official quant thread for cat08

[asghan]

Originally Posted by esh.nil

Using the Inverse Euler shld be the shortest method for this problem..
123 = 3*41
E(123) = 40
31^40%123 = 1
we have 31^78 = 31^(80-2)%123 = 31^-2%123
Let me give a small intro on inverse euler,
For Inverse Euler,
When u have a problem like this 7^94%1000,
u know the Euler of 1000 w.r.t to 7 is 100 ,
so we know 7^100%1000 = 1
but we have 7^94,
so we write 7^94%1000 as 7^100.7^-6%100 = 7^-6%1000
now we have n^-b%p
we need to find a no A such that a*n%p = 1
and this number a^b%p = 1
this is Inverse Euler theorem.
so in our problem, 143*7%1000 = 1
a= 143,
so 143^6%1000 = 1
this looks a lot easier and better than 7^94%1000..
This is Inv euler theorem, for more on this u can look the concepts total fundas thread and the thread posted my marijuana_user...

so here the problem becomes 31^-2%123
we need to find n where n*31%123 = 1
n = 4
so 4^2%123 = 16.
hence the answer

Hi esh I am getting the euler of 1000 as 400 and 123 as 80, I am following this method

for 123, 123(1-1/41)*(1-1/3), 123*40/41 *2/3 = 40 * 2 = 80, and the same method for 1000 also,

is there anything wrong in my approach ???

[guneet_01]

I had made a small mistake in the question posted earlier.. Missed on the highlighted part.. pls tell me if that changes the answer..

This is question no. 14 in LOD 2 of Time Speed Distance chapter in Arun Sharma's Quant book:
Two indian tourists in the US cycled towards each other, one from point A and the other from point B. The first tourist left point A 6 hrs later than the second left point B, and it turned out on their meeting that he travelled 12 km less than the second tourist. After their meeting, they kept cycling with the same speed, and the first tourist arrived at B 8 hours later and the second arrived at A 9 hours later. Find the speed of the faster tourist.
(a) 4 km/hr
(b) 6 km/hr
(c) 9 km/hr
(d) 2 km/hr
(e) 5 km/hr

[apatchofsnow]

Originally Posted by guneet_01

I had made a small mistake in the question posted earlier.. Missed on the highlighted part.. pls tell me if that changes the answer..

This is question no. 14 in LOD 2 of Time Speed Distance chapter in Arun Sharma's Quant book:
Two indian tourists in the US cycled towards each other, one from point A and the other from point B. The first tourist left point A 6 hrs later than the second left point B, and it turned out on their meeting that he travelled 12 km less than the second tourist. After their meeting, they kept cycling with the same speed, and the first tourist arrived at B 8 hours later and the second arrived at A 9 hours later. Find the speed of the faster tourist.
(a) 4 km/hr
(b) 6 km/hr
(c) 9 km/hr
(d) 2 km/hr
(e) 5 km/hr

I don't think anything changes................

[isimplyrock]

Hi Avinav,

I tend to disagree with your answer. The answer according to me is 660 Km.

Here is the explanation:

1) Since all the three run at constant speeds, the ratio of the distance traveled by the dog towards the thief and that traveled towards the police in one round trip should be constant.
Let’s find this ratio for the dog’s first round-trip.
Distance towards the thief during the first round trip (dt) = 360 km
Distance towards the police during the first round trip (dp) = 360/11 km

Therefore, dp = dt / 11
2) Now, the police caught the thief after 12 hrs. So the dog runs for a total of 12 hrs or 720 km.

Therefore,
Total distance traveled by dog towards thief + Total distance traveled by dog towards police = 720 km that is, dt + dp = 720km

Substituting dp = dt / 11 (from 1), we get dt = 660. Therefore the total distance traveled by dog towards the thief is 660 km.

[naga25french]

Originally Posted by vivek417

Thanks for such a useful post. I got stuck when I reached power of (-2) on 31.

Naga can you pls tell if the answer is correct?

@@@@@@@@@@@@

'Always see a glass as half full rather than half empty.'

@vivek

the answer is 16..

@esh.nil approach is right...

anyways i will explain how i worked out using inverse euler concept...

31*4=124=123+1....its clear..i know
=======> 31*4 =1mod123
31^78=31^80*31^-2
thus,
31^78=31^-2 mod123..
we can write it as:-
=31^-2 * 4^-2 *4^2 mod123
=(31*4)^-2*4^2 mod 123
=124^-2 * 4^2 mod123
=1*4^2mod123
=16mod123
so remainder = 16
hope its clear...
this is the shortest possible method i could work out...
@ jha june...

there is nothing in practising the remainder problem again... only practice makes man perfect... pls bear with me man.. dont mind to reply to those question..

[esh.nil]

Originally Posted by asghan

Hi esh I am getting the euler of 1000 as 400 and 123 as 80, I am following this method

for 123, 123(1-1/41)*(1-1/3), 123*40/41 *2/3 = 40 * 2 = 80, and the same method for 1000 also,

is there anything wrong in my approach ???

@asghan... when u are finding the euler of a non-prime number,
Use the LCM of the eulers of it's prime factors.
so 123 = 3*41
E(3)=2
E(41) = 40
LCM of 40,2 = 40
Hence the Euler of 123 though acc to euler theorem is 80,
the cyclicity of the remainders is exhibited even at 40..
Try that out..

[smartie]

Originally Posted by naga25french

is q =2 and s = 3 ?????

so answer according to my calculations
1. 3] 2
2. 1] 3

correct me if i am wrong..

please explain the solution....i dint understand the question itself..
In (pqr)q(qs) what is the relation between (pqr) and q and s?

[asghan]

Originally Posted by esh.nil

@asghan... when u are finding the euler of a non-prime number,
Use the LCM of the eulers of it's prime factors.
so 123 = 3*41
E(3)=2
E(41) = 40
LCM of 40,2 = 40
Hence the Euler of 123 though acc to euler theorem is 80,
the cyclicity of the remainders is exhibited even at 40..
Try that out..

So for 1000 = 125 * 8, we are finding euler of 125 which is 100 and 8 which is 4 hence LCM of these number 100, So i guess we can use this funda for a number like 5^3 * 2^3 and not for 5 and 3 which will give lcm of its euler number as 4....... Please comment @esh

[esh.nil]

Originally Posted by asghan

So for 1000 = 125 * 8, we are finding euler of 125 which is 100 and 8 which is 4 hence LCM of these number 100, So i guess we can use this funda for a number like 5^3 * 2^3 and not for 5 and 3 which will give lcm of its euler number as 4....... Please comment @esh

Sorry for my poor choice of words... It's not the prime factors,
instead split the given number into the product of co-prime numbers...
so 1000 shall be split into 125*8
the highest power of the prime factor available...

[satanica]

Originally Posted by esh.nil

Using the Inverse Euler shld be the shortest method for this problem..
123 = 3*41
E(123) = 40
31^40%123 = 1
we have 31^78 = 31^(80-2)%123 = 31^-2%123
Let me give a small intro on inverse euler,
For Inverse Euler,
When u have a problem like this 7^94%1000,
u know the Euler of 1000 w.r.t to 7 is 100 ,
so we know 7^100%1000 = 1
but we have 7^94,
so we write 7^94%1000 as 7^100.7^-6%100 = 7^-6%1000
now we have n^-b%p
we need to find a no A such that a*n%p = 1
and this number a^b%p = 1
this is Inverse Euler theorem.
so in our problem, 143*7%1000 = 1
a= 143,
so 143^6%1000 = 1
this looks a lot easier and better than 7^94%1000..
This is Inv euler theorem, for more on this u can look the concepts total fundas thread and the thread posted my marijuana_user...

so here the problem becomes 31^-2%123
we need to find n where n*31%123 = 1
n = 4
so 4^2%123 = 16.
hence the answer

This method rocks...

U made a mistake though..
I guess when U are refering to Euler, U mean Euler's Totient function of 1000 which will be 400 and also it is independent of 7, with the only thing to be kept in mind that HCF(1000,7) = 1.

I solved the same in a totally different manner.

123 = 3 * 41

31 mod 3 = 1
31^78 mod 3 = 1

31 mod 41 = -10
-10^3 mod 41 = -16
-16^2 mod 41 = 10
which means -10^6 mod 41 = 10
31^78 mod 41 = -10^78 mod 41 = (-10^6)^13 mod 41 = 10^13 mod 41 = 10^12 mod 41 * 10 = 10^3 mod 41 = 16

Now, the solution is 'x' where..
x mod 3 = 1
x mod 41 = 16

By observation..
x=16