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  1. #1
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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Here I start a thread for CAT quant, I will cover various topics with tips, examples and exercise problems ! Please keep your discussion related to this topic, do not post your own problems, unless they relate to a concept or problem already being discussed.

    Best Regards

    Implex

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Concept I Perfect Squares

    There has been a huge surge in the number of questions about perfect squares, in almost all mocks. The basic trick to any such question is assuming the number as a perfect square of an integer k and then using techniques of completion of square and then the formula of (a^2-b^2) and solving using divisibility theory

    Example I Find all n such that n(n+16) is a perfect square

    step 1 n(n+16)=k^2

    step 2 (n^2+2.8.n+8^2)-8^2=k^2

    step3 (n+8+k)(n+8-k)=64

    see now lhs and rhs both are integers then both of (n+8-k) and (n+8+) are divisors of 64. But note that we add the two equations we will get 2n+16, so teh sum of two divisors should be even hence both divisors even or both odd

    so n+8+k=32,16,8,4,2 and n+8-k=2,4,8,16,32

    but see this n is positive hence k is positive, thus n+8+k>n+8-k
    so only two options

    and solving we get 2n+16=34,20
    so n=9,2

    Note : The source of this problem is Pomona Wisconsin mathematics talent search exam!

    Practice problem!!
    Find the sum of all such positive integers m's such that m^2+25m+19 is a perfect square

    Now we will extend the method to other kinds of problems
    Basically what we used in the above problem is difference of square method

    lets take an example
    x^6=y^2+127, find the no of pairs of postive integers (x,y)


    first step in this problem is recognising that 127 is a prime
    then we move to
    (x^3+y)(x^3-y)=127
    so clearly 2x^3=128 x=4 and y=63

    so one pair (4,63)

  3. #3
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    Post QuantoLogic: Marathon Thread for CAT Quant!

    This is a great initiative be implex!

    I have made this a sticky thread.

    Moderator

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by implex
    Concept I Perfect Squares


    Practice problem!!
    Find the sum of all such positive integers m's such that m^2+25m+19 is a perfect square


    Hey Implex..
    I am still not able to solve this!!!!!
    Can you please explain it?

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    [quote=implex]Concept I Perfect Squares

    [b]Practice problem!!
    Find the sum of all such positive integers m's such that m^2+25m+19 is a perfect square

    m^2 + 25m + 19 = k^2 say
    m^2 + 2*12.5m + 19 = k^2
    (m^2 + 2*12.5m + 12.5^2) + 19 - 12.5^2 = k^2
    (m + 12.5)^2 - 137.25 = k^2
    (m + 12.5 + k)(m + 12.5 - k) = 137.25

    But now how do I divide 137.25 is not a perfect square. Then how do we solve this.....

  6. #6
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    Post QuantoLogic: Marathon Thread for CAT Quant!

    m^2+25m+19 =k^2
    m^2+2.(25/2)m+ (25/2)^2+19-(25/2)^2=k^2
    4m^2+100m+25^2-4k^2=25^2-76
    (2m+25)^2-(2k)^2=549
    (2m+25+2k)(2m+25-2k)=549.1, 183.3,61.9

    4m+50=550,186,70
    m=125,34,5
    so sum is 125+34+5=164

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by implex
    m^2+25m+19 =k^2
    m^2+2.(25/2)m+ (25/2)^2+19-(25/2)^2=k^2
    4m^2+100m+25^2-4k^2=25^2-76
    (2m+25)^2-(2k)^2=549
    (2m+25+2k)(2m+25-2k)=549.1, 183.3,61.9

    4m+50=550,186,70
    m=125,34,5
    so sum is 125+34+5=164
    Ok here we have multiplied the RHS by 4 to simplify it....

  8. #8
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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by Trinity

    Ok here we have multiplied the RHS by 4 to simplify it....
    yeah we have just used method of completion of squares, (a+b)^2=a^2+b^2+2ab!
    as we do not have 2 in the 2ab term that is 25m we assume a=m and b=25/2 and proceed
    so we get a factor of 4 in the denominator we multiply throughout by 4 to get rid of it !

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by implex
    Quote Originally Posted by Trinity

    Ok here we have multiplied the RHS by 4 to simplify it....
    yeah we have just used method of completion of squares, (a+b)^2=a^2+b^2+2ab!
    as we do not have 2 in the 2ab term that is 25m we assume a=m and b=25/2 and proceed
    so we get a factor of 4 in the denominator we multiply throughout by 4 to get rid of it !
    Great concept!!!
    Have there been any CAT problems based on this ...

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by Matrix
    Great concept!!!
    Have there been any CAT problems based on this ...
    yeah such problems keep coming in cat, xat and fms regularly, if not directly then the use of it in some or the other way!
    just scan old papers, you will fidn for sure !

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Let me give you people three more problems based on similar concepts !



    254) If x^2-y^2=a where x,y are positive integers and a is an odd integer. If (x,y) has only one solution then a is
    a) prime
    b) composite
    c) prime square
    d) a ,b,c
    e) a,c

    255) If m^2+24m+21 has 3 factors for a natural number m, find the sum of all such ms

    256) Find the number of pairs of positive numbers (x,y) such that x^3-y^3=100 and (x-y) and xy are integers.

    a) 0 b) 1 c) 2 d) 4 e) none of these

  12. #12
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    Post QuantoLogic: Marathon Thread for CAT Quant!

    254>

    x^2 - y^2 = a

    the answer is d

    put x=3, y=4 we get as prime
    put x=4, y=5 we get a as prime square and composite

    hence option c

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    option d or c, you have answered both ?

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    option d .. sorry dat was a typo

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by implex

    255) If m^2+24m+21 has 3 factors for a natural number m, find the sum of all such ms
    m^2 + 24m + 21 = (m+12)^2 -144 +21 = (m+12)^2 - 123 = (m+12)^2 - 121 - 2 = (m+12-11)(m+12+11) - 2 =
    (m+1)(m+23) - 2
    am i proceeding in the right direction....
    But now how do we get the number of factors from here.

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by implex
    256) Find the number of pairs of positive numbers (x,y) such that x^3-y^3=100 and (x-y) and xy are integers.

    a) 0 b) 1 c) 2 d) 4 e) none of these[/b]
    x^3 - y^3 = (x - y)(x^2 + xy + y^2) = 100
    Since x and y are positive (x - y) < (x^2 + xy + y^2)
    Now the factors of 100 are 1, 100, 2, 50, 4, 25 5 and 20
    Therefore the possible values of (x - y) and (x^2 + xy + y^2) is (1,100) , (2,50) (4,25) (5,20)

    Case I
    Let us consider (1,100)
    So x - y = 1 and (x^2 + xy + y^2) = 100
    After squaring the first term we get x^2 -2xy + y^2 = 1
    Using this with the other equation we get 3xy = 99 i.e. xy = 33
    Combining this with x - y = 1 we get x - 33/x = 1....
    x^2 - x - 33 = 0
    x = (1+?(133))/2. Similarly we can find y.

    Case II

    Let us consider (2,50)
    So x - y = 2 and (x^2 + xy + y^2) = 50
    After squaring the first term we get x^2 -2xy + y^2 = 4
    Using this with the other equation we get 3xy = 44 i.e. xy = 44/3 which is not allowed

    Case III
    Let us consider (4,25)
    So x - y = 4 and (x^2 + xy + y^2) = 25
    After squaring the first term we get x^2 -2xy + y^2 = 16
    Using this with the other equation we get 3xy = 9 i.e. xy = 3
    Since x - y = 4 we get x - 3/x = 4
    Hence x^2 - 4x - 3 = 0
    Solving we get x = (4 + ?28)/2 = 2 + ?7

    Case IV
    Let us consider (5,20)
    So x - y = 5 and (x^2 + xy + y^2) = 20
    After squaring the first term we get x^2 -2xy + y^2 = 25
    Using this with the other equation we get 3xy = -5 which is not possible.
    Hence only two pairs possible.
    Option c.

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    @streethawk your answer is wrong try again !

    @vickram, think again
    here is a hint if N=p^a.q^b.r^c
    where p, q , r distinct prime nos and a,b,c are non negative integers then
    no of factors of N is (a+1)(b+1)(c+1)

  18. #18
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    Post QuantoLogic: Marathon Thread for CAT Quant!

    @jain.mah absolutely correct!! though approach is a bit different, you have done correctly

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    Quote Originally Posted by implex
    @streethawk your answer is wrong try again !

    @vickram, think again
    here is a hint if N=p^a.q^b.r^c
    where p, q , r distinct prime nos and a,b,c are non negative integers then
    no of factors of N is (a+1)(b+1)(c+1)
    Since the number of factors are 3 that means the number is a perfect square of a prime number.
    so m^2 + 24m + 21 = n^2
    (m+12)^2 + 21 - 144 = n^2
    (m+12)^2 - n^2 = 123
    (m+12 +n)*(m+12 - n) = 123 = 41*3
    therefore m +12 + n = 41 and (m+12 - n) = 3
    hence m = 10 and n = 19

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    Post QuantoLogic: Marathon Thread for CAT Quant!

    @streethawk you forgot one solution
    2m+24=123+1=124
    m=50
    so n=123-50-12=61

    so two solution m=10,50
    sum of which is 60

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