Official Quant thread for CAT 2013

MBA CAT 2024
24588
@bodhi_vriksha said:
Evening Teaser: (Easy) LCM and HCF of two numbers is 11! and 7! respectively. How many such pairs of two numbers are possible?Team BV - Kamal Lohia
let two numbers be 7!k and 7!n

7!^2 * k * n = 7! * 11!

k * n = 11!/7! = 2^4 * 3^2 * 5^1 * 11^1

now , "k" and "n" should be co-prime

k = 2^a1 * 3^a2 * 5^a3 * 11^a4
n = 2^b1 * 3^b2 * 5^b3 * 11^b4

number of ways = 2*2*2*2 = 16

but we need to find unordered pairs here so , 16/2 = 8 such pairs
24589
@pavimai said:
OA 23.min value of k is 1 followed by 22 9s..so minimum is 23 digits
why don't you want to solve it further where I left it.

Anyways..f(k) has to be greater than 19 and has sum of digits 19. So smallest value of f(k) is 199.

Similarly k has to be greater than 199 but has sum of digits 199, so smallest value of k is 1 followed by 22 9's i.e. a 23 digit number. :)

Team BV - Kamal Lohia
24590
@pavimai said:
options 3 23111111111111
23 ryt?
24591
@bodhi_vriksha said:
One more: What is the largest n such that 2^n divides [10^30/(10^10 + 3)], where [] represents greatest integer function?Team BV - Kamal Lohia
Getting 1 😞 Going to re-check :splat:

regards
scrabbler
24592

Last for today from my side: Nimai and Nitai start running simultaneously from opposite ends of a road towards each other with speeds of 3m/s and 5m/s respectively. As soon as they reach some end, they turn without stopping and keep running towards other end till both of them reach their start position simultaneously. Nitai hands over a lotus flower to Nimai everytime while meeting him from the opposite direction.And whenever Nitai overtakes Nimai i.e. crossing while moving in same direction, Nitai takes a lotus flower back from Nimai.
What is the number of lotus flowers with Nimai at the end of the run?

Team BV - Kamal Lohia

24593
@scrabbler said:
Getting 1 Going to re-check regardsscrabbler
You should

Team BV - Kamal Lohia
24594
@bodhi_vriksha said:
One more: What is the largest n such that 2^n divides [10^30/(10^10 + 3)], where [] represents greatest integer function?Team BV - Kamal Lohia
is it 3?
24595
@bodhi_vriksha said:
You should Team BV - Kamal Lohia
3 rather? Bad handwriting, read 8 as 18 right at the end :splat: that's why I am better off trying sums orally...

Still not convinced of my answer though. Goes back to thinking :(

regards
scrabbler
24596

I am very delighted to be among great brains as @scrabbler and @ChirpiBird around.

Please post your approaches earthly gods :)

Team BV - Kamal Lohia

24597

@bodhi_vriksha said:
One more: What is the largest n such that 2^n divides [10^30/(10^10 + 3)], where [] represents greatest integer function?Team BV - Kamal Lohia

10^30 = 10^10 cube....so tried to express using a^3 and b^3 logic.

Now a^3 + b^3 = (a+b)(a^2 - ab + b^2)

So (10^30 + 27) / (10^10 + 3) would be (10^20 - 3*10^10 + 9)

Now, if we minus 27 from the numerator a very tiny fraction will be subtracted....so the Greatest Integer value will decrease by 1, giving (10^20 - 3*10^10 + 8) Highest power of 2 in this is 3.

Edited for +/- typos

@ChirpiBird

regards
scrabbler
24598
@bodhi_vriksha said:
Last for today from my side: Nimai and Nitai start running simultaneously from opposite ends of a road towards each other with speeds of 3m/s and 5m/s respectively. As soon as they reach some end, they turn without stopping and keep running towards other end till both of them reach their start position simultaneously. Nitai hands over a lotus flower to Nimai everytime while meeting him from the opposite direction.And whenever Nitai overtakes Nimai i.e. crossing while moving in same direction, Nitai takes a lotus flower back from Nimai. What is the number of lotus flowers with Nimai at the end of the run?Team BV - Kamal Lohia
1?
i guess it should be 5
24599
@scrabbler said:


10^30 = 10^10 cube....so tried to express using a^3 and b^3 logic.

Now a^3 + b^3 = (a+b)(a^2 - ab + b^2)

So (10^30 + 27) / (10^10 + 3) would be (10^20 - 3*10^10 + 9)

Now, if we minus 27 from the numerator a very tiny fraction will be subtracted....so the Greatest Integer value will decrease by 1, giving (10^20 - 3*10^10 + 8) Highest power of 2 in this is 3.
Edited for +/- typos@ChirpiBird regardsscrabbler

Good one scrabbler :)
Team BV - Vineet
24600
@bodhi_vriksha said:
I am very delighted to be among great brains as @scrabbler and @ChirpiBird around.Please post your approaches earthly gods Team BV - Kamal Lohia
approach..

10^30 = -27mod(10^10+3)=10^10+24 mod (10^10 + 3)
10^10 + 24 ...mein 2^3 is common..
2^3(2^8*5^10 +3) ... or simply.. 8(something odd)
10^30 has 30 2s.
and 10^30 = M(10^10 +3) + 2^3*(something odd) .... this we found in previous step.
(10^10 +3)doesnot have any 2. so.. Q needs an odd power of 2 for the above equation to make sense.. so 2^3.

24601
@jain4444 said:
9720000 = 2^6 * 5^4 * 7^3 x^3 = 2^3a * 5^3b * 7^3c y^2 = 2^2p * 5^2q * 7^2r z^4 = 2^4x * 5^4y * 7^4z 3a + 2p + 4x = 6 => 3 solutions 3b + 2q + 4y = 4 => 2 solutions 3c + 2r + 4z = 3 => 1 solution and we could have 2 values for "y" and "z" so , total solutions = 3*2*1*2*2 = 24

Bro one doubt here,

3a + 2p + 4x = 6 => 3 solutions
How did u get 3 solutions here?

24602
@nramachandran said:

Bro one doubt here,

3a + 2p + 4x = 6 => 3 solutions
How did u get 3 solutions here?

for x=0, 3a+2p =6 -> (a,p)=0,3 and 2,0
for x=1, 3a+2p =2 -> (a,p)=0,1
24603
@nramachandran said:
Bro one doubt here,3a + 2p + 4x = 6 => 3 solutions How did u get 3 solutions here?
odd+even+even = odd
so 3a = 6k form..

6k+2p+4x=6
3k+p+2x=3

p+2x = 3(1-k)
so p+2x = 3y form..

3y+3k = 3
y+k = 1

2 solutions..

y can be maximum = 1
or p+2x = 1
only 1 solution where p = 1

so total 3 solutions...

this is a longer approach for cases where RHS is a comparatively smaller number (like here)
so in cases like these u can directly make cases...

3a+2p+4x = 6

a=2,p=0,x=0
a=0,p=3,x=0
a=0,p=1,x=1

so 3 solutions...


24604
@bodhi_vriksha said:
Last for today from my side: Nimai and Nitai start running simultaneously from opposite ends of a road towards each other with speeds of 3m/s and 5m/s respectively. As soon as they reach some end, they turn without stopping and keep running towards other end till both of them reach their start position simultaneously. Nitai hands over a lotus flower to Nimai everytime while meeting him from the opposite direction.And whenever Nitai overtakes Nimai i.e. crossing while moving in same direction, Nitai takes a lotus flower back from Nimai. What is the number of lotus flowers with Nimai at the end of the run?Team BV - Kamal Lohia
is it 8?
24605

Guys, now try this: Medians of a triangle are of length 3,4 and 5 respectively. Find the area of the triangle.

Team BV - Vineet

24606

@bodhi_vriksha said:
Guys, now try this: Medians of a triangle are of length 3,4 and 5 respectively. Find the area of the triangle.
Area =4/3 * 6 * 3 * 2 * 1 = 48 ??
24607
@nramachandran said:
Area =4/3 * 6 * 3 * 2 * 1 = 48 ??
Apologies
4/3 * sqrt (36) = 8??