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Originally Posted by quixotic_me A number of 6digit numbers that can be made with the digits 0,1,2,3,4and 5 so that even digits occupy odd places,is
a)24
b)36
c)48
d)none of these 24 is the answer....
Even digits are 0,2 and 4. 0 cannot come into the first place , so there are two options for first digit that is 2 and 4 , so 2 ways , similarly for 3rd digit 2 ways and for 5th digit only 1 way.
So 2*2*1= 4 ways
now remaining can be filled in 3*2*1 = 6
total 4*6 =24 ways
so option (a) is the asnwer

The Following 2 Users Say Thank You to sumit2goody For This Useful Post:
anjalig (16Apr11), quixotic_me (16Apr11)

Originally Posted by quixotic_me Let A be the set of 4digit numbers a1a2a3a4 where a1>a2>a3>a4 then how many values of A are possible?
a)126
b)84
c)210
d)none of these
There are 10 digits {0,1,2,...,9}
From these any 4 can be selected in 10C4 ways.
Now as in every selection of 4 digits all are distinct,they can be arranged in descending order from R to L in only 1 way.
So 10C4 * 1 = 210 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 
The Following 4 Users Say Thank You to shaunak_87 For This Useful Post:
anjalig (16Apr11), maddy9021 (17Jul11), quixotic_me (16Apr11), sumit2goody (18Apr11)

Q24
The total number of integral solutions for (x,y,z) such that xyz=24 is
a)36
b)90
c)120
d)none of these


Topic surds
Hello Friends ...could u pls help me ...i have a problem related to surds & the query is:
If x=2+۟√3+√5 then find the value of:
x^{4}8x^{3}+8x^{2}+32x...?
what the shortest trick to solve this query??
pls give ur advice.....
thanx 4 being with me... 

Originally Posted by quixotic_me The total number of integral solutions for (x,y,z) such that xyz=24 is
a)36
b)90
c)120
d)none of these
Now for x * y *z = 24
We can have the abs values of (x,y,z) such that
Case1 : 2 are same and 3rd one distinct
1 * 1* 24
2 * 2 * 6
Arranged in 3!/2! = 3 ways each
So there are 3*2=6 ways.
Case2 : all 3 distinct
1*2*12
1*3*8
1*4*6
2*3*4
As each can be arranged in 3! ways, there are 4*3! = 24 ways
But as for all cases we can have any 2 out of x,y,z as ve and all 3 +ve, each case has (3C2 +1 ) = 4 ways
So total = (6+24)*4 = 120 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 
The Following 2 Users Say Thank You to shaunak_87 For This Useful Post:
maddy9021 (17Jul11), quixotic_me (17Apr11)

Originally Posted by Rajasekaran Q21:There are 1 to 500 questions..
In round 1,a student attempts Q1.
In round 2,he attempts(or checks) Q2,Q4,Q6.....
In round 3,he attempts(or checks) Q3,Q6,Q9....
Similarly for all the rounds.
What will be the no. of unattempted questions after round 90?
a)0
b)1
c)4
d)6 option c) 4. is it correct?


Q 26
Let S be the set of fivedigit numbers formed by digits 1,2,3,4 and 5 using each digit exactly once such that exactly two odd positions are occupied by odd digits.what is the sum of the digits in the rightmost position of the numbers in S?
a)228
b)216
c)294
d)192


Originally Posted by quixotic_me Let S be the set of fivedigit numbers formed by digits 1,2,3,4 and 5 using each digit exactly once such that exactly two odd positions are occupied by odd digits.what is the sum of the digits in the rightmost position of the numbers in S?
a)228
b)216
c)294
d)192 op b)216 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 



_ _ _ _ _
A B C D E
Case 1: E has a even no. and A and C have odd
E can be filled in 2 ways(2 or 4)
A & C have to be filled by odd nos.{1,3,5} in 3P2 = 6 ways.
remaining B & D can be filled in 2*1 ways
So there are 6*2*1 = 12 nos. each with unit's digit 2 or 4
So sum of unit place digit is 12*(2+4)=72
Case 2: E has odd no. and A has odd no. and C has even
E can have 1/3/5
For each selection of E, A can each be filled in 2 ways(with 2 remaining odd nos.) and thus C has to be an even no.(2 or 4) in 2 ways.
Remaining places can be filled in 2*1 ways
So there are 2*2*2*1 = 8 nos. having 1/3/5 at unit place
As another case can be E has odd no. and A has odd no. and C has even,the above case repeats
and so sum of unit place is 2*8*(1+3+5) = 144
So total sum = 72+144 = 216 CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 



Originally Posted by sumit2goody Q27)
One of the most concept clearing question for number system , want the newbies to try this ..so shaunak plz dont post solution first
find the remainder when 128^1000 is divided by 153? Ok Sumit,
I had PMed u....Please do check.
And kindly get the DILR thread up and running, also the Quant thread
I request Sumit, Tushar to take up initiative and speed up the QA/DI/VA threads even with qstns from past 2 yrs threads....plz, i request this to both of u....maybe this will be the only means of preps..... CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Available on PM
Originally Posted by shaunak_87 Ok Sumit,
I had PMed u....Please do check.
And kindly get the DILR thread up and running, also the Quant thread
I request Sumit, Tushar to take up initiative and speed up the QA/DI/VA threads even with qstns from past 2 yrs threads....plz, i request this to both of u....maybe this will be the only means of preps..... will try to do so..difficult to access net while in internships and will reply to ur PM


Originally Posted by sumit2goody Q27)
One of the most concept clearing question for number system , want the newbies to try this ..so shaunak plz dont post solution first
find the remainder when 128^1000 is divided by 153? The remainder is 52


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Originally Posted by anjalig The remainder is 52 method please


Originally Posted by anjalig The remainder is 52 Yes, Please do share some good probs, i request u.....please...
CAT2011 : 99.97%ile
XAT2012 : 99.07%ile
FMS 201214
Other Converts : IIML, MDI, NITIE, SJMSOM,IIM RRR 

Originally Posted by shaunak_87 Yes, Please do share some good probs, i request u.....please... ya sure .... 

Ques.28 Find the highest natural number N,less than 400,such that N can be written as sum of consecutive natural numbers in 11 ways but can't be written as sum of 11 consecutive natural numbers.


Ques 29. P and Q are two natural numbers,both greater than 1 such that
P * Q = 5! . If Q can be represented as xy where (x+y) is minimum and P is not a multiple of x, what is the value of P ?
A. 10
B. 30
C. 40
D. 60


Ques. 30 In how many zeroes does 2002 ! / (1001 !)^2 end in ?
A. 0
B. 1
C. 2
D. 3

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