Official Quant thread for CAT 2013

MBA CAT 2024
30758
@chillfactor said:

2 when divided by 6 remainder will be 2

2*6 when divided by 6*6 OR 12 when divided by 36 remainder will be 12 not 2

As per your method 12/36 = 2/6, so remainder will be 2 (but its incorrect)

While finding out remainders when you cancel something then you have to multiply it back to get the correct remainder

Lets take one more example:- What will be the remainder when 49 is divided by 42

Clearly answer is 7

But as you mentioned 49/42 = 7/6, so remainder 1 which is incorrect
Thats why you need to multiply back by the factor which you cancelled, hence answer will be 1*7 = 7

Solve this question:-
What will be the remainder when 2^96 is divided by 96
we can reduce the expression to 2^91 / 3
2^91 / 3 --> (-1)^91 / 3 --> -1
so remainder is -1*32 = -32 or 64
30759
@chillfactor said:

While finding out remainders when you cancel something then you have to multiply it back to get the correct remainder
Perfect explanation sir. :-)
@chillfactor said:

Solve this question:-What will be the remainder when 2^96 is divided by 96
64 ?

30760
@Dexian why we are taking 6 c2 ..can u explain the method
30761
@junefever said:
I know but kaise bhi karke maine answer nikal diya plz share ur aproach
bhai appraoch has been shared by vijay bhai i.e. 2^96 mod 96
=2^5( 2^91 mod 3)
=2^5(-1 mod 3)
=2^5( 2 mod 3)
=64 mod 96

30762
A "superfactorial" defined as S(f) is the product of factorials..for eg. s(5)=5!*4!*3!*2!*1!....which superfactorial will have 2n preceding zeroes if the number is denoted by n ??
A) 7
B)23
C)121
D) never
30763
@viewpt said:
bhai appraoch has been shared by vijay bhai i.e. 2^96 mod 96=2^5( 2^91 mod 3)=2^5(-1 mod 3)=2^5( 2 mod 3)=64 mod 96
using Eulers
E(96)=16

(2^16)^6 mod 96
1*2^6 mod 96
64 mod 96
hence answer=64 :)

30764
Q: 16^56 div 58 R?
30765
@junefever said:
using EulersE(96)=16(2^16)(2^6) mod 961*2^6 mod 9664 mod 96 hence answer=64
bhai eulers is applicable here X^n div m where x and m are coprime in case above these are not coprime 2 and 96 are even .
30766
@falcao said:
A "superfactorial" defined as S(f) is the product of factorials..for eg. s(5)=5!*4!*3!*2!*1!....which superfactorial will have 2n preceding zeroes if the number is denoted by n ?? A) 7 B)23 C)121 D) never
23 ?
30767
@fireatwill said:
@Dexian why we are taking 6 c2 ..can u explain the method
Check this link :
http://www.pagalguy.com/forums/quantitative-ability-and-di/cat-2010-concepts-fundas-tips-crack-quants-section-t-55747/p-2191164/r-2194411
30768
Find remainder when 33333333......86 times is divided by 13?
30769

Puys...i m new to this group.... is it cool for me to just dive right in or is there an initiation ceremony? 😛

30770
@viewpt said:
Q: 16^56 div 58 R?
24?
30771
@falcao said:
how?i don't know the answer.but wouldn't 1! *2! *3!... 23! end with 19 zeroes?
we get 5^46 till 23! which means it should have 46 zeroes. please let me know the answer once you come to know about it.
30772
@junefever said:
Find remainder when 33333333......86 times is divided by 13?
Apply fermi little theorem ..
prob reduces to 33 / 13 . remainder is 7
fermi little theorem explanation : http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-cat-2010-t-47731/p-1772936/r-1909113
30773
@falcao said:
A "superfactorial" defined as S(f) is the product of factorials..for eg. s(5)=5!*4!*3!*2!*1!....which superfactorial will have 2n preceding zeroes if the number is denoted by n ?? A) 7 B)23 C)121 D) never
Never
sorry
it will be 23 so total zero will be 46 nos

S(23)
30774
@junefever said:
Find remainder when 33333333......86 times is divided by 13?
Applying Fermi Little Theorem..Link has already been given by @naga25french sir.
33/13, remainder=7...??
30775
@falcao said:
A "superfactorial" defined as S(f) is the product of factorials..for eg. s(5)=5!*4!*3!*2!*1!....which superfactorial will have 2n preceding zeroes if the number is denoted by n ?? A) 7 B)23 C)121 D) never
23 will have 46 zeroes ?
30776
@naga25french said:
Apply fermi little theorem .. prob reduces to 33 / 12 . remainder is 9 fermi little theorem explanation : http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-cat-2010-t-47731/p-1772936/r-1909113
Is it a typo error??
shouldn't it be prob reduces to 33 / 13 . remainder is 7

thanks for the link, very helpful
30777
@viewpt said:
Q: 16^56 div 58 R?
30 hai kya?