ALL IMPORTANT POSTS BY INDER JEET of CAT prep 2014 -2015 on FB
#1Concept: Number of ways a number N can be written as product of three numbers
Let N= 360 = 2^3*3^2*5 =a^x*b^y*c^z
Now the primes 2,3,5 can be anyway arranged among a,b,c such that power ofeach primes remains same.
So for 2^3, x+y+z=3 => 5c2 For 3^2, x+y+z=2, => 4c2For 5, x+y+z=1, => 3c2
So total ordered solutions => 5c2*4c2*3c2= 10*6*3=180
Unordered: Here there are three possible triplets abc ie all distinct, whichis arranged in 3! Ways, aab which is arranged in 3!/2=3 ways and aaa which isarranged in only 1 way in its Ordered solution. In case of unordered we justwrite those repeated cases in only 1 way
for ex (123),(231),(312),(321),(132),
So how to convert into unordered: From total, remove aab and aaa type ifthey are present. Now we are left with ordered distinct triplets. Which can beconverted into unordered by dividing by 3! And then just add aab and aaa case
So as for 360=2^3*3^2*5, we got 180 as total, aab types : (1,1,N, ),(2,2,new), (3,3,n),(6,6,n). Total 4 which are allarranged in 3 ways. So 3*4=12
aaa type : not possible.
Unordered => (180-3*4)/3!+4= 32.When question says: in how many ways a number can bewritten as product of three numbers. You need to calculate unorderedWhen question say how many solutions for x*y*z= NThen you have to find ordered.
#2 BASE SYSTEM Suppose you have a 1 000 L tank to be filled with water. The bucketsthat are available to you all have sizes that are powers of 3, i.e. 1, 3, 9,27, 81, 243, and 729 L. Which buckets do you use to fill the tank in theminimum possible time?You will certainly tell me that the first bucket you will use is of 729 L. Thatwill leave 271 L of the tank still empty. The next few buckets you will usewill 243 L, 27 L and 1 L. The use of buckets can be shown as belowimage
We can say that1 000 = 729 + 243 + 27 + 1= 1*3^6 + 1*3^5 + 0*3^4 + 1*3^3 + 0* 3^2 + 0*3^1 +1*3^0 .The number 1 000 has been written in increasing powers of 3. Therefore, 3 isknown as the base in which we are expressing 1 000.
· Concept 1 : for a base system N to find last digit divide by N for last 2 digits divide by N^2 .find last digit of 17 in base 3 --->. 17/3 ..rem 2 .. Last digit 2
· IMPORTANTRULES ABOUT BASES
Rule1. A number in base N when written in base 10 isdivisible by N -1 when the sum of the digits of the number in base N isdivisible by N-1.EXAMPLE. The number 35A246772 is in base 9. This numberwhen written in base 10 is divisible by 8. Find the value of digit A.
Rule2. When the digits of a k-digits number written in base N are rearranged inany order to form a new k-digits number, the difference of the two numbers,when written in base 10, is divisible by N- 1.EXAMPLE44. A four-digit number N1 is written in base 13. Anew four-digit number N2 is formed by rearranging the digits of N1 in anyorder. Then the difference N1-N2 when calculated in base 10 is divisible by ??
· Concept 2 trailing zero in factorial in base N no of trailing zeroes in k! when k! Written in base 10 N=a*b*c ...a,b,c all are prime factors of N say c isgreatest so no of 0 k/c+k/c^2.... ex.4! In base 3 no of trailing 0 s 4/3 =1
· To check even andodd numbers in base system 1)if base b is even in number( N) with base bStep 1: Check last digit in (N) in that baseStep: if Last digit is even then (N) in base b iseven else oddEg:(23456) base 32 is evenAs base 32 is even and last digit in N is 6, is even2)if base is odd in number (N)Step:1 check sum of digitsStep:2 if sum is even, then number (N) is even elseoddEg:(234565) base 33 is evenAs base is 33 (odd) & sum of digits=20 is even
#3 Concept: Number of zeroes in a base N depends on the limiting prime containedin N. In decimal system ie base 10=2*5, we look for the number of 5's in it.
Zeroes in 100! => [100/5]+[100/5^2] = 20+4=24
In base 7 = 1*7, so number of 7
=> [100/7]+[100/7^2]=14+2=16
In base 33=3*11, so number of 11
=> [100/11]= 9
But in case if 12=2^2*3 or 24=2^3*3, Limiting primes are 2^2 and 2^3 respectively.
So number of zeroes in 100! In base 12=2^2*3
=> [100/2]+[100/2^2]+[100/
So number of 2^2 => [97/2]= 48
#4
=>For a cube of side n x n x n
Cube For 0 sides painted, => (n-2)^3
Cube For 1 side painted => 6(n - 2)^2
For cube with 2 sides painted, => 12(n-2)For the cubes with 3 sides painted, it will always be 8. The eight paintedcubes are the 8 corners of any and all cubes.
=>For cuboid of side a x b x c
0 side painted => (a-2)(b-2)(c-2)
1 side painted => 2[ (a-2)(b-2) + (b-2)(c-2) + (a-2)(c-2) ]
2 side painted => 4(a+b+c -6)
3 side painted => 8
#5 Short tricks (self made): In case of distributing n distinct balls into 3identical boxes, instead of making long cases you can just apply the formula(3^n -3)/3! +1
ex: 5 distinct balls in 3 boxes, => (3^5 -3)/3! + 1 = 41 4 balls, => (3^4-3)/3! +1 = 14 6 balls => (3^6 -3)/3! + 1= 122
(1). When is a function valid ? => Our definition of a function says that it is a rule mapping a number toanother unique number.So we cannot have a function which gives two differentoutputs for the same argument.so f(x)=root(x) isnt a valid function until youput a domain x>=0, as x=1,2..., y=+/-1 , +/-1.4... etc.
(2). Linear functions:=>Functions of the formf(x) = ax+b are linear, andthey are represented graphically by straight lines. The number a representsthegradient of the line, and the number b represents the y-axis intercept.
(3) . Polynomial Functions: .=> A polynomial is a function of theformf(x) = anxn +an−1xn−1 +...+a2x2 +a1x+a0 .The degree of a polynomial is thehighest power of x in its expression. Constant (non-zero)polynomials, linearpolynomials, quadratics, cubics and quartics are polynomials of degree 0,1, 2 ,3 and 4 respectively. The function f(x) = 0 is also a polynomial, but we saythat itsdegree is ‘undefined’. .=> A polynomial of degree n can have up to(n−1) turning points in it's graph. Eg. a quadratic having U shape has oneturn. .=> The number a is a root of the polynomial function f(x) if f(a) =0, and this occurs when(x−a) is a factor of f(x).If a is a root of f(x), and if(x−a)^m is a factor of f(x) but (x−a)^(m+1) is not a factor, then we say thatthe root has multiplicity m. At a root of odd multiplicity the graph of thefunction crosses the x-axis, whereas at a root of even multiplicity the graphtouches the x-axis.
(4.)Exponential and Logarithmic Functions:.=>Exponential:A function ofthe form f(x) = a^x (where a > 0) is called an exponential function.Thefunction f(x)=1^x is just the constant function f(x) = 1.The function f(x) =a^x for a > 1 has a graph which is close to the x-axis for negative x andincreases rapidly for positive x.The function f(x) = a^x for 0 Logarithmic: A function of the form f(x) = log(basea) x, (where a > 0 and a =/=1) is called a logarithm function.The functionf(x) = log(base a) x for a > 1 has a graph which is close to the negativef(x)-axis for x < 1 and increases slowly for positive x.The function f(x) =log(base a) x for 0
(5). Composite Functions: f(g(x) => In general gf(x) is not equal tofg(x). => The domain of a composed function is either the same as the domainof the first function, or else lies inside it.The range of a composed functionis either the same as the range of the second function, or else lies insideit..
(6).Maxima Minima: => We can locate the position of stationary points by looking for points where dy/dx= 0.As we have seen, it is possible that somesuch points will not be turning points.At a turning point dy/dx= 0.Not allpoints where dy/dx= 0 are turning points, i.e. not all stationary points areturning points. =>We can calculate d2y/dx2 at each point we find. =>Ifd2y/dx2 is positive then the stationary point is a minimum turning point.=>If d2y/dx2 is negative, then the point is a maximum turning point. =>Ifd2y/dx2= 0 it is possible that we have a maximum, or a minimum, or indeed othersorts of behaviour.So if d2y/dx2= 0 this second derivative test does not giveus useful information and we must seek an alternative method like AM-GM.
(7). AM-GM theory: Arithmetic mean is always greater than or equal togeometric mean. => (a+b)/2 >= root(a*b)
(8 ). Pascal's Triangle: Pascal’s triangle 11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 1
(9).Binomial expansion: => (a+b)^n = nC0*a^0*b^n + nC1*a^1*b^(n-1) +nC2*a^2*b^(n-2) + .... nCn*a^n*b^0
.(10)Number of functions:Set A has m elements Set B has n elements.
(1) If m>n One to One function =0 ; If n>m One to One function =P(n,m) ;
(2) Number of functions = n^m ; Number of relations = 2^(m*n) ;
(3) No of Many to One functions = =Number of functions - Number of One toOne =n^m -P(n,m) (4)
(4)If n>m Number of Onto functions =0 ; If m>n Number of Ontofunctions= sigma ( r = 1 to n) (-1)^(n-r) *C(n,r)*r^m;.
(0r) Number of FunctionsA = ( a,e,i,o,u ), B = ( 1,2,3)1. No. of Functions from A to B - Distributing 5 distinct Balls in 3 Distinctboxes.2. No. of Onto functions from A to B - Distributing 5 distinct Balls in 3Distinct boxes such that none of the box is empty.3. No. of Into functions from A to B - Distributing 5 distinct Balls in 3Distinct boxes such that atleast one of the box is empty.A = ( a,e,i ), B = ( 1,2,3,4,5)4. No. of One to One functions from A to B - Distributing 3 distinct Balls in 5Distinct boxes such that none of the boxes has more than 1 ball.5. No. of Many to One functions from A to B - Distributing 3 distinct Balls in5 Distinct boxes such that atleast one of the boxes has more than 1 ball
#7 #Alligations
1) Alligation
It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.
2) Mean Price
Mean price is the cost price of a unit quantity of the mixture
3) Suppose a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid = [x(1−y/x)^n] units.
4) Rule of Alligation
If two ingredients are mixed, then Ratio
(Quantity of cheaper/Quantity of dearer)= (C.P. of dearer - Mean Price)/(Mean price - C.P. of cheaper)
#8
#Orderedunordered funda...
Whenever we need to find the nos of triplets or two nos whose LCM is XYZ then we first factorize the no. say the no is 2^2 * 3^3 * 5^3.let two nos be a and b whose LCM is 2^2 * 3^3 * 5^3. so first pick up 2^2 . now a nd b can have power 0,1,or 2 power of 2 i.e 3 ways so 3 * 3 = 3^2 ways for them.but there will be a case when none of them will have 2^2 as their power. we need to subtract that coz then LCM wont have 2 with power 2.
so basically subtract the case when a and b have 0 and 1 only as power of 2. i.w 2*2 = 2^2 ways so for 2^2 ways is (3^2 - 2^2) = 5 which is same as (2*2 +1 ) similarly for 3^3 it is 4^2 - 3^2 = 7 which is same as ( 2*3 +1) and for 5^3 it is 4^2 -3^2 = 7 which is same as (2*3 +1)
so basically we are doing [ (a+1)^2 -a^2] [ (b+1)^2 - b^2] [ (c+1)^2 - c^2 ] = (2a + 1 ) ( 2b +1) (2c+1)ORDERED 2) THREE NOS a b and c similarly for three nos imply a b and c. pick 2^2 a b an c all have options of 0,1,2 i.i. 3*3*3 =3^3 ways.now subtract when both dont have 2 as their power i.e have only 0,1 as their power. so 2*2*2 = 2^3 so for 2^2 we have ( 3^3 - 2^3) for 3^3 we have ( 4^3 - 3^3) for 5^3 we have ( 4^3 - 3^3) and so on.
3) FOUR NOS : ( 3^4 - 2^4) ( 4^4 - 3^4)(4^4 - 3^4) SO FORMULA FOR ORDERED IS : [ (a+1)^N -a^N ] [ (b+1)^N - b^N ] [ (c+1)^N - c^N ] WHERE N = no of the numbers whose LCM is to be found.
Unordered. First of all let me explain what is unordered. coz most guys have asked me the difference bw ordered nd unordered. See ordered means (a,b) is different from (b,a). But in unordered (a,b) is same as (b,a) i.e 1,2 and 2,1 are different for ordered and 1,2 is same as 2,1
a) Two nos a and b . LCM is N = 2^2 * 3^3 * 5^3 ordered (2a + 1) ( 2b + 1)(2c+1) = 5*7*7 here one of the pair will be of form 2^2 * 3^3 * 5^3 and 2^2 * 3^3 * 5^3 ( coz LCM of these two is also 2^2 * 3^3 * 5^3)
So except for this pair all the nos have distincr pairs of a and b so we just need to subtract 1 from 5*7*7 and then divide by 2! (why 2! coz here we are finding UNORDERD .so divide by 2! . i.e UNPERMUTE them )so ( 5*7*7 -1) /2 = 122 +1 = UNORDERED ( why +1 coz we have removed the case of 2^2 * 3^3 * 5^3 and 2^2 *3^3 * 5^3. so +1 to add it back )
b) THREE NOS a b c ORDERED : ( 3^3 - 2^3) * ( 4^3 - 3^3) * ( 4^3 - 3^3) now let us look at hose nos in which all the tree no a b and c are not distinct. first let us look at a a b . here we are talking about nos like 1 1 X 2 2 X 4 4 X 3 3 X 5 5 X ... and so on.
so basically we need to subtract the cases where the no can be expressed as LCM of two nos ( a b ) 2^2 * 3^3 * 5^3 can be expressed as LCM of two nos in 5*7*7 = 245 -1 =244 ( why -1 coz here again N,N are the two nos in this pair which are same ) now this triplet a a b repeats 3 times. (not 3! times as the two nos are same. hence 3!/2 = 3 ways ) also N, N, N is another triplet whose LCM is N ( N = 2^2 * 3^3 * 5^3) and occures only once )
so our final answer is [ ORDERED - 3 * those occuring twice - those occurng once]/3! + those occuring twice + those occuring once = [ ( 3^3 - 2^3) * ( 4^3 - 3^3) * ( 4^3 - 3^3) - 3* 244 - 1 ]/ 3! + 244 + 1#9 #Geometry#Triangles#centres Triangles: A three sided polygon is a triangle. Properties 1) A+B+C=180 degrees .
2)exterior angle=sum of opposite interior angles
3)For any a,b,c where a,b,c are sides of a triangle a+b>c,a-b
4)Incentre=point of intersection of angular
bisectors.Incentre=((ax1+bx2+c
Excentre opposite to A=((-ax1+bx2+cx3)/(-a+b+c) ,(-ay1+by2+cy3)/(-a+b+c))
5)Angle made by incentre and any side=90+angle of opposite side/2
6)circumcentre of (0,0) (x1,y1) (x2,y2)=(y1-y2)(x1x2-y1y2)/
For any other triangle , transform one of the points to origin
7)In equilateral triangle -the circumcentre,centriod ,incentre and orthocentre all lie on median(Euler line)
8)In equilateral triangle -the circumcentre,centriod ,incentre and orthocentre coincide.circumradius=a/
1) All sides are equal. 2) Angles of every equilateral triangle are equal to 60° 3) Every altitude is also a median and a bisector.
4) Every median is also an altitude and a bisector.
5) Every bisector is also an altitude and a median.
6) If the length of a side is a the area of the equilateral triangle is ¼a2√3
7) The altitudes, medians and the bisectors of a equilateral triangle are equal to ½a√3
9)In right triangle , circumradius=C/2 where c is hypotenuse.
10) In right triangle Incentre=ab/(a+b+c)
11) In right triangle Area=ab/2=2ab/4=((a+b)^2-c2)/
triangle given a,b,c = sqrt[s(s-a)(s-b)(s-c)] when s = (a+b+c)/2 (Heron's formula)
In right angled triangle the euler line goes through the right vertex and midpoint of hypotenuse
12)Apollonius theorem:ab^2+ac^2=2(ad^2+dc^2)
13)The area K of any triangle is the product of its inradius (the radius of its inscribed circle) and its semiperimeter:
A= rs.
14)The area of a triangle can also be calculated from its semiperimeter and side lengths a, b, c using Heron's formula: K = root(s*(s-a)(s-b)(s-c))
15)The circumradius R of a triangle can also be calculated from the semiperimeter and side lengths: R = (abc) \(4*sqrt(s(s-a)(s-b)(s-c)).=ab
16)In any triangle a/sinA=b/sinB=c/sinc=2R
17)The inradius is r = Root((s-a)(s-b)(s-c)/s).
18)The law of cotangents gives the cotangents of the half-angles at the vertices of a triangle in terms of the semiperimeter, the sides, and the inradius.
19) In a right triangle, the radius of the excircle on the hypotenuse equals the semiperimeter. The semiperimeter is the sum of the inradius and twice the circumradius. The area of the right triangle is (s-a)(s-b) where a and b are the legs.
20) Right triangle properties The area of a right triangle is given by the formula: A = ½ab c2 = a2 + b2(Pythagoras' theorem) n.c = a2 m.c = b2 h.c = a.b a = c.sin(A) = c.cos(B) b = c.sin(B) = c.cos(A)#10CirclesDefinitions:Chord: A chord is a line-segment that is drawn between any two points on the circumference of the circle. The largest chord in the circle is the diameter.Arc: Arc is a curved line that is part of the circumference of the circle. A minor arc is one that is lesser than the semi-circle and a major arc is one that is greater than the semi-circle.Segment: The part of area of the circle that is cut by a chord is called the segment. The part that is smaller than the area of the semi-circle is the minor segment and the part that is larger than the area of the semi-circle is the major segment.Sector: The part of the area of the circle that is bounded by an arc and two radii is called the sector.Tangent: A line drawn from a point outside the circle, such that it touches the circle at only one point is called the tangent.Secant: A line drawn from a point outside the circle, such that it cuts the circle at two points on its circumference is called the secant.Central angle is the angle made at the center of the circle by a chord.Formulae:Consider a circle with radius 'r'.Area of the circle = π∗r2Circumference of the circle = 2*π∗rLength of an arc with central angle θ∘ = (θ/360)*2*π*rArea of a sector with central angle θ∘ = (θ/360)* π∗r2Circle Properties:1. The perpendicular drawn from the center of a circle to a chord bisects the chord. Conversely, the line joining the mid-point of a chord and the center of the circle, is perpendicular to the chord.2. Chords that are equal are equidistant from the center of the circle. Conversely, if the lengths of perpendiculars from the center of the circle to two chords are equal, the chords are equal.3. The length of a chord keeps increasing as it moves closer to the center of the circle.4. The angles subtended by an arc at any point on the circumference of the circle in the major segment, are equal.5. The angle subtended by an arc at the center is twice the angle subtended by the arc at any point on the circumference of the circle in the major segment.6. Equal chords subtend equal angles at the center.7. The angle subtended by a diameter on the circumference is a right angle.8. From an external point, there can be two tangents to a circle. The lengths of both these tangents are equal.9. The line joining the center of the circle and the point where a tangent touches the circle is perpendicular to the tangent.10. A chord is drawn from the point at which a tangent touches the circle. The angle made by this chord with the tangent is equal to the angle made by the chord at any point on the circumference in the alternate segment.11. Two chords AB and CD are drawn in such a way that they intersect at a point O. Then, AO*OB = CO*OD.12. From a point O outside a circle, two secants are drawn such that first secant cuts the circle at A and B and the second secant cuts the circle at C and D.Then, OA*OB = OC*OD.13. From a point A outside the circle, a secant is drawn such that it cuts the circle are points S and U. From the same point A, a tangent is drawn such that it touches the circle at T. Then, AS*AU = AT^2.Common Tangents:1. If two circles intersect have zero points of intersection, the number of common tangents to the two circles is four.2. If two circles touch each other at one point, and the circles do not lie inside each other, the number of common tangents to the two circles is three.3. If two circles intersect each other at 2 points, the number of common tangents to the two circles is two.4. If two circles touch each other at 1 point, and one circle lies completely inside the other circle, the number of tangents to the two circles is one.5. If one circle lies completely inside the other circle and there is no point of intersection between the two circles, the number of common tangents is zero.#11#Integral_Solutions
1) x+y+z+...(r terms) = n
Non negative solutions => (n+r-1)C(r-1)
Positive solns => (n-1)C(r-1)
2) ax+by= n
Non negative integeal solns => [n/lcm(a,b)]+1, if anyone of a or b is divisible by n, otherwise -1
For positive solns remove those including x/y=0 from non negative.
3) |x|+|y|= n has 4n solns
|x|+|y|+|z|=n has 4n^2+2 solns
where n => natural number
when n=0, only 1 soln For n
4) xy=N ,where N is natural number
Number of positive solns => Number of factors of N. Integral soln => 2*Number of factors of N
5) xy= N
Number of unordered positive solns such that x and y are coprimes is 2^(n-1), where n is number of primes containing N.
6) x + y = N, where N => Natural number
Number of positive solutions such that x and y are coprime is Euler of N. (If N=a^x*b^y, Euler(N)= N(1-1/a)(1-1/b)
7) i) a/x + b/y = 1/n
Number of positive integral solns => factors of a*b*n^2
Number of integral solns => 2*factors of a*b*n^2 - 1.
ii) a/x - b/y = 1/n
Number of integral solns => 2*factors of a*b*n^2 -1
Positive => factors of a*b*n^2 which lies below a*n
8) xyz= N = a^x*b^y
Number of ordered positive integral solns => (x+2)C2*(y+2)C2
Integqral solns => 4* (x+2)C2*(y+2)C2
9) x^2 - y^2 = N
Positive solns => Number of even*even or odd*odd factors if not perfect square otherwise -1
Integral = 4*positive solns
Zero solns if N = 4k+2 form as they are always odd*even form.
10) x^2 + y^2 = N
It will have integer solns only if N contains primes of the form 4k+1 type.
If all primes are 4k+1 type Then positive solns => number of factors of N, (Be careful in case of perfect square) it will be -1 than
If it contains some 4k+3 primes with odd power then 0 solns
If it contains even powers of 4k+3 primes, then positive solns = number of factors of 4k+1 form only.
Integral solns = 4*positive solns
#12
Concept : Rings and Fingers. (Both Distinct)
Suppose there are 'n' rings to be worn in 'r' fingers. Here we will first assume Rings to be identical. Distribute them,then arrange them all.
So a+b+c..(r terms)= n => (n+r-1)C(r-1) for distribution And and arranging those r rings in r! Ways So your answer will be (n+r-1)C(r-1)*r!
Ex:1 Let there be 5 rings to be worn in 3 fingers
So its just 7c2*5!.
Ex: 2 let there be 10 rings to be worn in 5 fingers such that each finger has at least 1 ring. Here just apply natural number solution instead. (n-1)C(r-1)
Then a+b+c+d+e= 10, => (10-1)C(5-1) => 9C4*10!.
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