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Thread: Quant Thread For CAT 2011

  1. #101
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    Quote Originally Posted by grimreaper View Post
    oh sorry about that.. inlcude 0 too ... greater than equal to zero...
    and yes there is a proper method to do this prob.....
    try it... wld post the solution then... just that noone tried it at all...apart from shaunak... so didnt post the solution as of yet......

    @all - give it a try
    Sir, Can u Please post the method as i had used hit n Trial only..
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    Quote Originally Posted by shaunak_87 View Post
    Sir, Can u Please post the method as i had used hit n Trial only..
    sir???? O.o

    am younger than you... :| ... lol.... well let others try... if any...
    wld post the method in sometime

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    Quote Originally Posted by shaunak_87 View Post
    @Grim:

    D should be >52 and the values of D will be D=64,128,256 i guess.

    Pardon me if i'm wrong...
    y D should be greater than 52.please explain

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    Quote Originally Posted by anjalig View Post
    y D should be greater than 52.please explain
    cause D gives remainder 52 with N
    now a number smaller than 52 cannot give 52 as a remainder... thats why... hope you got it now...
    as in 3 can give only remainders as 0 1 2
    while 4 can give 0 1 2 3

    same way for D to give 52 as remainder... it should be greater than 52 in the first place

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    Quote Originally Posted by anjalig View Post
    y D should be greater than 52.please explain
    Anjali
    when u divide N by D u get a rem of 52. Well the divisor D can't be less than 53....

    Regards:
    Raj( KKHH frame)


    @GRIM
    Yeah yeah, i'm old now, already 24...

    Well, Sir was meant to honour ur capabilities and not age..
    CAT2011 : 99.97%ile
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    Quote Originally Posted by grimreaper View Post
    oh sorry about that.. inlcude 0 too ... greater than equal to zero...
    and yes there is a proper method to do this prob.....
    try it... wld post the solution then... just that noone tried it at all...apart from shaunak... so didnt post the solution as of yet......

    @all - give it a try
    5^x will always end in 5 as unit digit
    11^x will always end in 1 as unit digit
    7's unit digit cycle is 7,9,3,1
    3*2^x will have unit digit cycle as 6,2,4,8
    5+1+x+y = 8 (8 from 2008)
    Addition of any numbers between (7,9,3,1) and (6,2,4,8) will never result in a even number. This decides the p has to be zero.

    Now 5^q + 7^r + 11^s = 2005.

    The last digit is five now this is possible with the unit digit combination of 5,1 and 9. This determines that r has to be even.

    Unable to proceed further

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    Quote Originally Posted by Rajasekaran View Post
    5^x will always end in 5 as unit digit
    11^x will always end in 1 as unit digit
    7's unit digit cycle is 7,9,3,1
    3*2^x will have unit digit cycle as 6,2,4,8
    5+1+x+y = 8 (8 from 2008)
    Addition of any numbers between (7,9,3,1) and (6,2,4,8) will never result in a even number. This decides the p has to be zero.

    Now 5^q + 7^r + 11^s = 2005.

    The last digit is five now this is possible with the unit digit combination of 5,1 and 9. This determines that r has to be even.

    Unable to proceed further
    you going fine
    see 5^x will always be 5 in unit place...
    while 11^x will have 1 in unit place...
    so these two would give 6...
    so all is left is to realise that 7^x shld have 9 as last digit

    this is only possi when x is even...that to of form 4x+2 hence def it has to be 7^2 cause else 7^6 is too big..
    now said that its simple to see the other two values....

    so thats yo question done

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    Quote Originally Posted by shaunak_87 View Post
    Anjali
    when u divide N by D u get a rem of 52. Well the divisor D can't be less than 53....

    Regards:
    Raj( KKHH frame)


    @GRIM
    Yeah yeah, i'm old now, already 24...

    Well, Sir was meant to honour ur capabilities and not age..
    hahahah grandpa :P :P eheh kiddin... :P :P.... shaunak dada

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    Quote Originally Posted by grimreaper View Post
    Pls follow the question numbering....

    Q5) How many integral values of x are possible such that 2^70+2^1039+2^x is a perfect square of a whole number

    a) two
    b) three
    c)infinite
    d) one
    e) four
    Quote Originally Posted by mohasan02 View Post
    option (d)

    The given expression has three terms and therefore it should be of the form (a+b)2 for it to be perfect square.

    270 = 22(35)
    2x = 22(y)
    21039 = 2. 235.2y
    Therefore x = 1003

    Thus x has a unique value
    Friends,

    Please let me know the solution for this problem.

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    Quote Originally Posted by Rajasekaran View Post
    Friends,

    Please let me know the solution for this problem.
    Sorry didn't notice the solution posted

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    Quote Originally Posted by shaunak_87 View Post
    Ans should be 3

    D={64,128,256}
    well the answer given in d book is 1, however gvn d high error rate ,m nt very sure abt its correctness. plz post d soln..

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    Quote Originally Posted by grimreaper View Post
    getting answer as 315 .... nice set of Qs...
    post ur soln !!

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    Quote Originally Posted by shaunak_87 View Post
    Getting 315 = 3^2*5*7 as the No.
    to u too !! soln plzz ..

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    Quote Originally Posted by grimreaper View Post
    you going fine
    see 5^x will always be 5 in unit place...
    while 11^x will have 1 in unit place...
    so these two would give 6...
    so all is left is to realise that 7^x shld have 9 as last digit

    this is only possi when x is even...that to of form 4x+2 hence def it has to be 7^2 cause else 7^6 is too big..
    now said that its simple to see the other two values....

    so thats yo question done
    so wats p+q+r+s... 8?

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    Default Q14

    The sum of 20 numbers is 801.what is their minimum LCM possible?
    a) the numbers may or may not be distinct
    b) when the numbers are distinct

    ans is 36 and 360 respectively.plz post ur solutions as well..

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    Quote Originally Posted by quixotic_me View Post
    so wats p+q+r+s... 8?
    no it wld be 9
    3*2^0+ 5^4 + 7^2 + 11^3 = 2008

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    Quote Originally Posted by quixotic_me View Post
    plz solve ds one ...
    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 cannot be written as sum of 11 consecutive numbers.
    See first thing... u gtta knw that a number can be written in the form of n consecutive numbers... .
    now how many ways depends on the number of odd factors of the number.
    say 15
    this has factor 3*5 = hence 4 odd factors...
    so it can be written in 3 ways where it is a sum of consecutive numbers.
    15= 7+8
    15= 4+5+6
    15= 1+2+3+4+5

    so, Odd factors - 1

    hence coming to the Q.... we need 11 ways... means that number has 12 factors in all.... as said before... i add one to the number of ways to get the number of odd factors...
    so only way i can get 12 factors for a number smaller than 400 is when it is of the form a^2*b^1*c^1 "note... here a,b,c are odd... i can have a number say N= a^2*b^1*c^1 * 2^x but then the number wld be more than 400"
    so well i cant have any even number..

    now said that... only way i can take 3 odd primes are in the form
    3^2*5^1*7^1 any other combo wld exceed 400

    hence your answer 315

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    Quote Originally Posted by shaunak_87 View Post
    Anjali
    when u divide N by D u get a rem of 52. Well the divisor D can't be less than 53....

    Regards:
    Raj( KKHH frame)


    @GRIM
    Yeah yeah, i'm old now, already 24...

    Well, Sir was meant to honour ur capabilities and not age..
    hey Shaunak ,thanks. I realized that it was a silly doubt on my part . sorry for troubling you.

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    Quote Originally Posted by grimreaper View Post
    no it wld be 9
    3*2^0+ 5^4 + 7^2 + 11^3 = 2008
    arghh !! took 1331 as 11^2 !!

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    Quote Originally Posted by grimreaper View Post
    See first thing... u gtta knw that a number can be written in the form of n consecutive numbers... .
    now how many ways depends on the number of odd factors of the number.
    say 15
    this has factor 3*5 = hence 4 odd factors...
    so it can be written in 3 ways where it is a sum of consecutive numbers.
    15= 7+8
    15= 4+5+6
    15= 1+2+3+4+5

    so, Odd factors - 1

    hence coming to the Q.... we need 11 ways... means that number has 12 factors in all.... as said before... i add one to the number of ways to get the number of odd factors...
    so only way i can get 12 factors for a number smaller than 400 is when it is of the form a^2*b^1*c^1 "note... here a,b,c are odd... i can have a number say N= a^2*b^1*c^1 * 2^x but then the number wld be more than 400"
    so well i cant have any even number..

    now said that... only way i can take 3 odd primes are in the form
    3^2*5^1*7^1 any other combo wld exceed 400

    hence your answer 315
    whoa !! wsnt aware of ds concept !!

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