The Official CAT 2007 Quant Thread

[the.bombardier]

Originally Posted by tatimatla

Solve this:
What is the remainder when 17^28820 is divided by 30030?

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.

[tatimatla]

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.

THere is a mistake in the bold line...
17^28820= 3mod13

[the.bombardier]

Originally Posted by tatimatla

THere is a mistake in the bold line...
17^28820= 3mod13

yeah ur correct..sorry bout that.the solution will be
17^28820=1mod330
17^28820=2mod7
17^28820=3mod13 ---------eqn 1
17^28820=331mod(330*7)=331mod2310 ---------------eqn2
from 1 and 2 we get solution
17^28820=9571mod(2310*13)=9571mod(30030)
therefore remainder is 9571
am i correct now?

[tatimatla]

Originally Posted by the.bombardier

yeah ur correct..sorry bout that.the solution will be
17^28820=1mod330
17^28820=2mod7
17^28820=3mod13 ---------eqn 1
17^28820=331mod(330*7)=331mod2310 ---------------eqn2
from 1 and 2 we get solution
17^28820=9571mod(2310*13)=9571mod(30030)
therefore remainder is 9571
am i correct now?

yes...absolutely...

[vikram_k51]

Originally Posted by tatimatla

Solve this:
What is the remainder when 17^28820 is divided by 30030?

Hi Tati,
30030=2*3*5*7*11*13
Thus E.N(30030)=L.C.M(2*4*6*10*12)=60
17^28820=17^60n*17^2mod30030
Thus ans=289

Pls tell me where m I wrong.

[tatimatla]

Originally Posted by tatimatla

yes...absolutely...

Ek aur solve karo...
What's the remainder when 2^2001 is divided by 2001?

[tatimatla]

Originally Posted by vikram_k51

Hi Tati,
30030=2*3*5*7*11*13
Thus E.N(30030)=L.C.M(2*4*6*10*12)=60
17^28820=17^60n*17^2mod30030
Thus ans=289

Pls tell me where m I wrong.

mistake in the bold part...
17^28820=17^60n*17^20mod30030

[vikram_k51]

Originally Posted by tatimatla

Ek aur solve karo...
What's the remainder when 2^2001 is divided by 2001?

2001=3*23*29
Thus 2^2001mod3=2
2^2001mod23=2
2^2001mod29=2
Hence 2^2001mod2001=8
Am I right?

[tatimatla]

Originally Posted by vikram_k51

2001=3*23*29
Thus 2^2001mod3=2
2^2001mod23=2
2^2001mod29=2
Hence 2^2001mod2001=8
Am I right?

recheck the stuff in bold...

[vikram_k51]

Originally Posted by tatimatla

recheck the stuff in bold...

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?