[thakarsagar]

Originally Posted by clsuresh

Guys here comes our next question.

Find the remainder when 1212121212..........1212 (300 digits) is divided by 999

Regards,
Suresh.

Ans:
666 ( the no of the beast....any iron maiden fans out here??? :-)
regards,
-sagar

[thakarsagar]

Originally Posted by clsuresh

Yes that's right.

Guys now let me explain the second problem.

2. A pious rich man goes to a temple everyday and distributes certain amount of money to the poor people. However he follows the following two rules

Rule1: He always takes money in the denominations of 1 Rupee only.
Rule 2: He distributes the money equally among the poor people irrespective of the no. of people he distributes.

On a particular day if he has taken with him Rs 8640/ find the probability of money being distributed among 27 people.


Sol: First let's calculate the total no.of ways in which he can divide the money among different no. of people such that each receives an integral amount.

If P persons are given Y Rs each then P x Y =8640.
i.e the total no.of ways in which the money can be divided is nothing but the total no. of ways in which 8640 can be written as product of two numbers.

We know that the total no. of ways in which a no. can be written as product of two numbers is 1/2(total factors).

N= 8640 = 2^5 x 3^3 x 5.
Total factors = (5+1)(3+1)(1+1) = 28.

But we should take it as 56 and the simple reason is this.

Now 8640 = 864 x 10
as well as 8640 = 10 x 864.

Since in the normal case we do not take both of them we get 28 ways.
But here we need to take both of them because
8640 = 864 x 10 means 10 Rs each to 864 people whereas
8640 = 10 x 864 means 864 Rs each to 10 people which are two different ways in our problem.

So the total no. of ways in which 8640 Rs can be distributed is 56 out of which there will be only one way of distributing the money among 27 people
( i.e 27 people getting 320 Rs each)

So the probability is 1/56

Regards,
Suresh.

hey suresh,
A mistake in the red lines shown in the beginning....
8640 = 2^6 * 3^3 * 5^1 and not 2^5
Total no of factors = 56...
Now suppose i have 12 rupees instead of 8640...
factors = 6...they are 1 2 3 4 6 12...

Rupees: Person
1:12
2:6
3:4
4:3
6:2
12:1
so no need to multiply it twice as you said....
Only 1 way possible => probability = 1/(56)...
suresh, if i am wrong please correct me...
-sagar

[zahoor3]

Originally Posted by clsuresh

Govi, Thanx for correction, I have edited the post

Regards,
Suresh

hi suresh
gr8 work pal i am also confused about this mod used in the forum plz PM and also PM your mail ID if its okay

[clsuresh]

Sagar,

Thanx for correction and I edited the post. ( I think from this time I have to be careful because many silly mistakes are creeping in)

Let me explain u here with your example 12 itself.

12= 1x12
2x6
3x4
4x3
6x2
12x1
In the above case if I want the no. ways in which 12 can be expressed as product of two factors the answer is 3 and not 6 since 1x12 as well as 12x1 is counted as only one way.
But coming to our problem we have to consider them as different because 12 people 1Rs each and second case is 1 person receivng 12 Rs .

That's what I said when it comes 8640.


By the way can u give in brief the explanation for the previous one. (Normal approach or something different)


Byeeeee
Suresh

[thakarsagar]

Originally Posted by clsuresh

Sagar,

Thanx for correction and I edited the post. ( I think from this time I have to be careful because many silly mistakes are creeping in)

Let me explain u here with your example 12 itself.

12= 1x12
2x6
3x4
4x3
6x2
12x1
In the above case if I want the no. ways in which 12 can be expressed as product of two factors the answer is 3 and not 6 since 1x12 as well as 12x1 is counted as only one way.
But coming to our problem we have to consider them as different because 12 people 1Rs each and second case is 1 person receivng 12 Rs .

That's what I said when it comes 8640.


By the way can u give in brief the explanation for the previous one. (Normal approach or something different)


Byeeeee
Suresh

Soln:
we know that a^n-1 is divisible by a-1,
999 = 10^3 - 1
10^(3n) = [10^3n - 1] + 1
So,
(10^3n)%999 = 1
(10^(3n+1))%999 = 10
(10^(3n+2))%999 = 100

The given number can be written as
[10^299 + 10^297 +.... 10^1] +
2*[10^298 + 10^296 +.... 10^0]

Each of the terms in the square bracket has 50 3n's 50 3n+1's and 50 3n+2's
So the remainder is
100*50 + 10*50 + 50 + 2*[100*50 + 10*50 + 50] = 16650
16650%999 = 666
-sagar

[ganeshiyer]

Originally Posted by clsuresh

Sagar,


By the way can u give in brief the explanation for the previous one. (Normal approach or something different)


Byeeeee
Suresh

hey sagar and suresh

i have given my method below. but it takes a lot of time to find the quotient and the remainders.



1212.......1212 %999

999=111*9
hence we find 121212.....1212%111 =0
the quotient here is
109200109200.....1092......298 digits

now 109200......1092 % 9= 6[ 1092 comes 50 times in the no. hence sum=12*50=600]

i.e 1092...001092=9k+6
i.e 109200...1092 *111 = 9*111k + 6*111

12121........1212= 999k +666

hence ans=666

[ganeshiyer]

Originally Posted by clsuresh

Sagar,




By the way can u give in brief the explanation for the previous one. (Normal approach or something different)


Byeeeee
Suresh

hey sagar and suresh

i have given my method below. but it takes a lot of time to find the quotient and the remainders.



1212.......1212 %999

999=111*9
hence we find 121212.....1212%111 =0
the quotient here is
109200109200.....1092......298 digits

now 109200......1092 % 9= 6[ 1092 comes 50 times in the no. hence sum=12*50=600]

i.e 1092...001092=9k+6
i.e 109200...1092 *111 = 9*111k + 6*111

12121........1212= 999k +666

hence ans=666


p.s suresh can u p.m me the mod wala funda???????

[clsuresh]

Sagar, What u said is fine.

I have a different approach here.
But before that let's look at a small problem here.

If I want the remainder when 5 x 36 / 34

Normally we can calculate the rem of 36 when divided by 34 and we get it as 2
So the final remainder is 5x2 = 10

Now 5x36/ 34
Since 2 is common in 36 and 34 let me cacel that 2 in 36 and 34
Now my problem becomes 5x 18/17
I think there will be no change in the quotient but my remainder will be altered.

Now the remainer of 5x18/17 is ofcourse 5x1 =5
Since I cancelled 2 earlier in both 36 and 34 multiply this remainder with 2 i.e
the final remainder is going to be 5x2=10

Another example:

Rem of (23 x 75)/ 45
It is reduced to (23 x 5) /3 ( cancelling 15 in both 75 and 45)
So the remainder becomes 1.
But since we cancelled 15 earlier the final remainder will 1x15=15

Now let's look at our problem

1212121212121212.................1212121212( 300 digits) /999

all of us know that a number of the form ABCD can be written as
A x 10^3 + B x 10^2 + C x 10^1 + D x 10^0 ( here our base is 10)
Similarly if I have no. of the form ABABABAB instead writting it in the base 10 let's write it in thebase 100 since two digits are same i.e

ABABABAB = AB X 100^3 + AB X 100^2 + AB X 100^1 + AB X 100^0

S our no. 12121212.......121212 (300 digits) can be written as

12 x 100^149 + 12 x 100^148 + .................+ 12 x 100^1 + 12 x 100^0.
[simply treat it as 150 digited no. where each digit is 12 now i.e 12,12,12,12,12,.....12 (150 digits)]

= 12 x (100^149 + 100^148 +.............+ 100^1 + 100^0)
= 12 x ( 10^298 + 10^ 296 +..............+ 10^2 + 10^0
= 12 x (10^150 -1)/(10^2-1) ( above sum is in GP where the first term is 10^0 i.e 1 and the common ration is 10^2 and no. of terms is 150)

= 12 x ( 10^150-1)/99
= 12 x [9999..........9999 ( 150 digits)]/99
= 12 x [10101010...........1010101 (149 digits)] (please remember this generalisation,don't deduce iteverytime)
i.e ABABABAB.....................ABAB ( 2n digited number)
is equal to AB x [ 101010...........101(2n-1 digits)]

Now remainder of 12 x [101010.....10101(149 digits)]/999
Again express 101010-------10101 and proceeding further

we get remainder as 666

This may seem little bit hectic but i want all of u to get the concepts.

The other faster way for this is as we have discussed earlier split 999 as 111 and 9 Clearly the given no. is exactly divisible by 9 as well as 111.
Now use your options given in the exam.

Regards,
Suresh.

[clsuresh]

Ok Folks, that's it from my side for today.

I am pushing off to my class and for the time being I pen off.

I'll catch up with u tommorrow.

Have a goodnight.

Byeeeeee
Suresh.

[chinmay686]

this thread rocks!!!!!!