Suresh's Corner: Miscellaneous Questions From Quants Part-II

[clsuresh1]

Dear friends, this is the continuation of my very popular old thread. You can visit that thread here.

http://www.pagalguy.com/forum/quanti...ns-quants.html (Suresh's Corner: Miscellaneous Questions From Quants)


Unfortunately, I have lost the controls of my login and hence I registered again with another username "clsuresh1" (my old one is "clsuresh"). This thread is a continuation of the old thread. Since the old thread is clustered with full of problems, I am starting this fresh where I will start with the topic "Number System" first and then will continue with P&C and Geometry. So I request all of you to support me by first discussing about the questions on the topic "Number System". I hope the moderator will allow me to continue this thread.

I want everyone to adhere to the following rules.

1. Please ensure that the questions already posted are
answered first before you post the new questions.

2. Please stick to the topic area (First Number System and once
I think we are done with it then I will announce the next
topic. So till then discuss problems only from the Number
System)

3. Please ensure that you are acquainted with the basics of the
topic that we discuss here because as I have mentioned in
my earlier thread we will discuss the problems of CAT
standard and advanced concepts needed to answer these
questions.


Wishing u all the best........

Regards,
Suresh

[clsuresh1]

Here is our first question......

Let X = 1111…….111 (where the number of 1’s is ‘m’)
Let Y = 1000…….005 (where the number of 0’s between 1 and 5 is ‘m-1’)

If m > 20, find the remainder when sqrt(XY + 1) is divided by 16.

[kamnashish]

What a welcome thread again

[aayush_lakshmi]

Dear Sir,

I wacked my brain over the sum but could not crack it ...can u please send in the solutions

[clsuresh1]

Dear Aayush, let's wait for some more time.......I want some more students to try this out.....O.K I will give u a small clue......

Try to write the numbers X and Y in the form of 10^k .......something like this should help u to solve this.........

[kamnashish]

Hmmm

the question can be written as

(10^(m+1) - 1) (10^(m+1) +5)/9 +1

so,
simplifying ( i am omitting some steps)

we get 100*10^2m+ 40*10^m +4 / 9

taking 10^m as y

we get the equation as 100y^2 + 40^y +4 /9

we need to find the sqrt of this

this is nothing but (10y+2/3)^2

so we need to find the remainder of (10^m+1 +2)/3 when it is divided by 16

correct me if I am going wrong newhre!!

[kamnashish]

and hence we get the remainder to be 6.

if m> 2 the remainder is always 6....

[shivam_01]

Here is our first question......

Let X = 1111…….111 (where the number of 1’s is ‘m’)
Let Y = 1000…….005 (where the number of 0’s between 1 and 5 is ‘m-1’)

If m > 20, find the remainder when sqrt(XY + 1) is divided by 16.

going on the foot steps of kamna we get

from this we get

(10^(m+1) - 1) (10^(m+1) +5)/9 +1

as (10^m+1 +2)/3

so we need to find the remainder when (10^m+1 +2)/3 is divided by 16.
as m>20 say take the value 21

we get (10^22 +2)/3 as we are concerned with the remainder when this term is divided by 16 so we need to take the last four digits only AND THAT WILL BE 333.... 4
hence the remainder would be 6 on division with 16

@ kamna how cum u are getting as 6 and also m>20 not m>2 please check and correct me if i ma wrong anywhere ........

[kamnashish]

Originally Posted by shivam_01

going on the foot steps of kamna we get

from this we get

(10^(m+1) - 1) (10^(m+1) +5)/9 +1

as (10^m+1 +2)/3

so we need to find the remainder when (10^m+1 +2)/3 is divided by 16.
as m>20 say take the value 21

we get (10^22 +2)/3 as we are concerned with the remainder when this term is divided by 16 so we need to take the last four digits only hence the answer would be 2

@ kamna how cum u are getting as 6 and also m>20 not m>2 please check and correct me if i ma wrong anywhere ........

Hie shivam!!

Ah.. okie. what u are doing is that u are taking the number to be1000...2.

but then theres the denominator of 3. this no will be divisible by 3 as the sum will always be three and the no after dividing will be of the form 33333...4 so if you take the last 4 digits i.e. 3334 so diving by 16 will give u 6 as the answer.
I hope I am able to convince you..

Lets c what suresh sir has to say on this..

[shivam_01]

Hello kamna

ya realised my mistake

Thanks a lot.
corrected my post

donot know whats happening these days with me.. need a break