Official Quant thread for CAT 2013

MBA CAT 2024
14763
@PURITAN said:
remainder when 128^1000 is divided by 153 ??
153=9*17

128^1000mod 9= 2^1000=(2^3)^333*2=-2 or 7

128^1000 mod 17=9^1000mod 7=(9^4)^250 =1

9x+7=17y+1
y=3
52 remainder
14764
@pirateiim478 said:
13x+13y+13z=117x+y+z=9 =>11C2 =55 ordered solutionsWhat is the largest sum of rupees which can never be paid using infinite number of coins of denominations Rs. 5, Rs. 7 and Rs. 11?
Every number greater than 13 can be formed

5=4k+1
7=4k-1
11=4k-1
14765
@PURITAN said:
remainder when 128^1000 is divided by 153 ??
153 = 9*17
128^1000%9 = 2^1000%9
e(9) = 6
100%6 = 4
=> 2^4%9 = 7
similarly -> 9^8%17 = (-4)^4%17 = 1
hence, remainder -> 9x+7=17y+1
so 52....
14766
@ravi.theja said:
The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
13(x+y+z) = 117
x+y+z = 9
x'+y'+z' = 6
8c2 = 28
but we need to take out x=y=z=3
hence, total = 28-1 = 27 cases...
14767
@TootaHuaDil said:
(3^1024-1)/2^nthen find the highest value of n which will divide this..
(3^1024)-1

(1+2)^1024-1

1+1024C1*2+1024C2*2^2+1024C3*2^3+------------------+2^1024-1

1024C1*2+1024C2*2^2+1024C3*2^3+-----------------------+2^1024

Minimum is 1024*2=2^11
14768
@pirateiim478 said:
13x+13y+13z=117x+y+z=9 =>11C2 =55 ordered solutionsWhat is the largest sum of rupees which can never be paid using infinite number of coins of denominations Rs. 5, Rs. 7 and Rs. 11?
Is it 14??
14769
@ravi.theja said:
10000! = (100!)^K — P, where P and K are integers. What can be the maximum value of K?a) 103b) 104c) 102d) 105
10000!=2^9995
100!=2^97

9995/97=103
14770
@TootaHuaDil said:
(3^1024-1)/2^nthen find the highest value of n which will divide this..
@gnehagarg

answer wld be 12...

3^1024 - 1 = 9^512 - 1 = (8+1)^512 - 1 = 8^512 + ..... + 512c2*8^2 + 512c1*8 + 1 - 1 = 8^512 + ..... + 512c2*8^2 + 512c1*8

the last term has 2^12
14771

Q1

14772

Q2

14773
@vbhvgupta said:
Q1
let n =1
so A = 1 and B = 3/2
hence, A/B = 2/3 (option C)
14774
@vbhvgupta said:
Q2
10 ?
14775
@vbhvgupta said:
Q2
10??
sigma (n=1 to 120) 1/(rt(n) + rt(n+1))

14776
@vbhvgupta said:
Q1
2/3 ?
14777
@vbhvgupta said:
Q2
Rationalise the denominator by multiplying,
You you get
[ root(1)-root(2)+root(2)-root(3)+....+root(120)-root(121) ] / (-1
= root(1) - root(121) / -1 = 10
14778
@vbhvgupta said:
Q1
A=4/3[1-1/4^n]
B=2(1-1/2^2n]=2(1-1/4^n]
So A/B=2/3
14779
@vbhvgupta said:
Q1
2/3

A=(4^n-1)*(4^(1-n))/3

B=(2^2n-1)*(2^(1-2n))

A/B=2/3
14780
@vbhvgupta said:
Q2
10?
-(1-rt2+rt2-rt3+.......rt120-rt121)
-(1-rt121)
-(1-11)=10
14781
@vbhvgupta said:
Q2
rt2-rt1+rt3-rt2+rt4-rt3+------------------------+rt121-rt120

-rt1+rt121

10
14782

hw many 6 digit no are dre having 3 odd & 3 even digits?