The Official CAT 2007 Quant Thread

[tatimatla]

Ahem,

2001=3*23*29
Thus 2^2001=2^(3*23*29) = 2^(3*29) * 2^22(3*29) and not 2*2^22(3*29)
same with this step:
2^2001=2*2^28(3*23)=2*2^28kmod29=2,since e.n(29)=28

[hanuman]

Originally Posted by vikram_k51

Tati 2001=3*23*29
Thus 2^2001=2*2^22(3*29)=2*2^22kmod23=2,since e.n(23)=22
Likewis 2^2001=2*2^28(3*23)=2*2^28kmod29=2,since e.n(29)=28
Right?

vikram could not get ur approach
I believe this is what u r saying
2001 = 3*23*29

2^2001 mod 3 = 2 mod 3 -(1)
2^2001 mod 23 = 2^21 mod 23 = 2*(2^10mod 23)*(2^10 mod 23)=288 mod 23=12 mod 23 -(2)
2^2001 mod 29 = 2^13 mod 29 =14 mod 29-(3)

Apply Chinese remainder theorem for (1),(2) and (3) u will get the ans..will update the ans.

Added later
Getting 71 as the ans. (Possible calculation errors)

[getintoiimb]

Hi all
I have a question to be solved
The question is
Arrange the following fractions in ascending order

Q) 17/25, 18/27, 23/33, 47/59, 61/71

My question is how do u solve this question is less time without calculating the actual value of each fraction by division? Is there any shortcut or some basic funda to do this?
Im wasting time by calculating value of each and every fraction and then arranging them. Please help.

[vikram_k51]

Originally Posted by getintoiimb

Hi all
I have a question to be solved
The question is
Arrange the following fractions in ascending order

Q) 17/25, 18/27, 23/33, 47/59, 61/71

My question is how do u solve this question is less time without calculating the actual value of each fraction by division? Is there any shortcut or some basic funda to do this?
Im wasting time by calculating value of each and every fraction and then arranging them. Please help.

Is the answer:

61/71>47/59>23/33>17/25>18/27
See the %change in the numerator and denominator.

[mohit_353]

Originally Posted by the.bombardier

30030 = 2*3*5*7*11*13
all the factors are coprime
17=1mod2 17^28820=1mod2
17=2mod3 17^28820=1mod3
17=2mod5 17^28820=1mod5
17=3mod7 17^28820=2mod7
17=6mod11 17^28820=1mod11
17=4mod13 17^28820= 1mod13
therefore 17^28820=1mod(2*3*5*11*13)=1mod4290
also 17^28820=2mod7
from these conditions v get 17^28820=25741mod4290 & 17^28820=25741mod7
therefore 17^28820=25741mod 30030
therefore the remainder is 25741.

Hi;
can any body pls explain this problem while considering the euler's or fermat;s rule step by step......i tried to go through both the rules but can't understand completely....

thanks in advance...

[the.bombardier]

Originally Posted by mohit_353

Hi;
can any body pls explain this problem while considering the euler's or fermat;s rule step by step......i tried to go through both the rules but can't understand completely....

thanks in advance...

17^28820/30030
the factors of 30030 are 2,3,5,7,11 and 13
euler's no of 30030 = 30030(1-1/2)(1-1/3)(1-1/5)(1-1/7)(1-1/11)(1-1/13)
=1*2*4*6*10*12
=5760
therefore 17^5760=1mod30030
which implies (17^5760)^5=17^28800=1mod30030
so v need to find the remainder of 17^20/30030
17^20 is odd.
so 17^20=1mod2 -------------------1
using fermat's little theorem
17^2=1mod3
so 17^20=1mod3 -------------------2
17^4=1mod5
so 17^20=1mod5 --------------------3
17^6=1mod7
17^20=17^18 * 17^2=289mod7=2mod7 --------------4
17^10=1mod11
so 17^20=1mod11 -----------------5
17^12=1mod13
17^20=17^12 * 17^8 =(17^8 )mod13=(3^4)mod13=3mod13 --------------6
from eqns 1,2,3,5
17^20=1mod330 -----------7
from 4 and 7
17^20=331mod2310 --------------8
from 6 and 8
v get remainder as 9571.
is it clear now??go through the link tatimatla has posted for remainder problems..it will be very helpful.

[gpat]

Just joined in today...
plz can ne1 clearly explain the concept of permutations and combinations with a few examples...
plz keep it simple

[mohit_353]

Originally Posted by the.bombardier

17^28820/30030
the factors of 30030 are 2,3,5,7,11 and 13
euler's no of 30030 = 30030(1-1/2)(1-1/3)(1-1/5)(1-1/7)(1-1/11)(1-1/13)
=1*2*4*6*10*12
=5760
therefore 17^5760=1mod30030
which implies (17^5760)^5=17^28800=1mod30030
so v need to find the remainder of 17^20/30030
17^20 is odd.
so 17^20=1mod2 -------------------1
using fermat's little theorem
17^2=1mod3
so 17^20=1mod3 -------------------2
17^4=1mod5
so 17^20=1mod5 --------------------3
17^6=1mod7
17^20=17^18 * 17^2=289mod7=2mod7 --------------4
17^10=1mod11
so 17^20=1mod11 -----------------5
17^12=1mod13
17^20=17^12 * 17^8 =(17^8 )mod13=(3^4)mod13=3mod13 --------------6
from eqns 1,2,3,5
17^20=1mod330 -----------7
from 4 and 7
17^20=331mod2310 --------------8
from 6 and 8
v get remainder as 9571.
is it clear now??go through the link tatimatla has posted for remainder problems..it will be very helpful.

thanks for ur reply...bhai baki sab to samaj aa gaya ..but how did u merge eqn..4 & 7 ....and...eqn. 6&8...i think my quants just sucks...

[the.bombardier]

Originally Posted by mohit_353

thanks for ur reply...bhai baki sab to samaj aa gaya ..but how did u merge eqn..4 & 7 ....and...eqn. 6&8...i think my quants just sucks...

il explain eqn 4 and 7,u can try out the other one urself...
eqn 4 is 2mod7
eqn 6 is 1mod330
when u r combining such eqns u have to find the smallest no which satisfies both the conditions.in this case when u look at eqn 6,the series of nos which are 1mod330 is:
1,331,661.... that is these numbers leave a remainder 1 when divided by 330.
out of this series u have to find the smallest no which leaves a remainder 2 when divided by 7.
then u can say that this number satisfies eqn 4 and 7 both.
in this case the no is 331...
therefore v get 331mod(7*330) which is 331mod2310

[the.bombardier]

Originally Posted by the.bombardier

therefore v get 331mod(7*330) which is 331mod2310

i hope u knw that when a no 'n' divided by two different nos a & b gives the same remainder r, then the remainder when n is divided by lcm(a,b) is also r. that is the justification for this step.