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1) SIGNS OF ax2 + bx + c =0:
A) IF ROOTS ARE IMAGINARY, THEN SIGN IS SAME AS THAT OF a, FOR ALL REAL
VALUES OF x.
B) IF ROOTS ARE REAL AND DISTINCT, SIGN IS SAME OR OPPOSITE TO SIGN OF a
ACCD. AS IN ACCORDANCE OF INTERVAL IN WHICH x LIES.
C) IF ROOTS ARE EQUAL, THEN SIGN IS SAME AS THAT OF a.
D) IF ax2 + bx + c > 0 ALWAYS, THEN ITS D <=0 AND a>0
E) IF ax2 + bx + c < 0 ALWAYS, THEN ITS D <=0 AND a<0
SOME IMPORTANT SHORTCUTS:
f(x) = ax2 + bx + c , a>0,
then,
A) CONDITIONS FOR BOTH ROOTS OF f(x) = 0, TO BE GREATER THAN NUMBER K
ARE D>=0, f(K) > 0, ( -b / 2a) > K.
B) THE NUMBER K LIES BETWEEN ROOTS OF f(x)=0 IF f(K)<0.
C) EXACTLY ONE ROOT LIES BETWEEN NUMBERS D AND E, IF f(D)*f(E) < 0.
D) BOTH ROOTS CONFINED BETWEEN NUMBERS M AND N, IF 
1) THE NO. OF FUNCTIONS FROM FINITE SET A TO FINITE SET B = 
2) THE NO. OF ONE-ONE FUNCTIONS DEFINED FROM FINITE SET A TO FINITE SET B IS = 
0 OTHERWISE.
3) THE NO. OF ONTO FUNCTIONS DEFINED FROM FINITE SET A CONTAINING N ELEMENTS TO FINITE SET B = 
4) THE NO. OF ONTO FUNCTIONS DEFINED FROM FINITE SET A ONTO A FINITE SET B IS EQUAL TO NUMBER OF WAYS OF DIVIDING
n(A) THINGS INTO n(B) GROUPS SO THAT NO GROUP IS EMPTY, IF n(A) >= n(B). AND IS 0, OTHERWISE.
5)THE NO. OF BIJECTIONS FROM FINITE SET A ONTO A FINITE SET B IS [ n(A) ] ! IF n(A) = n(B) AND IS 0, OTHERWISE.
1)THE NO. OF WAYS OF DIVIDING 2M DISTINCT OBJECTS INTO TWO GROUPS OF M OBJECTS EACH:
A) IS 
, WHEN DISTINCTION CAN BE MADE BETWEEN TWO GROUPS.
B) IS 
, WHEN NO DISTINCTION CAN BE MADE BETWEEN TWO GROUPS.
2) THE TOTAL WAYS OF SELECTING P ALIKE OF ONE KIND, Q OF OTHER KIND, R OF OTHER KIND , OUT OF ( P + Q +R ) THINGS IS
( P + 1 )( Q + 1 )(R + 1 ) - 1.
WE CAN USE THESE FOR MORE ALIKE CASES ALSO.
1)THE NO. OF PERMUTATIONS OF N THINGS TAKEN TOGETHER WHEN P ARE ALIKE OF ONE KIND, Q OF OTHER TYPE, R OF OTHER
KIND AND REST OF ALL DIFFERENT IS 
. WE MAY CONTINUE IT FOR ANY NUMBER OF ITEMS.
2)THE NO. OF PERMUTATIONS OF N DISSIMILAR THINGS, TAKEN ALL AT A TIME IS (N-1)!
3)THE NUMBER OF WAYS OF ARRANGING N DISTINCT OBJECTS ALONG A CIRCLE, WHEN CLOCKWISE AND ANTICLOCKWISE
ARRANGEMENTS ARE CONSIDERED SAME IS 
4)THE NUMBER OF NECKLACES FORMED WITH N BEADS OF DIFFERENT COLOURS OR
GARLANDS FORMED WITH N DIFFERENT FLOWERS
IS .
5)THE NUMBER OF WAYS OF DIVIDING ( M + N ) THINGS INTO TWO GROUPS OF M THINGS AND N THINGS IS 
A) SUM OF NUMBERS FORMED BY TAKING ALL N DIGITS = [( SUM OF ALL N DIGITS ) * [ (N-1) ! ] * ( 1111... N TIMES ) ].
B) THE NO. OF WAYS IN WHICH M OF ONE KIND, N OF OTHER KIND, ARRANGED IN A ROW, SO THAT ALL OTHER TYPE OF THINGS
COME TOGETHER = [ (N) ! * (M+1) ! ].
C) FOR N OF ONE KIND, ( N-1 ) OF OTHER KIND, NO. OF WAYS OF ARRANGING IN A ROW, SO THAT NO TWO THINGS OF SAME TYPE
OCCUR TOGETHER = [ (N) ! * ( N - 1 ) ! ]
D)NO. OF WAYS FOR M (OF ONE TYPE), N(OF OTHER TYPE), WHERE ( M >= N ), CAN BE ARRANGED IN A CIRCLE, SO THAT TWO THINGS
OF SAME TYPE DO NOT OCCUR TOGETHER = 
AND WHEN THINGS OF SECOND TYPE COME
TOGETHER = ( M) ! * ( N ) !
A) THE NO. OF DISTINCT POSITIVE INTEGRAL DIVISORS OF 
WHERE P1, P2, ...., PR ARE PRIMES
OF ASCENDING ORDER IS 
B) THE SUM OF DISTINCT POSITIVE INTEGRAL DIVISORS OF WHERE P1, P2, ...., PR ARE PRIMES
OF ASCENDING ORDER IS 
C) IF N ITEMS ARE TO ARRANGED IN A ROW, THEN NO. OF WAYS IN WHICH THEY CAN BE REARRANGED SO THAT ONE OF THEM
OCCUPIES THE PLACE ASSIGNED TO IT IS 
D) IF THERE ARE M ITEMS OF ONE KIND, N ITEMS OF OTHER KIND, AND SO ON, THEN NO. OF WAYS OF SELECTING R ITEMS OUT
OF THESE ITEMS IS 
E) IF THERE ARE M ITEMS OF ONE KIND, N ITEMS OF OTHER KIND, AND SO ON, THEN NO. OF WAYS OF SELECTING R ITEMS OUT
OF THESE ITEMS, SUCH THAT AT LEAST ONE OF EACH KIND IS INCLUDED
IS 
F) THE NO. OF WAYS OF SELECTING R ITEMS FROM N THINGS, P ARE IDENTICAL, = 
AND =
A) IF GIVEN N POINTS IN A PLANE,NO 3 OF WHICH ARE COLLINEAR, THEN THE NO. OF LINE SEGMENTS FORMED = 
BUT IF M (M >=3) POINTS IN THIS N SIDED POLYGON ARE COLLINEAR, THEN NO. OF LINE SEGMENTS FORMED =
B) IN A N SIDED CLOSED POLYGON, THE NO. OF DIAGONALS = 
C)IN A PLANE HAVING N DISTINCT POINTS, NO 3 OF WHICH ARE COLLINEAR, THE NO. OF TRIANGLES FORMED = 
NOW , IF M POINTS OUT OF THSE N POINTS ARE ON THE CIRCUMFERENCE OF CIRCLE,
1) NO. OF QUADRILATERALS = 
2)RESTRICTED PERMUTATIONS:
THE NO. OF PERMUTATIONS OF N OBJECTS TAKEN R AT A TIME, IN WHICH M PARTICULAR OBJECTS ARE ALWAYS:
1) EXCLUDED = 
2) INCLUDED = 
3)RESTRICTED COMBINATIONS:
1) EXCLUDED = 
2) INCLUDED = 
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1) 2^2n-1 Is Always Divisible By 3
2^2n-1 = (3-1)^2n -1
= 3M +1 -1
= 3M, thus divisible by 3
2) What Is The Sum Of The Divisors Of 2^5.3^7.5^3.7^2?
ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6
Funda : if a number 'n' is represented as
a^x * b^y * c^z ....
where, {a,b,c,.. } are prime numbers then
Quote:
(a) the total number of factors is (x+1)(y+1)(z+1) ....
(b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c)....
(c) the sum of relatively prime numbers less than the number is n/2 * n * (1-1/a) * (1-1/b) * (1-1/c)....
(d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x*y*...)
3) What Is The Highest Power Of 10 In 203!
ANS : express 10 as product of primes; 10 = 2*5
divide 203 with 2 and 5 individually
203/2 = 101
101/2 = 50
50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1
thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198
divide 203 with 5
203/5 = 40
40/5 = 8
8/5 = 1
thus power of 5 in 203! is, 49
so the power of 10 in 203! factorial is 49
4) If x + y + z = 7 And xy + yz + zx = 10,
Then What Is The Maximum Value Of x? ( Cat 2002 Has Similar Question )
ANS:
49-20 = 29, now if one of the y,z is zero, then the sum of other 2 squares shud be equal to 29, which means, x can take a max value at 5
5) In How Many Ways Can 2310 Be Expressed As A Product Of 3 Factors?
ANS: 2310 = 2*3*5*7*11
When a number can be expressed as a product of n distinct primes,
then it can be expressed as a product of 3 numbers in (3^(n+1) + 1)/2 ways
6) In How Many Ways, 729 Can Be Expressed As A Difference Of 2 Squares?
ANS: 729 = a^2 - b^2
= (a-b)(a+b),
since 729 = 3^5,
total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27.
So 4 ways
Funda is that, all four ways of expressing can be used to findout distinct a,b values,
for example take 9*81
now since 9*81 = (a-b)(a+b) by solving the system a-b = 9 and a+b = 81 we can have 45,36 as soln.
7) How Many Times The Digit 0 Will Appear From 1 To 10000
ANS: In 2 digit numbers : 9,
In 3 digit numbers : 18 + 162 = 180,
In 4 digit numbers : 2187 + 486 + 27 = 2700,
total = 9 + 180 + 2700 + 4 = 2893
8 ) What Is The Sum Of All Irreducible Factors Between 10 And 20 With Denominator As 3?
ANS :
sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 + 13.33 + 13.66…….
= 21 + 23 + ……
= 300
9) If n = 1+x Where x Is The Product Of 4 Consecutive Number Then n Is,
1) an odd number,
2) is a perfect square
SOLN : (1) is clearly evident
(2) let the 4 numbers be n-2,n-1,n and n+1 then by multing the whole thing and adding 1 we will have a perfect square
10) When 987 And 643 Are Divided By Same Number 'N' The Reminder Is Also Same, What Is That Number If The Number Is A Odd Prime Number?
ANS : since both leave the same reminder, let the reminder be 'r',
then, 987 = an + r
and 643 = bn + r and thus
987 - 643 is divisible by 'r' and
987 - 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43
hence 'r' is 43
11) When A Number Is Divided By 11,7,4 The Reminders Are 5,6,3 Respectively. What Would Be The Reminders When The Same Number Is Divided By 4,7,11 Respectively?
)
ANS : whenever such problem is given,
we need to write the numbers in top row and rems in the bottom row like this
11 7 4
| \ \
5 6 3
( coudnt express here properly
now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 + 6) + 5
that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by 4,7,11 respectively are,
302 mod 4 = 2
75 mod 7 = 5
10 mod 11 = 10
12) a^n - b^n Is Always Divisible By a-b
13) If a+b+c = 0 Then a^3 + b^3 + c^3 = 3abc
EXAMPLE: 40^3-17^3-23^3 is divisble by
since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus, the number is divisible by 3,5,8,17,23 etc.
14) There Is A Seller Of Cigerette And Match Boxes Who Sits In The Narrow Lanes OfCochin
a) 49
b) 81
c) 84
d) 92
ANS : (d)
since 77.28 = 92 * 84, and since price of cigarette is less than 85, we have (d) as answer
Quote:
i have given this question to make the funda clear
15) What Does 100 Stand For If 5 X 6 = 33
ANS : 81
SOLN : this is a number system question,
30 in decimal system is 33 in some base 'n', by solving we will have n as 9
and thus, 100 will be 9^2 = 81
16) In Any Number System 121 IsA Perfect Square
SOLN: let the base be 'n'
then 121 can be written as n^2 + 2*n + 1 = (n+1)^2
hence proved
17) Most Of You People Know These, Anyways, Just In Case
Quote:
(a) sum of first 'n' natural numbers - n*(n+1)/2
(b) sum of the squares of first 'n' natural numbers - n*(n+1)*(2n+1)/6
(c) sum of the cubes of first 'n' natural numbers - n^2*(n+1)^2/4
(d) total number of primes between 1 and 100 - 25
18) Converting Recurring Decimals To Fractions
let the number x be 0.23434343434........
thus 1000 x = 234.3434343434......
and 10 x = 2.3434343434.........
thus, 990 x = 232
and hence, x = 232/990
19) Reminder Funda
(a) (a + b + c) % n = (a%n + b%n + c%n) %n
EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder when you divide 3993 with 9?
THE REMINDER WOULD BE (6 + 8 + 1) % 9 = 6
(b) (a*b*c) % n = (a%n * b%n * c%n) %n
EXAMPLE: What is the remainder left when 1073 * 1079 * 1087 is divided by 119 ? ( seen this kinda questions alot
) 
1073 % 119 = ?
since 1190 is divisible by 119, 1073 mod 119 is 2
and thus, "the remainder left when 1073 * 1079 * 1087 is divided by 119 " is 2*8*16 mod 119 and that is 256 mod 119 and that is (238 + 18 ) mod 119 and that is 18
Glossary : % stands for reminder operation
find the number of zeroes in 1^1* 2^2* 3^3* 4^4.............. 98^98* 99^99* 100^100
the expresion can be rewritten as (100!)^100 / 0!* 1!* 2!* 3!....99!
Now the numerator has 2400 zeros
the formular for finding number of zeros in n! is
[n/5]+[n/5^2]...[n/5^r]
where r is such that 5^r<=n<5^(r+1)>
and [..] is the grestest integer function
for the numerator find the number of zeros using the above formulae..
for 0!...4! number of zeros ..0
5!...9!.number os zeros ..1
9!...14!... 2
15!..19!..................3
20!..24!..................4!
now at 25! the series makes a jump to 6
25!...29!.................6
30!...34!.................7
this goes on and again makes a jump at 50!
and then at 75!
so the number of zeros is...
5(1+2....19) + 25+ 50+ 75
the last 3 terms 25 50 and 75 are because of the jumps..
this gives numerator has 1100 zeros
now total number of zeros in expression is no of zeros in denominator - no of zeros in numerator
2400 - 1100
the Answer 1300