I found this method, which i found, to be pretty useful, but I can't prove it... but is It has never let me down... So the method is this... All of you might remember the method of balancing a beam in school days. So here's what you can do. Consider line AC such that F is the fulcrum and AF=2 units and FC=7units. Now to balance this beam, we need to put loads in the inverse ratio, ie 7:2. So now a load of 7x will act on point A and 2x will act on point C and a net load of 9x will act on point F. Similarly create a beam balance for BC also. And you have loads at point B & F, so when you consider

the fulcrum at O, where AE divides BF, you can calculate the length and that will give you the ratio. OK this was not the best of explanations, but if you can just check out the attached image, I believe it will be pretty clear. Attached thumbnails.(later pg 5 post 50) Difference of Squares : The concept is if there is any N = a^2 - b^2 and a and b are integers. So many ways a and b are chosen? If N = a^2 - b^2 = (a+b) (a-b), if you want to find integers a and b, then (a+b) and (a-b) should be both either odd or even

For example if we take N = 9 , in that case 9 = 9*1 = 3*3 (In this case both Odd) if a+b = 9 and a-b = 1 therefore a = 5 and b = 4, thats why we have 25-16 = 9 and also if a+b = 3 and a-b = 3, a = 3 and b = 0 and thats why we 9-0 =9 so we see that only two ways we can represent the number 9 as DIFFERENCE OF SQUARES OF INTEGERS numbers.

If we extend this concept. For odd Numbers, 1) Any odd prime numbers can be represented in ONLY 1 way. For example : 3 = 3*1 ONLY so only 1 way 2) Other than this any composite Odd numbers can be represented in More than 1 way

For Even Numbers: 1) Any number of the form 4K+2 CANNOT be represented at all, hence 0 ways. For example Number 6 = 2*3 = 1*6 (So we don't get any combination for either Both Even or Both Odd), hence Integer numbers IS NOT POSSIBLE. 2) Any Prime factor Multiple of 4 is ALWAYS 1 way. For example : Number 8 = 2*4 = always 1 way. The reasoning behind is for prime number 2 when mutplied with 2 we can always break up into factors of Even numbers. Hence 4*2 we can always break up. And for odd prime numbers, we already have 2 twos in 4, so also we can break up into 1 ways into both even factors.

3) Any other even numbers more than 1 way. Source : Byju's class Notes last year Its great to keep all the concepts at a single place.

find the sum of all the numbers greater than 10000 formed by digits 1,3,5,7,9 ? No repetition. (1 + 3 + 5 + 7 + 9)*4!*11111 = 6666600 find the sum of all the numbers less than 10000 formed by digits 1,3,5,7,9 ? No repetition ---it means all single digit,double,triple,four digit nos. (1 + 3 + 5 + 7 + 9)(1 + 11*4 + 111*4*3 + 1111*4*3*2) = 701025 jus form the general formula for other number The general formula is (n-1)! * (111...n times) (Sum of the terms). So for the 1st Case where the number of digits are always same thats is always a 5 digit number, then 4! * (11111) * 25 = 600 * 11111 = 6666600.

The Second case is when you take one digit, two digit etc. For one digit. Sigma (Numbers ) = 25 = (1 + 3 + 5 + 7 + 9) For two digit = 1! * 11 * (Sum of the numbers taking two at a time) = 1! * 11 * 4C1 * (1 + 3 + 5 + 7 + 9) For three Digits : = 2! * 11 * 4C2 * (1 + 3 + 5 + 7 + 9) For four Digits : = 3! * 11 * 4C3 * (1 + 3 + 5 + 7 + 9) Adding all (1 + 3 + 5 + 7 + 9) * (1 + 4C1 * 11 + 2! * 4C2* 111 + 3! * 4C3*1111 + 4! * 4C4 * 11111) Hence General terms can be Sigma (Terms) * (1 + 1! * (n-1)C(1) (11) + 2! * (n-1)C2 * 111 + ... (n-1)! * (n-1)C(n-1) * (1111...n times)) Good one even I learnt this concept For the first question(explanation) How many numbers will be there having 1 at the first place - 4! there contribution to sum = 1*24*10000 Similarly for other numbers it will be 3*24*10000, 5*24*10000 and so on => Contribution of first place to the sum = (1 + 3 + 5 + 7 + 9)*4!*10000 Similarly for 2nd place it will be (1 + 3 + 5 + 7 + 9)*4!*1000 Similarly for other places => Final sum comes out to be (1 + 3 + 5 + 7 + 9)*4!*11111 In the similar manner you can try next one There are some questions where a person A and person B starts at different time and meets at a specific time and then reaches the destination at same time from where the other has started. This type of questions appear frequently in Mocks and other exams. Lets say person A and B starts from points P and Q at different times and meets at point S and reaches the points Q and P at the same time. P----------------------S----------------------Q -----------------------------------------------Time taken by person A is say t1 and speed being s1. Time taken by person B is say t2 and speed being s2.

hence PS = t1*s1. SQ = t2 * s2. As they reach the destination at the same time. hence (t1 * s1)/(t2) = (t2*s2)/(t1) hence s1/s2 = root(t2/t1) This is the principle and depending upon the questions, use options to keep either the time ratio same or speed ratio same. Find the number of ways of selecting 4 books from a set of 12 books such that no 2 books are adjacent to one another. The usual method involves finding the total number of combinations and eliminating the invalid ones. Here's a little short cut.. To pick 4 books we'd require at least 7 books.. ie x-x-x-x (where x is book and - is space between books) There are 3 hpyens. subtract this from the total no. of books. Ans is 9c4 (12-3) I normally use A + B + C + D+E = 12 where + are books and A, B , C , D and E are the separators. Hence Sum of A, B, C, D, E can take 12-4 (4 Books) = 8 B, C, D >=1 and A and E can be 0. Hence Subtract 8 -3 which is 5 hence 9C4, How we can find different pythagorus triplets. The Formula is a mixed fraction n[n/(2n+1)] hence N can take values from 1,2..... hence the first mixed fraction becomes 1[1/3] = 4/3 One side of the pythagorus theorem is denominator which is 3 and other side is numerator which is 4, the hypotenuse is 4+1 = 5 Hence the triplet is 3,4,5. Lets take n = 3 2n+1 = 7 Mixed fraction becomes = 24/7 the first side becomes 7, the other side is 24 and hypotenuse is 24+1 = 25 hence 7,24,25. all the other remaining pythagorus triplet is multiples of the triplets given by the above formula n + [n/2n+1] if not mixed fraction a) In a plane if there are n points of which no three are collinear, then

The number of straight lines that can be formed by joining them isnC2. The number of triangles that can be formed by joining them is nC3. The number of polygons with k sides that can be formed by joining them is nCk.

(b) In a plane if there are n points out of which m points are collinear, then

The number of straight lines that can be formed by joining them isnC2 mC2 + 1. The number of triangles that can be formed by joining them is nC3 mC3.

The number of polygons with k sides that can be formed by joining them is nCk mCk.

(c) The number of diagonals of a n sided polygon are nC2 n = n (n 3)/2. (d) The number of triangles that can be formed by joining the vertices of a n-sided polygon which has,

Exactly one side common with that of the polygon are n (n 4). Exactly two sides common with that of the polygon are n. No side common with that of the polygon are n (n 4) (n 5)/6.

Cauchy-Schwartz Equation: If a , b , c , d are four real numbers, they always satisfy the relationship (a^2+b^2)(c^2+d^2)>=(ac+bd)^2 This can be generalized to a large number of variables as (a1^2+a2^2+a3^2+.....)(b1^2+b2^2+b3^2+....)>=(a1b1 +a2b2+a3b3+....)^2 Questions: 1>Find the least value of X^2+Y^2+z^2 if X+2y+3Z=14 Sol:-->(X^2+Y^2+z^2)(1^2+2^2+3^2)>=(X*1+Y*2+z*3)^2 hence , min value= 14^2/14=14 every odd no is a part of pythagorian triplet... the other two nos are two consecutive no.s addin up to the odd nos squre... fr 3...4+5=9=3^2 fr 5.....12+13=25=5^2 .......fr 9...40+41=81=9^2 fr 25=312+313=625....i hope so... in geometry....we no that in a rt angld isosceles triangle the hypotenuse is tr2 times the equal sides... in an isosceles triangle with enclosed angle=120 degree the larger side is rt3 times the equal sides... Few Concepts: 1> For any prime number p, (p-1)times same digit is repeated , then that number formed is exactly divisible by p eg. 666666 is divisible by 7. 2>All perfect squares of the form 3K+N where N=0,1 4K+N where N=0,1 8K+N where N=0,1,4 9K+N where N=0,1,4,7 3>For any two integers satisfying, 3 b^A 4>If three circles touch Each other in a row and they have two direct common tangents, then their radii are in GP series

5>Product of factors of a number = (Number )^(total no of factors/2) 6>All odd natural numbers can be represented as the difference of two perfect squares. All even numbers which are multiple of 4 can be also written as the difference of two perfect squares. 7>Suppose we get a question like(I) A and B start at same time towards each other, meet at a point after time T then reach their respective destinations after time T1 and T2, or (II)their start at different time ,meet each other at a time where A takes T1 to reach and B takes T2 to reach, and then reaches their destination after time T at the same time , Then VA/VB =Root(T2/T1) And T = Root(T1*T2) where T is the same they take either to reach their meeting point in first or Destination in second case 8>If a Right Angled Triangle is rotated about its Axis to generate a cone, then the cone will have maximum volume when rotated about the smallest side as an axis and minimum when rotated about the side perpendicular to the smallest side as an axis. 9>If a n digit number is multiplied by an (n+1) digit number , the product has 2n or (2n+1) digits. 10> Difference between Principal Interest and Compound Interest for the second year=Pr^2 And Difference between PI and CI for the third year=Pr^2(r+2)where r=(R/100) and P is the amount and R is the rate of interest. 11>If A and B takes K days when working together and X+K and Y+K days when working alone respectively to complete a work then (X/K)=(K/Y) or XY=K^2. 12>The number of times the sign of the coefficients of an equation changes, gives the number of roots of a quadratic or higher degree equation. 13>The number of ways of writing any number as a sum of two or more consecutive positive numbers = number of odd factors of that number-1. Any number that doesnt have an odd factor cant be expressed as a sum of two or more consecutive numbers. eg. 50 has two odd factors. So, it can be written in only one way as a sum of two or more consecutive positive numbers which is 11+12+13+14. Adding some more fundas: 1) Every isosceles Trapezium is a cyclic quadrilateral. 2) if in Trapezium, AB and CD are Parallel sides, then AC^2 + BD^2 = AD^2 + BC^2 + 2 * AB * CD. 3 points are required for a circle. And there are 17 points. hence 17C3 circles can be formed. But 7 points are on the same circle. To Find out Pythagorean triplets There is a formula (N2-1)/2,( N2+1/2) of any number N. For Example if the number is 3, the pythagorean triplets are (3)2-1/2 = 4 and (3)2+1/2 = 5. For any other number , i.e. !5, the triplets are (15)2-1/2 = 112 and (15)2+1/2 = 113.

please solve one question from quadratic equations: if a and b(not equal to 0) are the roots of the eqation x^2+ax+b=0, then the least value of x^2+ax+b is 1)9/4 2)-9/4 3)-1/4 4)1/4a + b = -a => a = -b/2 ab = b => a = 1

=> b = -2 So, the eq becomes x^2 + x - 2 x^2 + x - 2 = (x + 1/2)^2 -9/4 => Least value will be -9/4 the min. would exist at x= -b/2*a.. and the value is = -D/4*a b is coeff of x a is coeff of x^2 hence min val = -9/4 How to find the number of factor of form (3n + 0/1/2) of number N. Its easier for the factors of kind 3n. Just find the number of factors of N/3 and that will be the answer. For factors of form (3n + 1) or (3n + 2) We know that, (3n + 1)(3n + 1) = (3k + 1) (3n + 1)(3n + 2) = (3k + 2) (3n + 2)(3n + 2) = (3k + 1) Lets consider the example of N = 3136 = (2^6)*(7^2) i) (3n + 1) We know that 2 is of form (3k + 2), but 4 is of form (3k + 1). 7 is also of form (3k + 1). So all the factors of form (3k + 1) are expressed as (2^2k)*(7^n) k van take 4 values(0, 1, 2, 3) and n can take 3 values (0, 1, 2) => Number of such factors = 4*3 = 12 ii) (3n + 2) We know 2 is of form (3n + 2) So, all factors of form (3n + 2) can be expressed as (2^(2k + 1))*(7^n) k can take 3 values (0, 1, 2) and b can take 2 values => Number of such factors = 3*3 = 9 Similarly, we can go for (4n + 1) or (4n + 3) Or some other form see all prime numbers (except 2 and 3) can be expressed in the form on 6k+1 or 6k1,though the reverse is not true..i.e. if a number can be expressed in the form of 6k+1 or 6k-1 it is not necessarily a prime number... for example- 13=6*2+1,5=6*1-1 another way of checking whether a number is prime or not is this... i m explaining it with the help of an example.. say the number is 171 and you want check whether the number is prime or not.. so find the nearest square root of the number,in this case it would be 13(13^2=169)..so divide 171 with all prime number less than or equal to 13..in this case check if 171 is divisible by 13,11,7,5,2,3...if it divisible then the number is not prime else the number will be prime... Equations: 1) Quadratic Equation ax^2+bx+c = 0, has maxima = -D/4a at x = -b/2a [D = b^2-4ac] 2) All polynomials of degree 1 would be a straight line. hence if a area under the curve is asked, solve for x = 0 and y =0 and we can get the two points and draw straight lines. This helps in elimininating some options. 3) if the Polynomial is given as ax^n + bx^(n-1) +...+ z = 0 then sum of the roots (taken one at a time) = -b/a

sum of roots taken two at a time = +c/a similarly it continues with alternate -/+ signs. 4) For Quadratic equations, if a>0, then the graph would be a parabola with opens upward and if a now since 12 is less than 100, let us find power of 12 which is nearer to 100.we get 12^2. step2. So the expr become 144^300/100 --> divide 144/100 rem = 44 step3. The expr now become 44^300/100 --> repeat step1. Like this go on and ultimately you will end with a term less than 100 in numerator. That will be the remainder. this is a bit time consuming but with practice you will get it. similarly... if numerator was 12^601, then you can write it as 12^600 * 12. Now you would have to find rem(12/100) * rem(12^600/100) In the end if numerator is greater than 100, again divide by 100. You always have to continue dividing by 100 until the numerator doesnt become less than denominator.