Official Quant thread for CAT 2013

MBA CAT 2024
19046
@Logrhythm said:
can you pls explain your method??i would have done it like this.. x = -3+/-{sqrt(9-36)}/2 = -3+/-{sqrt(-25)}/2 = (-3+/-5i)/2 = (-3-5i)/2 or (-3+5i)/2and then calculated x^3 by applying (a+/-b)^3 which is kind of arduous...your method seems less cumbersome...
x^3 + 3x^2 + 9x = 0 -------1

and x^2 = -9 - 3x

put value of x^2 in eq. 1
19047
@Logrhythm said:
can you pls explain your method??
I simplified the expression (x^3)

x^3 = (x^2)*x

Now, I replaced x^2 from the question itself i.e. x^2 + 3x + 9 = 0 => x^2 = -3x-9

then, I get => (-3x-9)*(x) = (-3)*(x^2 + 3x) , here I use the question equation again, to get 27
19048
@jain4444 said:
x is a number such that x^2 + 3x + 9 = 0. What is the value of x^3?
@Logrhythm said:
can you pls explain your method??i would have done it like this.. x = -3+/-{sqrt(9-36)}/2 = -3+/-{sqrt(-25)}/2 = (-3+/-5i)/2 = (-3-5i)/2 or (-3+5i)/2and then calculated x^3 by applying (a+/-b)^3 which is kind of arduous...your method seems less cumbersome...

x = -3 +/- _/(9 - 36) / 2

=> x = (-3 +/- 3_/3 i )/2

=> x = -3 e^(pi/3)

So, x^3 = -3^3 * e^(pi) = -27 * -1 = 27
19049
@jain4444 said:
x^3 + 3x^2 + 9x = 0 -------1and x^2 = -9 - 3xput value of x^2 in eq. 1
@YouMadFellow said:
I simplified the expression (x^3)x^3 = (x^2)*xNow, I replaced x^2 from the question itself i.e. x^2 + 3x + 9 = 0 => x^2 = -3x-9then, I get => (-3x-9)*(x) = (-3)*(x^2 + 3x) , here I use the question equation again, to get 27
awesome...
19050
@Aizen said:
x = -3 +/- _/(9 - 36) / 2=> x = (-3 +/- 3_/3 i )/2=> x = -3 e^(pi/3) So, x^3 = -3^3 * e^(pi) = -27 * -1 = 27
bhai abb ye strikethrough wala samjhao....aap sabb loag toh bohot advanced methods use kar rae ho...

#novice
19051
@Logrhythm said:
bhai abb ye strikethrough wala samjhao....aap sabb loag toh bohot advanced methods use kar rae ho... #novice
x = (-3 - 3_/3 i)/2

=> x = -3(1/2 + _/3/2 i)

=> x = -3 ( cos(x') + i sin(x') )

So, here x' = pi/3

WKT by Complex numbers :

e^(iz) = cos z + i sin z

So, x = -3 ( e^ (pi/3) )
19052
find the remainder

Q1. 888222888222. . . . . up to 9235 digits divided by 53

ďťż
19053
@joyjitpal said:
find the remainderQ1. 888222888222. . . . . up to 9235 digits divided by 53ďťż
remainder 27 ??
19054
@joyjitpal said:
find the remainderQ1. 888222888222. . . . . up to 9235 digits divided by 53
is the answer 52?
19055
@joyjitpal said:
find the remainderQ1. 888222888222. . . . . up to 9235 digits divided by 53
i am not sure abt this one....

is it 27??

9325%6 = 1
hence we would have a spare 8...

now the number can be written as -> 8+888222*10^1 + 888222*10^7 + 888222*10^13 +....
= 8+8882220(1+10^6+10^12+....+10^9234)
= 8+3{(1-4+4^2-4^3+...-4^1539)}
= {8+3(4^1540-1)/5}%53 ---- (1)
now we need to find {(4^1540-1)/5}%53
e(53)=52
1540%52 = 32
4^32%53 = 15...
so, it reduces to {14/5}%53
=> (5x - 14)%53 = 0
x = 24

putting in (1)
8+(3*24) % 53 = 80 % 53 = 27.......

pls confirm, thoda mushil ques tha...

19056
@Logrhythm said:
i am not sure abt this one.... is it 27??9325%6 = 1hence we would have a spare 8...now the number can be written as -> 8+888222*10^1 + 888222*10^7 + 888222*10^13 +....= 8+8882220(1+10^6+10^12+....+10^9234)= 8+3{(1-4+4^2-4^3+...-4^1539)} = {8+3(4^1540-1)/5}%53 ---- (1)now we need to find {(4^1540-1)/5}%53 e(53)=521540%52 = 324^32%53 = 15...so, it reduces to {14/5}%53=> (5x - 14)%53 = 0x = 24putting in (1)8+(3*24) % 53 = 80 % 53 = 27.......pls confirm, thoda mushil ques tha...
=> (5x - 14)%53 = 0

ye kaise huya bhai ?


19057
@Logrhythm said:
i am not sure abt this one.... is it 27??9325%6 = 1hence we would have a spare 8...now the number can be written as -> 8+888222*10^1 + 888222*10^7 + 888222*10^13 +....= 8+8882220(1+10^6+10^12+....+10^9234)= 8+3{(1-4+4^2-4^3+...-4^1539)} = {8+3(4^1540-1)/5}%53 ---- (1)now we need to find {(4^1540-1)/5}%53 e(53)=521540%52 = 324^32%53 = 15...so, it reduces to {14/5}%53=> (5x - 14)%53 = 0x = 24putting in (1)8+(3*24) % 53 = 80 % 53 = 27.......pls confirm, thoda mushil ques tha...
bro could you elaborate a lil more on your approach as i couldnt understand anything after the line which i have made bold
19058

if A= 8888^8888 B= sum of digits of A , C=sum of digits of B , D=sum of digits of C

...In this series there will be a point where you will get sum of digits of X=X..Find X
19059
@pavimai said:
if A= 8888^8888 B= sum of digits of A , C=sum of digits of B , D=sum of digits of C...In this series there will be a point where you will get sum of digits of X=X..Find X
ans 7
19060
@pavimai said:
if A= 8888^8888 B= sum of digits of A , C=sum of digits of B , D=sum of digits of C...In this series there will be a point where you will get sum of digits of X=X..Find X
this is a digit sum question so it should be approached this way
8888^8888/9 == 5^8888/9
now apply Euler-totient
you would get E(9)=6
8888/6 == 2
5^2/9 = 7(answer)
19061
@pavimai bro/sis can you explain your approach to the previous question 888222......... repeated 9325 times find the remainder when this number is divided by 53
19062

how many integers between 1000 and 10000 have their sum of the digits equal to 9???

19063
@pavimai said:
how many integers between 1000 and 10000 have their sum of the digits equal to 9???
is the answer 888 or 889?
19064
@iLoveTorres said:
is the answer 888 or 889?
options-
112
132
126
165

i dont know the OA
19065

hey guys,

mine is a very basic question.
i just wanna know the algorithm behind euclid's long divison method for finding hcf.i just want to know the logic behind this method of first dividing the bigger number by smaller number(in case of two numbers) and then dividing the divisor by the coming remainder and so on.......y do we divide the divisor by the remainder(what's the logic behind this!?!) and how come we get hcf by this continuous subtraction and division?? (though hcf comes out by finding factors of the numbers!!!!! )

p.s. i know its unnecessary but i just want to understand the concept in detail!!
please explain...