Hi all… After getting PG puyscar for most helpful in quants section this year , i really thought i should do something this year to atleast justify the puyscar… i jus thought to start a thread and post the concepts i learned through PG… m…
Hi all...
After getting PG puyscar for most helpful in quants section this year , i really thought i should do something this year to atleast justify the puyscar.. i jus thought to start a thread and post the concepts i learned through PG.. many legendary persons like vineet,implex,doc mod,maskedmenace,ankurb and lot many people (forgive me for not mentioning the names) have thought me many things which i feel invaluable .. so in coming days , i would like to share it to u as well :)
So , this thread is dedicated for posting quant concepts to tackle specific type of problems..lets post well known ways as well as self-discovered ways to tackle problems we're comfortable with...this way, we'll get to learn through each other...so if u have a specific way of dealing with a particular problem...that makes a difficult problem look easy...please let everybody know ur way...
we can also send our old posts, posts of other puys,concepts from internet , or any means...the idea is to share concepts...no matter where they come from or when were they posted....as concepts are never old...those unknown are always new to us!
newbies can find lots of info here...and i'd sincerely request the quant stalwarts to come up with concepts gained through their vast experience in this lovely forum...
hope this would help us work as a good team and we'll end up being better together.
one of the quantasaurus of PG "MaskedMenace"
has told he would contribute a lot to this thread.. i request all other quant champions to share their knowledge and enlighten us ..
few pointers:
1) i request doc mod and raghav to moderate this thread
2) avoid spamming in this thread so that thread is not diluted
3) kindly pm the concept contributor for specific queries :)
good luck ! lets rock on !
with warm regards
naga:cheers:
ps1 : @mods, if u find the thread to be irrelevant , pls take necessary action
After getting PG puyscar for most helpful in quants section this year , i really thought i should do something this year to atleast justify the puyscar.. i jus thought to start a thread and post the concepts i learned through PG.. many legendary persons like vineet,implex,doc mod,maskedmenace,ankurb and lot many people (forgive me for not mentioning the names) have thought me many things which i feel invaluable .. so in coming days , i would like to share it to u as well :)
So , this thread is dedicated for posting quant concepts to tackle specific type of problems..lets post well known ways as well as self-discovered ways to tackle problems we're comfortable with...this way, we'll get to learn through each other...so if u have a specific way of dealing with a particular problem...that makes a difficult problem look easy...please let everybody know ur way...
we can also send our old posts, posts of other puys,concepts from internet , or any means...the idea is to share concepts...no matter where they come from or when were they posted....as concepts are never old...those unknown are always new to us!
newbies can find lots of info here...and i'd sincerely request the quant stalwarts to come up with concepts gained through their vast experience in this lovely forum...
hope this would help us work as a good team and we'll end up being better together.
one of the quantasaurus of PG "MaskedMenace"
has told he would contribute a lot to this thread.. i request all other quant champions to share their knowledge and enlighten us .. few pointers:
1) i request doc mod and raghav to moderate this thread
2) avoid spamming in this thread so that thread is not diluted
3) kindly pm the concept contributor for specific queries :)
good luck ! lets rock on !
with warm regards
naga:cheers:
ps1 : @mods, if u find the thread to be irrelevant , pls take necessary action
Lets keep the thread flowing with new and innovative concepts.
we shall start with Geometry,as most of the people neglect geometry while atleast one or two questions are sure to be presented from this area.We shall learn the basic theorems first and then move on to applying these in various cat problems.
(I am not going to prove any theorems here).
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.
Hi all...
__________________
few pointers:
1) i request doc mod and raghav to moderate this thread
2) avoid spamming in this thread so that thread is not diluted
3) kindly pm the concept contributor for specific queries :)
good luck ! lets rock on !
with warm regards
naga:cheers:
ps1 : @mods, if u find the thread to be irrelevant , pls take necessary action
thread made sticky
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. distribute 1o chocolates aming 6 children such that no children is empty handed.
2. find whole number solutions for X+Y+Z = 22
3. find natural number solutions for X+Y+Z= 22
4. 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.
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)
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
Posted by vineet.nitd in NCR Dream Team Thread
:cheers:
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)
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
Posted by vineet.nitd in NCR Dream Team Thread
:cheers:
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 :)
P.s I wll keep on updating posts with new formulae 😃
Now problems -
1. 4 points out of 8 points are collinear . Number of different quadrilaterals that can be formed 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
--------------------------------
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
Basic funda : Given in funda list , above
--------------------------------
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
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 a
now example....
f(x) = X^2 + 4X + 3.
if 2X+4 = 0
X= -2
here a>0, 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
for more examples and explanation with graphs visit the link-
First derivative Test
second derivative test
2. f(x) = l x-a l + l x-b l + l x-c l
for such questions, either at x=a or x= b or x= c or x=avg(a,b,c) will f(x) minimum.
f(x) = l x-2 l + l x-8 l + l x-11 l
here x= 2+8+11/3 = 7 will give the minimum value ( f(7) = 11)
f(x) = l x-2 l + l x-5 l + l x-11 l
here x= 5 will give the minimum value. ( f(5) = 9)
3. when sum of any quantities is constant, there product is maximum when they are equal.
example. if 3x+5y=15. find maximum value of x^2*y^3.
here 3x+5y=15
=> 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
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 a
now example....
f(x) = X^2 + 4X + 3.
if 2X+4 = 0
X= -2
here a>0, 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
for more examples and explanation with graphs visit the link-
First derivative Test
second derivative test
2. f(x) = l x-a l + l x-b l + l x-c l
for such questions, either at x=a or x= b or x= c or x=avg(a,b,c) will f(x) minimum.
f(x) = l x-2 l + l x-8 l + l x-11 l
here x= 2+8+11/3 = 7 will give the minimum value ( f(7) = 11)
f(x) = l x-2 l + l x-5 l + l x-11 l
here x= 5 will give the minimum value. ( f(5) = 9)
3. when sum of any quantities is constant, there product is maximum when they are equal.
example. if 3x+5y=15. find maximum value of x^2*y^3.
here 3x+5y=15
=> 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.
hope this helps....
Bases
Problems
- A number is base N is divisible by N-1, when the sum of digits in base N is divisible by N-1
- 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.
- If a number in base N has even number of digits and that number is a palindrome, then the number is divisible by N+1
Problems
- A number 2342a121 is in base 8 and it is divisible by 7. Find the value of a.
- A palindromic number in base 16 will always be divisible by which number?
- 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 ??
Found it somewhere in PG only... thanks to a puy....posting it here so that it might help people who may not have seen it over there.
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 WORD
This 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
This 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
Basic Formulae for TSD
Basic Formulae for Sequences and Series
Some more formulae-
1. 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
3. Tn = Sn-S(n-1)
Some more formulae-
1. 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
3. Tn = Sn-S(n-1)
1.The perimeter of a right angle triangle=(2r+2R)
And the area =r^2+2rR
Where r=inradius
R=circumradius
2.For an right angle triangle
R>=r