   
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


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^2b^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+8k)=64 see now lhs and rhs both are integers then both of (n+8k) 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+8k=2,4,8,16,32
but see this n is positive hence k is positive, thus n+8+k>n+8k
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^3y)=127
so clearly 2x^3=128 x=4 and y=63
so one pair (4,63)


QuantoLogic: Marathon Thread for CAT Quant!
This is a great initiative be implex!
I have made this a sticky thread.
Moderator


Aficionado
QuantoLogic: Marathon Thread for CAT Quant!
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?


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.....


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+25^24k^2=25^276
(2m+25)^2(2k)^2=549
(2m+25+2k)(2m+252k)=549.1, 183.3,61.9
4m+50=550,186,70
m=125,34,5
so sum is 125+34+5=164


Intern
QuantoLogic: Marathon Thread for CAT Quant!
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+25^24k^2=25^276
(2m+25)^2(2k)^2=549
(2m+25+2k)(2m+252k)=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....


QuantoLogic: Marathon Thread for CAT Quant!
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 !


QuantoLogic: Marathon Thread for CAT Quant!
Originally Posted by implex 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 ...


QuantoLogic: Marathon Thread for CAT Quant!
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 !


QuantoLogic: Marathon Thread for CAT Quant!
Let me give you people three more problems based on similar concepts ! 254) If x^2y^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 m’s
256) Find the number of pairs of positive numbers (x,y) such that x^3y^3=100 and (xy) and xy are integers.
a) 0 b) 1 c) 2 d) 4 e) none of these 

Aficionado
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


QuantoLogic: Marathon Thread for CAT Quant!
option d or c, you have answered both ?


Aficionado
QuantoLogic: Marathon Thread for CAT Quant!
option d .. sorry dat was a typo


QuantoLogic: Marathon Thread for CAT Quant!
Originally Posted by implex
255) If m^2+24m+21 has 3 factors for a natural number m, find the sum of all such m’s
m^2 + 24m + 21 = (m+12)^2 144 +21 = (m+12)^2  123 = (m+12)^2  121  2 = (m+1211)(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.


QuantoLogic: Marathon Thread for CAT Quant!
Originally Posted by implex 256) Find the number of pairs of positive numbers (x,y) such that x^3y^3=100 and (xy) 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.


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)


QuantoLogic: Marathon Thread for CAT Quant!
@jain.mah absolutely correct!! though approach is a bit different, you have done correctly


Aficionado
QuantoLogic: Marathon Thread for CAT Quant!
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


QuantoLogic: Marathon Thread for CAT Quant!
@streethawk you forgot one solution
2m+24=123+1=124
m=50
so n=1235012=61
so two solution m=10,50
sum of which is 60

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