Official Quant thread for CAT 2013

MBA CAT 2024
8321
@MikelArteta said:
It should be 0. Only odd values of 'm' need to be checked and even those odd values show a diverging trend.
@ScareCrow28 said:
0 pairs hai kya sir?? Pta ni 0 se kya pyar h mje
ham bhi thoda hath paer mare the ..kuch nhn aaya..0 ke siwaye
8322
@chillfactor said:
Find the natural number n such that product of all positive divisors of n is 2011^2016
Is the answer 2011^63
8323
@sujamait said:
I dont have OAFind all pairs of positive integers (n; m) satisfying 3n^2 + 3n + 7 = m^3
3n(n + 1) + 7 = m^3

remainder when n(n + 1) is divided by 9 are 0, 2, 3, 6
So, remainder when 3n(n + 1) + 7 is divided by 9 are 6 or 7

But when m^3 is divided by 9 remainders are -1 or 0 or 1

So, not possible
8324
@sujamait said:
Find all triples (a; b; c) of natural numbers such that lcm(a; b; c) = a + b + c.
a=kp
b=kt
c=kn
k(p+t+n) = k
p+t+n = 1 nt possible :O
oops lcm i jagah hcf kar diya
8325
@sujamait said:
a=kpb=ktc=knk(p+t+n) = kp+t+n = 1 nt possible
Find all triples (a; b; c) of natural numbers such that lcm(a; b; c) = a + b + c.
1k,2k,3k K=1,2,3,4...so on?


8326
@sujamait said:
Find all triples (a; b; c) of natural numbers such that lcm(a; b; c) = a + b + c.
a = pxy
b = pyz
c = pzx, where x, y, z are cop-rime to each other

pxyz = pxy + pyz + pzx
xyz = xy + yz + zx
1/x + 1/y + 1/z = 1

without the loss of generality x ≥ y ≥ z
so, 1/z + 1/z + 1/z ≥ 1/x + 1/y + 1/z = 1
z ≤ 3

When z = 1 (not possible)

z = 2
1/x + 1/y = 1/2
(x - 2)(y - 2) = 4
(x, y) = (3, 6), (4, 4)

z = 3
1/x + 1/y = 2/3
(2x - 3)(2y - 3) = 9
(x, y) = (3, 3), (2, 6)

(2, 4, 4) and (3, 3, 3) are not possible as x, y, z are not coprime

So, only solution is (2, 3, 6)

(a, b, c) = (6p, 12p, 18p) and permutations
8327
@hedonistajay said:
x,y and z are positive integers such that x ,y and z are positive integers such that x
x=100,y=101,z=300
x+y+z=501??
8328
@chillfactor Sir \\___0// How do you make it so easy on the brain!
8329
@chillfactor said:
a = pxyb = pyzc = pzx, where x, y, z are cop-rime to each otherpxyz = pxy + pyz + pzx xyz = xy + yz + zx1/x + 1/y + 1/z = 1without the loss of generality x ≥ y ≥ zso, 1/z + 1/z + 1/z ≥ 1/x + 1/y + 1/z = 1z ≤ 3When z = 1 (not possible)z = 21/x + 1/y = 1/2(x - 2)(y - 2) = 4(x, y) = (3, 6), (4, 4)z = 31/x + 1/y = 2/3(2x - 3)(2y - 3) = 9(x, y) = (3, 3), (2, 6)(2, 4, 4) and (3, 3, 3) are not possible as x, y, z are not coprimeSo, only solution is (2, 3, 6)(a, b, c) = (6p, 12p, 18p) and permutations
sir 1,2,3 also satisfy so, shouldn't it be 1p,2p,3p?
8330
@chillfactor said:

Find the natural number n such that product of all positive divisors of n is 2011^2016
Product of the divisor for any number N(=x^a*y^b*z^c)=(N)^(a+1)(b+1)(c+1)/2

Here, 2011 is a prime number.
So, it's the case of N=x^a
Product of the divisors=N^(a+1)/2=x^a(a+1)/2

now, a(a+1)/2=2016
or a^2+a-4032=0
=> a=63, -64(not possible)

=> number would be 2011^63
8331
@sauravd2001 said:
Two poles of height 2 meters and 3 meters, are 5 meters apart. The height of the point of intersection of the lines joining the top of each poles to the foot of the opposite pole is,
1.2m?
8332
@sujamait said:
sirji..n is +tive integer..so with only D perfect square..se baat ban jayegi kya ? shouldn it be D shall be a perfect square = k and k > 90 and odd
D is automatically odd (if its integer), as D = 4k ˛ - 119

@sujamait said:
6 ke divi ..1,2,3,6 product = 6^2,so formula is N^nooffact/2N^#/2 = 2011^2016is it possible ?
@sujamait said:
ham bhi thoda hath paer mare the ..kuch nhn aaya..0 ke siwaye
@MikelArteta said:
Is the answer 2011^63
Ya 2011^63 is correct

2011 is prime, so number has to be of form 2011^n

Product of factors = 2011^(0 + 1 + .... + n) = 2011^{n(n + 1)/2}

n(n + 1)/2 = 2016
=> n = 63

No is 2011^63
8333
@sujamait said:
I dont have OAFind all pairs of positive integers (n; m) satisfying 3n^2 + 3n + 7 = m^3
0 hai kya bhai?
8334

Please someone help!
Find the range of a
Given that the roots of quadratic equation ax^2 + bx + c =0 lie in the interval (0,3).

8335
@chillfactor said:

only solution is (2, 3, 6)(a, b, c) = (6p, 12p, 18p) and permutations
Sir, for triplet 4, 6, 12 LCM=12

sum of the numbers=4+6+12=22

Edit: sir got it :embarrassed: :banghead:

It has to be p : 2*p : 3*p type triplet
8336
@chillfactor said:
Yeah, 0 is correct.n² + 89n + 2010 = k²n² + 89n + 2010 - k² = 0For n to be integer, D should be perfect sq89² - 8040 + 4k² = a²(2k - a)(2k + a) = 119 = 1*119 = 7*17k = 30 and 6n² + 89n + 2010 = 30²n = -74, -15n² + 89n + 2010 = 6²n = -47, -42So, given expression can not be a perfect sq for any positive n.Find the natural number n such that product of all positive divisors of n is 2011^2016
n(n+1)/2=2016
n=63
2011^63
PS: pagalguy par hi sikha hai aise Q solve karna..
8337
@vijay_chandola said:
Sir, for triplet 4, 6, 12 LCM=12sum of the numbers=4+6+12=22
Number should be of form 1k, 2k, 3k
4, 8, 12 and not 4, 6, 12
8338
@Torque024 said:
Please someone help!Find the range of aGiven that the roots of quadratic equation ax^2 + bx + c =0 lie in the interval (0,3).
Koi options doge to asani hogi bhai..
8339
8340
@Torque024 said:
Please someone help!Find the range of aGiven that the roots of quadratic equation ax^2 + bx + c =0 lie in the interval (0,3).
Range of a..? I guess a could be anything.

For example, roots of the equation k*(x-1)*(x-2)=0 lie in interval (0, 3).
but coefficient of x^2 could be anything, as 'k' is an independent term.

What are the options?