Official Quant thread for CAT 2013

MBA CAT 2024
22027
@albiesriram said:
OA is 1925
sir sol batao na..........
22028
@jain4444 said:
let say our answer = A money till 12th = A (A + x)*11/12 = x => x = 11A so , money till 11th = 11A + A = 12A like this we need to multiply resultant number by 5 at 7th minister when (12 - 7) = 5 , then by 7 at 5th minister(12 - 7 = 5) , then by 5 at 2nd minister(12 - 2 = 10) and then at last by 11 at 1st minister(12 - 1 = 11) so , 5*7*5*11 = 1925
bhai kuch nahi samajh mein aaya.. can you support your answer with a better explanation pls.. it would be really helpful
22029
what is the sum of all factors of (2^97-1)???
22030
@Dexian said:
if we tend to size down the problem for 3 ppl thsn acc to ur approach... 3rd guy will take away 3^2-(3-1)/3^3....if the total sum is 3^3 = 27 then its 7....but 27/3=9 I Guy18/3=6 II guy12 III guy....kuchh to missing hai na......is assumption of 27 as total right?????
yeah thats the reason i asked @scrabbler bhai to look into it. If my logic was all right i would have ended up with the answer :P
22031
@Jackson1 said:
what is the sum of all factors of (2^97-1)???
1?
22032
@iLoveTorres said:
bhai kuch nahi samajh mein aaya.. can you support your answer with a better explanation pls.. it would be really helpful
its very hard to explain bhai

you can go by scrabbler's approach
22033
@iLoveTorres said:
1?
don't know the ans....plz explain???
22034
@iLoveTorres said:
the first person will take 1/12th of the whole.. the second person will take away 11/144th part .. third will take away 133/1728th part..meine isse thoda analyse kiya toh it is falling in a pattern like from second person onwards the part the guy takes is 12-1/12^23rd person (12^2-(12-1)/12^3so for the 12th person it will be (12^11-(12^10-(12^9-(12^8-(12^7-(12^6-(12^5-(12^4-(12^3-(12^2-(12-1))/12^12(12^11-12^10+12^9-12^8+12^7-12^6+12^5-12^4+12^3-12^2+12-1)/12^12(12^10(11)+12^8(11)+12^6(11)+12^4(11)+12^2(11)-1)/12^12(12^8(11)(144+1)+12^4(11)(144+1)+1583)/12^12iske baad i was wondering what have i done.. so would love to kno any mistake or if i have applied wrong logic
Start to sahi lag raha hai...but the numbers become too complex too fast, remember you need to calculate and answer and your method will require you to calculate 12^12 ...so alternative approach needed...(unless you're a masochist who loves these heavy calculations )instead focus on what is left behind...first guy takes 1/12, leaves 11/12. Next takes 2/12 of this, leaves 10/12 of 11/12, similarly third guy leaves 9/12 of 10/12 0f 11/12 and so on...

So what is left for the last guy is 11/12 *10/12*9/12*8/12 *....1/12 and he takes 12/12 i.e. all of this. Hence his share is 12!/12^12.

Edit: Your approach, it is galat thoda...see post below!

regards
scrabbler
22035
@Jackson1 said:
what is the sum of all factors of (2^97-1)???
its a prime number

so , sum of factors = 2^97 - 1 + 1 = 2^97
22036
@jain4444 said:
its a prime number so , sum of factors = 2^97 - 1 + 1 = 2^97
how do you know that (2^97-1) is a prime number???
22037
@Jackson1 said:
how do you know that (2^97-1) is a prime number???
2^n - 1 is prime when n = prime number except when n = 11
22038
@iLoveTorres said:
the first person will take 1/12th of the whole.. the second person will take away 11/144th part .. third will take away 133/1728th part..meine isse thoda analyse kiya toh it is falling in a pattern like from second person onwards the part the guy takes is 12-1/12^23rd person (12^2-(12-1)/12^3so for the 12th person it will be (12^11-(12^10-(12^9-(12^8-(12^7-(12^6-(12^5-(12^4-(12^3-(12^2-(12-1))/12^12(12^11-12^10+12^9-12^8+12^7-12^6+12^5-12^4+12^3-12^2+12-1)/12^12(12^10(11)+12^8(11)+12^6(11)+12^4(11)+12^2(11)-1)/12^12(12^8(11)(144+1)+12^4(11)(144+1)+1583)/12^12iske baad i was wondering what have i done.. so would love to kno any mistake or if i have applied wrong logic
Oops no....mistake hai re, missed it...second guy will take 22/144 as he is taking 2/12 of 11/12, similarly 3rd guy will also change...

regards
scrabbler
22039
@jain4444 said:
2^n - 1 is prime when n = prime number except when n = 11
Sorry, not true....

http://en.wikipedia.org/wiki/Mersenne_prime

Is list mein 2^97 - 1 nahin hai btw...

http://en.wikipedia.org/wiki/Mersenne_prime#List_of_known_Mersenne_primes

regards
scrabbler
22040
@Jackson1 said:
what is the sum of all factors of (2^97-1)???
In fact the prime factorisation is 11447 x 13842607235828485645766393* (2 distinct prime factors) and so the sum of factors would be some weird number.

Yeh CAT question ho nahin sakta, koi bhi exam mein nahin aa sakta. Kahaan mila yehwaala?

* I didn't factorise, in case you're wondering 😉 looked it up on wolframalpha...

regards
scrabbler

@jain4444
22041
@jain4444 said:
its a prime number so , sum of factors = 2^97 - 1 + 1 = 2^97
@jain4444 said:
2^n - 1 is prime when n = prime number except when n = 11
not true bhai...

check this out..

http://www.wolframalpha.com/input/?i=factorize+2%5E97+-+1


22042
@Dexian said:
sir sol batao na..........

Suppose X is the initial number of scam amount.

12th pirate share is = X*11/12*10/12*9/12*..*1/12

= X* 11*5^2*7*3^4*2^8 / (2^22 * 3^ 11)

= x* 11* 5^2*7 / (2^14*3^7)

Now for X should be minimum and the above share is to be a whole number ->
(x= 2^14*3^7)

Which implies 12 th minister share is 1925.

We must also check whether every one receives the whole number share as well. 1925 works fine.

22043
@scrabbler @Logrhythm
in that 12 corrupt ministers wala sum...
wat will be the min sum required...
shud it not be 12^11........... assuming last guy walks away with watever is left ..(since he is a politician ...... lol)
PS its working for 4 ppl and 5 and 3
22044
@albiesriram said:
Suppose X is the initial number of scam amount. 12th pirate share is = X*11/12*10/12*9/12*..*1/12= X* 11*5^2*7*3^4*2^8 / (2^22 * 3^ 11)= x* 11* 5^2*7 / (2^14*3^7)Now for X should be minimum and the above share is to be a whole number -> (x= 2^14*3^7)Which implies 12 th minister share is 1925.We must also check whether every one receives the whole number share as well. 1925 works fine.
arey but the qus says 1/12th of the remaining sum left...
which i read as ....
if 1st guy takes 1/12th of T
then 2nd guy takes 1/12 * (1-1/12th) = 11/144...

m i right??
or kuch chemical locha hai ???
22045
@Dexian said:
arey but the qus says 1/12th of the remaining sum left...which i read as ....if 1st guy takes 1/12th of Tthen 2nd guy takes 1/12 * (1-1/12th) = 11/144...m i right??or kuch chemical locha hai ???
They agree that the kth minister will walk away with k/12 of the money remaining.

How is that 1/12?
@Dexian said:
@scrabbler@Logrhythmin that 12 corrupt ministers wala sum...wat will be the min sum required...shud it not be 12^11........... assuming last guy walks away with watever is left ..(since he is a politician ...... lol)PS its working for 4 ppl and 5 and 3

It will work. It just won't be the minimum sum possible. Any multiple of the minimum sum will also work (minimum sum being 12^7 - see my earlier post for the logic).

See if I ask you "A earns 11/87 of the total, and it is a positive integer, find the minimum value of total earnings" then A getting 11 out of 87 is the minimum case. But 22 out of 174 or 110 out of 870 or 550 out of 4350 will also work fine. So if we find that 4350 works, that by itself does not mean it is the minimum! 87 would be the answer.

regards
scrabbler
22046
@Dexian said:
arey but the qus says 1/12th of the remaining sum left...which i read as ....if 1st guy takes 1/12th of T.
. No. Kth minister takes k/12 of the remaining.

@albiesriram said:
Suppose X is the initial number of scam amount. 12th pirate share is = X*11/12*10/12*9/12*..*1/12= X* 11*5^2*7*3^4*2^8 / (2^22 * 3^ 11)= x* 11* 5^2*7 / (2^14*3^7)Now for X should be minimum and the above share is to be a whole number -> (x= 2^14*3^7)Which implies 12 th minister share is 1925.We must also check whether every one receives the whole number share as well. 1925 works fine.