PG Premier League - Chennai Dream Team

[krsh.vik]

Originally Posted by naga25french

superb question ..........

10^6 = 10 * 10^2 * 10 ^3

10 = 2*5

100 = 2*2*5*5

1000 = 2*2*2*5*5*5


the number of triplet product is ( 2 * 4 * 6 * 3) = 144

correct me if i am wrong..

explain more clearly please I couldn't understand much

[shivam_01]

Another problem.

How many ways are there to express 1000000 as a product of exactly three integers greater than 1? (For the purpose of this problem, abc is not considered different from bac, etc.)

My take:::

10^6 can be written as (2*5) * (2*2*5*5) * (2*2*2*5*5*5)

as we know that 10^6 = 10 * 10^2 * 10 ^3 and hence

no of ways of triplet product is ((2*4*6)* 3)
No of terms * 3 as three integers = 144 ways

[shivam_01]

@Nishant
till what time the team will be finalised
I am too much eager to join the team
please illustrate.......

Looking forward for a +ve reply

[nishant_rungta]

Hi guys,

On behalf on RMBT, I'm here to announce a change in the team

First, I apologize to Shivam for the fact that we cannot consider him cuz he's not from Chennai else we would love to have given him a place in DT deservingly.

Now coming on to the team shuffling thing

We have just one change this time after lots of deliberation:

Hantan (Ananth) moves out of the time because he would not be in a position to stay active long cuz of work related issues. We wish him all the best for his preps in the coming days and would wait for him to come back into the team with the same zeal.

Shweta.sahu comes into the team now as we want more and more newbies to step in and gain tremendously. All the best to her too

Now comes the time for a few caveats

We had some 1-2 oldies on our radar who haven't been contributing to the team in the manner required of them and also they haven't been showiing the requisite interest to gain out of the team as well. By all this we mean, active participation.

We would keep observing all of them so that we do not dilute the interest and the purpose for which the team was formed.

So please step in guys and make it a mutual process of knowledge enhancement.

Here's wishing u all good luck for the coming days


Nishant

[shivam_01]

No issuses nishant

Everything that happens happens for gud .. I will remain a supporter of the chennai team and will eargely luk forward for the gyan shared here

[rmbt]

Originally Posted by shivam_01

No issuses nishant

Everything that happens happens for gud .. I will remain a supporter of the chennai team and will eargely luk forward for the gyan shared here

Shivam, as already said by Nishant , we would loved to have you in the team but anyways, keep posting on thread.

I would suggest :why dont you inform rohit aka estranged_gnrs and lets see what happens. If he gives a thumbs up to this, then you can be surely in the team.

@All....Team is doing well and in this aspect I really respect ravi, he is active, participate but still know that if he cant contribute its better not to get into team.

The point I am trying to make is: we want everyone to do well in CAT and other exams, and help each other to motivate to perform better.However, if people take things for granted, then they will be dropped.

I am very happy with the way Chennuys, are performing in mocks and help each other to with loads of discussion and practice questions.

Offlate, Satanice, aditya, Krish.vik, shivam and not to forget Soham and Guhan for being active and helping Chennuys time to time.

Keep it up.

PS: Hope all of you have submitted CAT/IIFT forms.

PS2: Shweta welcome to DT and Hantan, all the best for the mysore trip and your prepartions.

[murty001]

Just back after taking Mock 5 of CL. The paper was very good and Verbal was of a nice level. Was getting bored with the TIME verbals. Anyway, I was able to attempt very less questions . Out of 75 i attempted only 32

My scores and percentiles:

DI - 21 (6C 3W) - 81.6

VA- 40 (11C 4W) - 97.9

QA- 12 (4C 4W) - 62.3

OA- 73 (21C 11W) - 91.90

Let's see what happens in AIMCAT tomorrow.

Cheers..

[venCat08]

Hei guys here i am posting in Chennai PL after a long time and
any guess its again for a quants problem

10^6 = 2^6 * 5^6
Let
a = 2^x1 * 5^y1,
b = 2^x2 * 5^y2 and
c = 2^x3 * 5^y3 be the three integers
Now
x1 + x2 + x3 = 6 and
y1 + y2 + y3 = 6
Lets take x1 + x2 + x3 = 6
The following are its distributions
a) All three same
i) (2,2,2) -> 1*1 = 1
b) Two of them same
ii) (0,0,6) -> 1*3 = 3
iii) (1,1,4) -> 1*3 = 3
iv) (3,3,0) -> 1*3 = 3
c) All different
v) (1,2,3) -> 1*3! = 6
vi) (0,2,4) -> 1*3! = 6
vii) (0,1,5) -> 1*3! = 6
[The above distribution has been come up w.r.t to (n+r-1)C(r-1) formula. If you total the values you can find the sum to be equal]

The distribution holds for y1+y2+y3 also
Also given
i) (a,b,c) ! = 1
If x = 0 then y should not be equal to zero
else y can take any value
Cases (ii),(iv),(vi) and (vii) come under this case
ii) (a,b,c) = (b,a,c) = ..............
Distributions in which there is repetition of numbers this case
comes into picture
> three elements are repeated.
In the cross product no arrangement is considered.
> two elements are repeated
In the cross product arrangement between the
repeated items is avoided


Now we have to find the cross product for the cases (i) to (vii) for the x series vs (i) to (vii) for y series with the above two conditions (I) and (II) in perspective

For (2,2,2) = 7 possibilities
For sets with 2 elements repeated - 3*3 + 3*2+ 1*1 - 16
possibilities * 3 = 48 possibilities are there
For no repeat - 6*3 + 3*3 + 1*1 = 28 possibilities * 3 =
84 possibilities

Hence total = 84 + 48 + 7 = 139 ways in all

phew that was one hell of a problem[if not the problem atleast the method i followed does warrant serious review] and still after solving cant say if i am correct
Not in a mood to re check it too

@ satanica hei dude is there any shortcut for the sum?
i have a strong feeling that I have arrived at a wrong answer

[venCat08]

Originally Posted by shivam_01

My take:::

10^6 can be written as (2*5) * (2*2*5*5) * (2*2*2*5*5*5)

as we know that 10^6 = 10 * 10^2 * 10 ^3 and hence

no of ways of triplet product is ((2*4*6)* 3)
No of terms * 3 as three integers = 144 ways

Hei can you explain your soln
cannot understand the 2*4*6 thing;
How can it be the no. of terms?

[maverick_srikan]

error-infested CL mock scores

VA-45 (99.0)
QA-11 (59.5)(should have been 16 but I marked it wrong in the e-omr)
DI-19 (77.6)
---------------
OA - 75(92.77)
---------------

Hope to have a better outing tomorrow.

Srikanth