The Official CAT 2007 Quant Thread

[doomsayer]

I have a serious doubt regarding reminders. I want to know is there any rule such that
if we want to find the reminder of A^n/B where A=ax and B=bx
then the answer will be x * (the reminder of a^n/b)

I will explain it through 1 question which has already posted in this thread.
whats the reminder of 2^643/96 ?
many people answerd it like that 2^5*( reminder of 2^638/3), since 96= (2^5 )*3

Till yesterday i thought that there is such a rule exists,1 time faculty taught me like that and I have seen that in many posts of PG.But right now i doubt the validity of the rule.

Quant gurus plz help..

[the.bombardier]

Originally Posted by doomsayer

I have a serious doubt regarding reminders. I want to know is there any rule such that
if we want to find the reminder of A^n/B where A=ax and B=bx
then the answer will be x * (the reminder of a^n/b)

yes its valid.

[vikram_k51]

Originally Posted by doomsayer

I have a serious doubt regarding reminders. I want to know is there any rule such that
if we want to find the reminder of A^n/B where A=ax and B=bx
then the answer will be x * (the reminder of a^n/b)

I will explain it through 1 question which has already posted in this thread.
whats the reminder of 2^643/96 ?
many people answerd it like that 2^5*( reminder of 2^638/3), since 96= (2^5 )*3

Till yesterday i thought that there is such a rule exists,1 time faculty taught me like that and I have seen that in many posts of PG.But right now i doubt the validity of the rule.

Quant gurus plz help..

Hi doomsayer,
We can cancel out the common factor,but the remainder so obtained has to be multiplied by the factor which is cancelled out.
So basically the question reduces to:2^638mod3---->1
But the remainder will not be 1,u need to multiply 2^5 to this remainder and take the mod once again.Finally the situation reduces to 2^5mod3=2mod3hence Ans=2

[doomsayer]

Originally Posted by vikram_k51

Hi doomsayer,
We can cancel out the common factor,but the remainder so obtained has to be multiplied by the factor which is cancelled out.
So basically the question reduces to:2^638mod3---->1
But the remainder will not be 1,u need to multiply 2^5 to this remainder and take the mod once again.Finally the situation reduces to 2^5mod3=2mod3hence Ans=2

How do u explain the below prob?
reminder of 72^2/15
as per the law it reduces to 3*( reminder of 24^2 /5)) { Here we cancel out the common term 3}
>>>3 * 1 >>> ans is 3 .But the actual answer is 9 not 3.
Where it went wrong?

[the.bombardier]

Originally Posted by vikram_k51

Hi doomsayer,
But the remainder will not be 1,u need to multiply 2^5 to this remainder and take the mod once again.Finally the situation reduces to 2^5mod3=2mod3hence Ans=2

that last step is not correct.once u multiply with 2^5,u dont need to take mod again.
the remainder will be 32.

[the.bombardier]

Originally Posted by doomsayer

How do u explain the below prob?
reminder of 72^2/15
as per the law it reduces to 3*( reminder of 24^2 /5)) { Here we cancel out the common term 3}
>>>3 * 1 >>> ans is 3 .But the actual answer is 9 not 3.
Where it went wrong?

u cant do that..u r cancelling out 2 3's from the numerator n just one from the denominator.
actually u need to do this:
72^2/15 = 72*72/(3*5)
=rem of(24*72/5) *3
=rem of(24/5) * rem of(72/5) *3
=(4*2)%5 * 3
=9
remember that u can cancel out only common terms..

[tatimatla]

An interesting prob:
Two villages lie on the opposite sides of a river, whose banks are parallel lines. A bridge is to be built over the river, perpendicular to the banks. Where should the bridge be built so that the path from one village to the other is as short as possible? (the villages need not lie on a perpendicular to the river)

I am looking for a generic answer...if someone can attach an image, that'd be great

[doomsayer]

Originally Posted by the.bombardier

that last step is not correct.once u multiply with 2^5,u dont need to take mod again.
the remainder will be 32.

@bombardier,
That's ur right.But how do u explain my above problem..(72^2/15)

[doomsayer]

Originally Posted by doomsayer

@bombardier,
That's ur right.But how do u explain my above problem..(72^2/15)

Got now...Thanks..

[the.bombardier]

Originally Posted by tatimatla

An interesting prob:
Two villages lie on the opposite sides of a river, whose banks are parallel lines. A bridge is to be built over the river, perpendicular to the banks. Where should the bridge be built so that the path from one village to the other is as short as possible? (the villages need not lie on a perpendicular to the river)

I am looking for a generic answer...if someone can attach an image, that'd be great

is this the correct answer? (c attachment)
A and B are the villages,ACDB is the shortest routes,n CD is the bridge..
i think u need to knw how far the villages r respectively from the banks..in this case i have assumed that A is farther away from the river than B.