Official Quant thread for CAT 2013

@hexagon said:
i got this quest in this years cat i was not able to solve it1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 find the sum of the series
19481?

summation from 1 to 21 (n(n+1)(2n+1)/6)
= S1-21 ((n^2 + n)(2n+1)/6)
= S1-21 ((2n^3 + 3n^2 + n)/6)
= 1/3 * S1-21(n^3) + 1/2 * S1-21(n^2) + 1/6 * S1-21(n)
= 1/3 * (21*11)^2 + 1/2 * (21*22*43)/6 + 1/6 * 21*11
= 17787 + 1655.5 + 38.5
= 19481
@hexagon said:
i got this quest in this years cat i was not able to solve it1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 find the sum of the series
19481 imo

solve this challenge !!!!!!!!!!!!

Two players play a game using the interval [0,33] on the x-axis. The first player randomly chooses a square of side length s∈Z+, which has a side that lies entirely on the interval. The second player randomly chooses a circle with radius r∈Z+, which has a diameter that lies entirely on the interval. After repeating choosing random squares and circles in this fashion, the players realize that the probability that the circle and square intersect is 12. Let S={(s,r):probability of intersection is 12}. Determine βˆ‘(s,r)∈S(s+r).
Clarification of notation: The set S is the set of all ordered pairs of integers, (s,r), such that the probability that a square of side length s and a circle of radius r will intersect is 12.

Two players play a game using the interval [0,33] on the x-axis. The first player randomly chooses a square of side length s∈Z+, which has a side that lies entirely on the interval. The second player randomly chooses a circle with radius r∈Z+, which has a diameter that lies entirely on the interval. After repeating choosing random squares and circles in this fashion, the players realize that the probability that the circle and square intersect is 12. Let S={(s,r):probability of intersection is 12}. Determine βˆ‘(s,r)∈S(s+r).
Details and assumptions
Clarification of notation: The set S is the set of all ordered pairs of integers, (s,r), such that the probability that a square of side length s and a circle of radius r will intersect is 12.
@hexagon
1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 can be written as

21*(1^2) +20*(2^2)+19*(3^2)+......+1*(21^2)

the general term being (22-n)*(n^2)..sum up this from n=1 to 21 and the answer follows.

@hexagon said:
i got this quest in this years cat i was not able to solve i1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 find the sum of the series
19481
Tn=n(n+1)*(2n+1)/6
Sn=[n^2*(n+1)^2+n(n+1)(2n+1)+n(n+1)]/12

@hexagon said:
i got this quest in this years cat i was not able to solve it1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 find the sum of the series
19481 ?

summation (n(n + 1)(2n + 1)/6
1/6 summation (2n^3 + 3n^2 + n)
1/6(2(n(n+1)/2)^2 + 3n(n + 1)(2n + 1)/6 + n(n + 1)/2)
1/6(2*(21*22/2)^2 + 3*21*22*43/6 + 21*22/2)
= 17787 + 1655.5 + 38.5 = 19481
What is the highest possible value of 'n' for which 3^1024 – 1 is divisible by 2n?
@INDRESH90 said:
What is the highest possible value of 'n' for which 3^1024 – 1 is divisible by 2n?
2048 ???
@vasanth1 said:
Two players play a game using the interval [0,33] on the x-axis. The first player randomly chooses a square of side length s∈Z+, which has a side that lies entirely on the interval. The second player randomly chooses a circle with radius r∈Z+, which has a diameter that lies entirely on the interval. After repeating choosing random squares and circles in this fashion, the players realize that the probability that the circle and square intersect is 12. Let S={(s,r)robability of intersection is 12}. Determine βˆ‘(s,r)∈S(s+r).
Details and assumptions
Clarification of notation: The set S is the set of all ordered pairs of integers, (s,r), such that the probability that a square of side length s and a circle of radius r will intersect is 12.
probability is 12 ????
@INDRESH90 said:
What is the highest possible value of 'n' for which 3^1024 – 1 is divisible by 2n?
expand the term as (3^512+1)*(3^256+1)*(3^128+1)*(3^64+1)*(3^32+1)*(3^16+1)*(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)
these terms are divisible by 2 and (3+1)*edited* is divisible by 4
Answer 12
@junefever said:
expand the term as (3^512+1)*(3^256+1)*(3^128+1)*(3^64+1)*(3^32+1)*(3^16+1)*(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1)
these terms are divisible by 2 and (3^256+1) is divisible by 4
Answer 12
(3^256+1) is divisible by 4 .......... ye kaise
aur (3+1)is divisible by 4........
@Dexian said:
(3^256+1) is divisible by 4 .......... ye kaiseaur (3+1)is divisible by 4........
sorry... my mistake

Find the number of different ordered quadruples (a,b,c,d) of complex numbers such that⎧⎩⎨βŽͺβŽͺβŽͺβŽͺβŽͺβŽͺβŽͺβŽͺβŽͺβŽͺa^2=1

b^3=1
c^4=1
d^6=1
a+b+c+d=0



4 is not the answer
@junefever said:
expand the term as (3^512+1)*(3^256+1)*(3^128+1)*(3^64+1)*(3^32+1)*(3^16+1)*(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1) these terms are divisible by 2 and (3^256+1) is divisible by 4Answer 12
answer cant be 12 bro!!!!
coz if n=12, then 2n=24 which means that the number 3^1024-1 should be divisible by 24, which in turn means that it shold be divisible by 3
But we can see that 3^1024-1 can never be divisible by 3 as it will always leave a remainder of 2 when divided by 3

@INDRESH90 said:
answer cant be 12 bro!!!!
coz if n=12, then 2n=24 which means that the number 3^1024-1 should be divisible by 24, which in turn means that it shold be divisible by 3
But we can see that 3^1024-1 can never be divisible by 3 as it will always leave a remainder of 2 when divided by 3

i think he meant 2^12
@Dexian ya! may be....but it is 2*n
@INDRESH90 said:
answer cant be 12 bro!!!!coz if n=12, then 2n=24 which means that the number 3^1024-1 should be divisible by 24, which in turn means that it shold be divisible by 3But we can see that 3^1024-1 can never be divisible by 3 as it will always leave a remainder of 2 when divided by 3
good observation but I think the question is something like this "What is the highest possible value of 'n' for which 3^1024 – 1 is divisible by 2^n?" kissi zamaane mein AIMCAT mein aaya tha :P
If this is not the case, then my solution is incorrect :)
@junefever said:
good observation but I think the question is something like this "What is the highest possible value of 'n' for which 3^1024 – 1 is divisible by 2^n?" kissi zamaane mein AIMCAT mein aaya tha If this is not the case, then my solution is incorrect
kk...but here it is 2*n not 2^n.....now solve it
@hexagon said:
i got this quest in this years cat i was not able to solve it1^2 +1^2+2^2+1^2+2^2+3^2+...........1^2+2^2+3^2.......+21^2 find the sum of the series
19481