The Official CAT 2007 Quant Thread

[kondapalli]

Originally Posted by junoonmba

Konda bhai How did u get 72 when u compute 777^777%125?????????

777%125 = 27

E(125) = 100 and 777%100 = 77

So eff becomes (27)^77%125

(25+2)^77%125 = (...+77.25.2^76+2^77)%125...

so proceed from here.

[esh.nil]

777^777%1000... The worst remainder problem ever...
all conventional Euler, reverse Euler, CRT were useless.. The best method was using a calculator , but here is the next best method,
777^777%1000,
E(125)=100
E( = 4
777^100%125 = 777^4%8 = 1
hence taking lcm, 777^100%1000 = 1
so the problem reduces to 777^77%1000
777(603729)^38%1000 =1
777*(729)^38 = 1
777*3^228%1000 = 1
=777*3^28
=777*(3^7)^4
=777*(2187)^4
=777*187^4%1000
=777*969^2 = 697
had to do a lot of multiplication.. phew!!!!
@sachin.. buddy this is the correct answer...

@konda... pls suggest if there is any other better method?? using the Inverse Euler I was stuck at 777.x%1000 = 1 or 223.x%1000 = 999
the calue of x in both the cases exceeded the number itself(i gave up after that)..

[kondapalli]

Originally Posted by esh.nil

@sachin.. dude assume the fruits be a,b,c,o
now we have the eqn as 1< =a+b+c+o <= 11
So typically we have to calculate for, a+b+c+o = 1
a+b+c+o = 2 etc.... until a+b+c+o = 11
Though this is an approaach, it is not doable in cat and I did not do it..
The approach specified in the soln was awesome to say the least...
assume a dummy variable say x,
now consider a+b+c+o+x = 11
this will include all the above mentioned cases,
i.e when the dummy variable x=0 we have a+b+c+o = 11
x = 1, a+b+c+o = 10 and so on,
so the total cases possible = 15C4-1 (1 for the case when x=11)
= 1365 - 1 =1364
Hope my expl was helpful...
Pls revert incase of qns....

Awesome Solution....Learning some mind-blowing concepts everday at PG

Here is one which falls into the same category

I have 4 watches and I met 10 people. I sold atleast one watch. In how many ways I could have done that?

[kondapalli]

Originally Posted by esh.nil

777^777%1000... The worst remainder problem ever...
all conventional Euler, reverse Euler, CRT were useless.. The best method was using a calculator , but here is the next best method,

=777*187^4%1000
=777*969^2 = 697
had to do a lot of multiplication.. phew!!!!
@sachin.. buddy this is the correct answer...

@konda... pls suggest if there is any other better method?? using the Inverse Euler I was stuck at 777.x%1000 = 1 or 223.x%1000 = 999
the calue of x in both the cases exceeded the number itself(i gave up after that)..

So eff becomes (27)^77%125

(25+2)^77%125 = (...+77.25.2^76+2^77)%125...


77.25.2^76%125 = 77.2^76%5 = 2.25 = 50

2^77%125 = 2^77%(2^7-3) = 3^11%125 = 243.243.3%125 = 49.3%125 = 147%125 = 22


Hence 27^77%125 = 72

Is it any better than yours.??? ......

[esh.nil]

Originally Posted by kondapalli

Awesome Solution....Learning some mind-blowing concepts everday at PG

Here is one which falls into the same category

I have 4 watches and I met 10 people. I sold atleast one watch. In how many ways I could have done that?

@konda.. completely agree with the statement..
the answer would be 14c4-1.(14c10-1)
Regarding the cat and dog problem, I am little skeptical abt the soln I gave yesterday.. I will discuss more abt it.. right now little work to catch on, will post when I get time..

[abhrabumba]

Originally Posted by mr.s.k.abhi

Sachin ur was a fundoo solution.. but i interpreted and solved it like this... any problems with my approach buddy.. plz point out..

number xy=10x+y
number ab=10a+b
similarly, pqr=100p+10q+r
xy*ab=100xa+10(ax+by)+by = 100p+10q+r

equating coefficients of various powers of 10
xa=p => xa=2x => a=2
by=r => by=2y => b=2

therefore ab=10*2+2=22

@m.s.k.abhi Sorry to post late bhai...but, see, once we reach this equation, xy*ab=100xa+10(ax+by)+by = 100p+10q+r.
we don't need to look anything further because, r=by(comparing coefficients). and also given, r=2y. So, b=2.
Only one option is there which has b=2 in the number ab.
Am I correct??

[abhrabumba]

Originally Posted by kondapalli

Awesome Solution....Learning some mind-blowing concepts everday at PG

Here is one which falls into the same category

I have 4 watches and I met 10 people. I sold atleast one watch. In how many ways I could have done that?

@hey Kondapalli For the question taht u posted (watches and all), it should be the number of non - zero solutions for the equation a1+a2+...+a9+a10 = 4.
i.e.(10+4-1)C(10-1) -1 = 13C9 - 1.
Please correct me if I am wrong.

[esh.nil]

Originally Posted by abhrabumba

@hey Kondapalli For the question taht u posted (watches and all), it should be the number of non - zero solutions for the equation a1+a2+...+a9+a10 = 4.
i.e.(10+4-1)C(10-1) -1 = 13C9 - 1.
Please correct me if I am wrong.

@abhrabumba... the condition is 1<=a1+a2+...+a9+a10 <= 4.
so u are missing a1+a2+...+a9+a10 = 3, a1+a2+...+a9+a10 = 2 and a1+a2+...+a9+a10 = 1
all these added up must give 14c4-1..

[kondapalli]

How many natural numbers n are there such that n^2+3n+5 is divisible by 121?
523abc is divisible by 8,7,9. What is the value of a.b.c?
Which is greater 99^11+100^11 or 101^11 ?

[sudiptabanerjee]

Hi,

Can someone please explain any shortcut method for solving the problems like "which is greater between a^b and b^a" (a^b denotes a to the power b) (e.g. which is greater between 9^13 or 13^9)?