how to find the sum of consecutive factorials..?
eg: 1!+2!+3!+4!+.......+21!

M not getting the correct answer by this method for 1st Feb 1984.
May b i m doing something wrong.:splat:...but kindly check!!!!

It is Wednesday.

hey bro is it friday

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.

M not getting the correct answer by this method for 1st Feb 1984.
May b i m doing something wrong.:splat:...but kindly check!!!!

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

Bro but with the approach above yours
Last Non.Zero digit for 40!
10!- 8
40! -8^4= 6(last non zero digit)
Now for 36!x7x8x9x4=6
36!*6=6(so wat wud be the the last non -zero digit nw???)
But you have written 36!*6=2 Hw???..tell me if m doing something wrong!!!!

Re: Rectangles Cut By a Diagonal Line

The general rule: If the lengths of sides (a x b) of the rectangle are
mutually prime, the number of squares cut is a+b-1

Thanks for this useful post but what do u mean by squares cut exactly...cud u pls elaborate??

Pigeon hole theoram

if n+1 pigeons fly to n holes, there must be a pigeonhole containing at least two pigeons


Well this theorem, look apparently simple and trivial, but its extremely powerful. Lets take a test of it.


Example= Let A be any set of nineteen integers chosen from the arithmetic progression 1,4, . . . ,100. Prove that there must be two distinct integers in A whose sum is 104.

Now how do we go about this? remember n and n+1. The hint is to make n+1=19. Something clicked?


see we have 34 numbers of the form 3k+1, from 1 to 100. If we do not want a sum of 104 , we will break them in the sets of 2 integers whose sum is 104
{4,100},{7,97}..{49,55} and {1}, {52}. Clearly we have 16 two element sets and 2 one element set.



So if we make a set of 19 integers, we will have to pick both the integers from atleast one of the two element sets, which will give us a sum of 104.
We are done here.


If you still have doubts, let me explain again, suppose you are four friends ( boys) and there are three girls. And each one of you like a girl out of the three. So at least one of the girls will be liked by two boys.


Lets solved a more involved example, wherein we need not prove a thing, but find a thing. Some people may be feeling cat does not want us to prove but find. Here is how we do that.


Example = Let there be n balls with Ram. he decides to colour one ball with colour 1, two balls with colour 2 and so on upto, fifty balls with colour 50. At the end of it , all n balls are used, and no ball is coloured twice. Ram then draws balls from the lot at random, without replacement. What is is the minimum number of balls that he must draw in order to gurantee drawing 10 balls of the same color?

What the hell is his problem. Why coloring and then taking out. Stupid chap. Let us help him with the math now.
see if he picks all the balls with colors which are less than 10 it will come upto (1+2+3..+9)=45.
Now for the worst case he will pick 9 balls each from rest of the balls, which is 41*9
so total is 41*9+45=41*10+4=414. ( avoid multiplying, be watchful)
now if he picks one more ball, atleast one of the set will be of 10. so we are done
he needs to draw 414+1=415 balls.


Practice Problem 1 = A circular table has exactly 60 chairs around it. There are N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest value of N?


Practice Problem 2 = We call a set "Sum-free" if no two elements of the set add upto a third element of the set. What is the maximum size of the "sum-free" subset of {1,2,3...2n-1}?

Type # 1.

find smallest no. other than k, that leaves remainder k when divided by w,x,y...

to solve such questions, take lcm of w,x,y...and add k to it.

e.g. find Smallest no. other than 4, that leaves remainder 4 when divided by 6,7,8 or 9...

take lcm of 6,7,8,9 and add 4

i.e. 504 + 4 = 508

Type # 2

find smallest no. that leaves remainder 3,5,7 when divided by 4,6,8 respectively.

unlike last case, this time the remainder is not constant. but if u see carefully, difference b/w divisor n remainder is constant. i.e. 4-3=6-5=8-7=1

in such questions, take lcm of divisors n subtract the common difference from it

here, the answer wud be lcm of 4,6,8 i.e 24 - 1 = 23

Type # 3

Smallest no. that leaves remainder 3,4,5 whn divided by 5,6,7 respectively and leaves remainder 1 with 11,

we have just seen a way to tackle the first 3 conditions...the no. wud be lcm of 5,6,7 - 2 = 208

now we have one more condition...remainder 1 with 11.

concept => to a no. if v add lcm of divisors...the corresponding remainders dont change.

i.e to 208, if v keep adding 210 ... the first 3 conditions will continue being fulfilled.

so, let 208 + 210k be the no. that will satisfy the 4th condition...viz (208 + 210k)% 11 = 1

208%11 = 10

210k%11 = k

therefore, 10 + k shud leave remainder 1 when divided by 11.

hence, k = 2. and the no. is 208 + 210 x 2 = 628


e.g. find the smallest no. that leaves remainder 2 when divided by 3,4 or 5 and is divisible by 7

for first 3 conditions....no. is 120 + 2 = 122

hence, 122 + 120k is the required no. which reduces to 3 + 2k when divided by 7...now 3+2k shud be a multiple of 7...easily, k=2 and the required no. is 122 + 120 x 2 = 362

Type # 4

What if there is no relation between divisors n remainders?

e.g. find the smallest no. that leaves remainders 1 with 5, 4 with 7, 6 with 11 and 7 with 13.

we can c...there's no relation among these divisor-remainder sets...neither is the remainder constant...nor is the difference b/w divisor n remainder a constant.

in such cases...take 1 case n target another case...
e.g. i take the case 7 with 13...and target 6 with 11.

which is the smallest no. that leaves 7 with 13? 7 itself...right? so all nos of the form 7 + 13k will give 7 rem with 13.

now am targeting 6 with 11...so i divide 7 + 13k by 11...i get remainder 7 + 2k...now 7 + 2k = 6,17,28,39,50...so that the remainder with 11 is 6.

a no. that gives integral value of k is 17 i.e. 7 + 2k = 17. hence, k =5 and the no. that satisfies these two conditions is 7 + 13 x 5 = 72

now that 2 conditions are fulfilled, lets target a third condition...say 4 with 7.

to 72, if v add lcm of 11, 13 i.e 143, 2 conditions awready satisfied wud continue being satisfied...

hence the no. is of the form 72 + 143 k.

72 + 143k % 7 = 2 + 3k

now 2 + 3k shud be = 4,11,18,25,32... to satisfy the condition of 4 rem with 7..

a no. that gives integral soln is 11..i.e. 2 + 3k = 11, k = 3.

hence, the no. that satisfies all 3 conditions is 72 + 143 x 3 = 501.

now if v see carefully...4th condition...remainder 1 with 501 has already been satisfied...so the no. v have been looking for is 501.

For ease of calculation, start from biggest divisor n gradually move to smaller ones...u'll always see that last 1-2 conditions will be satisfied automatically.

there are theorems for solving above questions...viz chinese theorem etc...but i solve such questions by the way i've suggested...i find this approach very practical as the flow of nos. is very much visible...n i believe i can tackle any twist in the question devised by cat makers thru this method...there are lotsa other questions based on this concept which i'll soon post but the basic concept remains the same...


ORIGINALLY POSTED BY MAXIMUS

Re: Rectangles Cut By a Diagonal Line

The general rule: If the lengths of sides (a x b) of the rectangle are
mutually prime, the number of squares cut is a+b-1

Thus, your example: (3 x 5) gives 3+5-1 = 7

Other examples: (8 x 5) gives 8+5-1 = 12
(9 x 4) gives 9+4-1 = 12
(9 x 5) gives 9+5-1 = 13

BUT (9 x 6) DOES NOT give 9+6-1 = 14. Instead you must proceed as
follows:

First divide (9 x 6) through by common factor 3 to get (3 x 2)
Then apply the rule to (3 x 2) to give 3+2-1 = 4
Now multiply by the factor 3 again to get 12 (which is correct).

Let's do a square figure, say (5 x 5). We divide through by 5. This
gives (1 x 1). Applying the rule gives 1+1-1 = 1. Now multiply up by
the factor 5 again to get 5. We know this is correct because in ANY
square figure the number of squares that are cut will be equal to the
side of the square.

The general procedure for a rectangle (a x b) is as follows:

If a and b are relatively prime the answer is a+b-1

If a and b have a common factor c, first divide through by c to get
(a/c x b/c).
Then apply the rule to get a/c + b/c - 1.
Finally, multiply through again by c to get (a+b-c).

For example, with (9 x 6) the correct answer is 9+6-3 = 12.

:splat::splat:

Attached JPEG

Can anyone pour how to factorize a quadratic equation with high value of C

in Ax2 + Bx + C = 0????

Kindly refer JPEG

In this particular question,by looking at the signs of B and C,you can say that the factorization would be of type (x-a)(x-b).Now you can check C and try some divisibility rules over it.Like 4293 is divisible by 9(sum of 4293 is 18].Similarly,we will see that 81 divides 4293. 81*53= 4293. Also 81+ 53 =134,which is B.I hope someone comes up with a particular technique rather hit and trial method.

Attached JPEG

Can anyone pour how to factorize a quadratic equation with high value of C

in Ax2 + Bx + C = 0????

Kindly refer JPEG

Remainders
Questions regarding finding remainders have appeared a lot of times in CAT. Here are few ways by which u can find the remainders quickly
1)EULER'S THEOREM:
If M and N are two co-prime nos. and N=a^P*b^q*c^r. where (n)=N*(1-1/a)*(1-1/b)*(1-1/c).
Then remainder M^(n)/N=1
For example:
Find the remainder when 5^37 is divided by 63
5 and 63 are co primes,
63= 3^2*7
So, (63)= 63*(1-1/3)*(1-1/7)
=36
Therefore , remainder( 5^37/63) = remainder (5^36*5/63)
=1*5
=5
It's a very easy concept, apply it on other nos and try to find the remainder

2)FERMAT'S THEOREM
If P is a prime number and N is prime to P, then (N^P-N) is divisible by P

e.g: what is the remainder when (N^7-N) is divided by 7
ans. 0
3)WILSON'S THEOREM
If P is a prime no. then remainder (P-1)!+1/P=0, which means that remainder when (P-1)! Is divided by P is P-1
For example: remainder when 40! Is divided by 41
= 41-1
= 40

1).Method to find the last non zero digit of n!

Z(n)= last two digit
Z(n1)= 4, if the tens digit is odd
6, if the tens digit is even
Z(n)= Z(n1)*Z(n/5)!*z( factorial of units digit)
e.g: find the last non zero digit of 36!
Z(36)= 4*Z(36/5)!*z(6)!
=4*7!*6!
=4*(last non zero digit of 7!)(last non zero digit of 6!)
=4*4*2
=2 ans

2). Number of numbers less than and prime to a given number:
If N is a natural number such that N= a^p*b^q*c^r, where a, b, c are different prime nos. and p,q, r are different positive integers then, the number of positive integers less than and prime to N = N(1-1/a)(1-1/b)(1-1/c)
e.g: how many first 1200 nos. are not divisible by 2,3 and 5??
1200= 2^4*3^1*5^2
Therefore number of numbers not divisible by 2, 3 and 5 will be equal to the numbers prime to 1200
=1200(1-1/2)(1-1/3)(1-1/5)
=1200*1/2*2/3*4/5
=320 ans
Thus there are 320 such nos which are not divisible by 2, 3 and 5

Hey,
can some one just tell me or paste a link or upload a pdf file as the tips, concepts, fundas in geometry and mensuration??

This is one chapter which I am not able to do not matter how much or how hard I am practising.. plz help

Thanks
Alok Biyani, Kolkata

Great thread Puys!!!!

pirateiim478 Says

Moderator, please update the thread name to CAT 2011 and if possible prepare a list of similar fundas,posts in other threads(like maximus's marathon thread on Concepts &so;) in this for easier reference..Truly this thread & its funda rocks and it helps Mba aspirants a lot

Important posts of this thread compiled into a simple 28 page pdf file....


22311

Also it can be found here
Concepts.pdf - 4shared.com - document sharing - download

divisibility rules :

we all perhaps know what is divisiblity rule of 2,4,6,8,3,9 etc...

but very few ppl know the logic behind it..

this post is jus to enlighten on this .. for those who know, its jus revisit ..

for others , learn it

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

hope this helps

thanks for reading

cheers
naga

hi can you please elaborate the divisibility of 7 part.
how you have subtract the last 3 digit from the first 3 digit and checked that with the 7? i did not get that. Also can we use this for any 6 digit no. and with any divisor???

Maximum value of an expression (a^m)(b^n)(c^p)... will occur when (a/m)=(b/n)=(c/p)....

Great work guys.. this thread really helped a lot..

i hope this material helps all puys
:cheerio:

dude that book is awesome.... where did you get it from? are there books like this on other topics also?

If two squares of dimension 1*1 are selected from a chess board of size 8*8. What is the probability that both squares share a common side?

Can anyone plz explain Fermat's Theorem and Euler's theorem and its application with exampls

Thank you thank you so much. I understood

Thanks... Though II still seems confusing, I i have followed

madhul90 Says

hello chill factor.. you told us a way to find the max(X1, X2)=X and there are 2X+1 ways to find that..?? can you elaborate on this please...

Since MAX(x1, x2) = x, we can say that at-least one of them should be x.

So, we can find the number of ways in 2 different manner:-

I) If x1 is x, then x2 can take (x + 1) values, i.e., 0 to x.
Similarly when x2 is x, x1 can take (x + 1) values.
So, total 2x + 2 ways, but we need to remove the case when both are x which is counted twice.

So, 2x + 1cases

II) We can choose (x1, x2) in (x + 1)^2 ways, as both x1, x2 can vary from 0 to x, but we need to remove those cases when none of x1, x2 is x.
So, x^2 such cases, as x1, x2 can vary from 0 to (x - 1)

So, (x + 1)^2 - x^2 = 2x + 1 ways

I hope it makes sense.

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

hello chill factor.. you told us a way to find the max(X1, X2)=X and there are 2X+1 ways to find that..?? can you elaborate on this please...

Concepts for Cyclic quadrilateral and how to calculate side of a hexagon if circumscribed circle's radius is given?????????


Plz pour in

properties of cyclic quadrilaterals..

1) opposite angles are supplementary...

2) if a cyclic quadrilateral is given with sides of length a,b,c,d and diagonals of length e,f, then
ac+bd=ef( this is known as Ptolemy's Theorem)

3) area= root under (s-a)(s-b)(s-c)(s-d)
where s= a+b+c+d/2, semiperimeter of cyclic quadrilateral..

(to hell with herons, this is from our own brahmgupta's formula )

There is only one cyclic quadrilateral concept I have come across.

Opposite angles in a cyclic quadrilateral are supplementary.

For examples:
http://jwilson.coe.uga.edu/EMAT6680Su06/Byrd/Assignment%20Nine/CyclicQuad.pdf

Concepts for Cyclic quadrilateral and how to calculate side of a hexagon if circumscribed circle's radius is given?????????


Plz pour in

Apun ka 2 paisa :D

1)Every even number > 2 can be expressed as the sum of 2 primes.
4=2+2, 6=3+3, 8=3+5, 10=5+5, 12=5+7, .. , 100=3+97

2) All these are prime numbers...

31
331
3331
33331
333331
3333331
333333331

next number 333333331 is not prime :D

Gone through all the posts in this thread..guys whats with this 'subscribed' and thanks thingy?? Just go to the ''thread tools'' and subscribe with it..let the thread be crisp and concise...also there is a GREEN button to thank the post..even if the desire to write that 5 letter word persists..you can always add a formula or two..and in the end can say thanks for the post:D

Thanks .See, I lead by example :D
:cheers:

rohan101=a then the area of the region encompassed is given by 2a^2.

plot the graph of x + y=a. it will be a square with length of side 2^1/2a... hence area will be 2a^2..

gurupdeshcheema Says

When we cut a cube of say a*a*a from it , then what way would you say we should proceed ??

brilliant analysis of the cubes......
i would request everyone to visualise and understand all the above concepts and not to learn these formulas by heart since it will only create confusion during exams..

Moderator, please update the thread name to CAT 2011 and if possible prepare a list of similar fundas,posts in other threads(like maximus's marathon thread on Concepts &so;) in this for easier reference..Truly this thread & its funda rocks and it helps Mba aspirants a lot

gr8 goin puys....keep up the good work,....

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:

When we cut a cube of say a*a*a from it , then what way would you say we should proceed ??

please update it...as CAT2011....

If x +y=a then the area of the region encompassed is given by 2a^2.

Can anyone explain how to compare fractions?
For eg a set of fractions :

117/229,143/291,123/241,109/219

How to arrange the above fractions in ascending order?

The approach I apply for this kind of prob is of %age and u really need to be very fast and calculative for this.

Consider 117/229. Now its obvious from numerator(N) and denominator(D) that the value is somewhere around 50%.

1%=2.29 and 10%=22.9 =>50%=114.5
now whats left is (117-114.5)/2.29
2.5/2.29
Again it comes out around 1
So the value would be 51.XXX

Now to find that .XXX u need to do some calculations finally

(2.5-2.29)/2.29 =0.31/2.29 = 0.135(upon solving)

Here too the same approach can be followed but be cautious while dealing with decimals.
(.31-.229)/2.29 = .1XX
Ando so on.

U can at times approximate decimal part coz generally the fractions given dont vary that minutely. But then again u need fast calculation.

And it comes out to be 51.135%

Similarly you can find the value for others too. See half the work is done as u see through the N & D. Just make out a division result in %age.

Hope this helps..!!

Can anyone explain how to compare fractions?
For eg a set of fractions :

117/229,143/291,123/241,109/219

How to arrange the above fractions in ascending order?