[ashu_vats]

actually yaar it should be

30*1 + 23 = 53 & 30*2 + 23 = 83 respectivey ( not 30 *2+3)
got it?

ashu

Originally Posted by ganeshiyer

hi suresh
i did not get the part in bold:huh:
can u xplain how did u get 30*1+23 and 30*2+3


ganesh

[ashu_vats]

bhai log pls. explain in detail..

Mod wala funda bata do ??? i hv seen it used here at PG but not able to understand..help me
not joking though...
ashu

Originally Posted by Stefan R

1980 = 99*20 = 9*11*20

N = 0 mod (20)
N = 0 mod (9) sum from 19 to 80 is div. by 9
N = 0 mod (11), club (19,80), (20,79)

Hence the remainder by 1980 is 0.

[clsuresh]

Hi Ashu,

This is Suresh who stared the thread. Thanx for correcting me. Dear Ashu please let's have continuation here. I have been posting queries and they are being followed up by the solutions. Once a question is posted please do not post other questions untill that questions are answered completely. I hope you will understand this.

I HAVE SENT THE EXPLANATION FOR MOD TO UR PERSONAL MAIL

Thanking you,
Suresh.

[thakarsagar]

Originally Posted by clsuresh

Hi Folks,

Good morining, and here comes our next set of questions.

1. If x = (0.12345678910111213............) and y = 1/x find x + y

Please give answer in the form of p/q

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.

Please give answer in the form of p/q


Regards,
Suresh

1) x = 0.12345....
10*x = 1.2345...
9x = 1.11111...
x = 10/81...
x + y = x+ 1/x
= 6661/810
-sagar

[thakarsagar]

Originally Posted by clsuresh

Hi Folks,

Good morining, and here comes our next set of questions.

1. If x = (0.12345678910111213............) and y = 1/x find x + y

Please give answer in the form of p/q

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.

Please give answer in the form of p/q


Regards,
Suresh

2) 8640 = 2^6 * 5^1 * 3^3
Probability = 1/(6+1)(1+1)(3+1)
= 1/56....
suresh, please confirm...
-sagar

[clsuresh]

Yes, Sagar is right.

Let me make it more clear.

X = 0.1234567........... (a)
10X = 1.2345678............ (b)

Now (b) - (a) is 9X = 1.111111111.......... (c)
90X = 11.111111111.......... (d)

Now (d) - (c) is 81X = 10.00000000

So X is 10/81 and Y= 81/10

X + Y =6661/810

Guys if u have any other approaches (using PROGRESSIONS)
please post them
Byeeeeee
Suresh.

[talkysandeep]

first of all 9x aint 1.11111.....

1.23456789101112131415.....
-.12345678910101213141....
1.11111110191010918274

so the soln is wrong, though idea was a bit good.....

[thakarsagar]

Originally Posted by talkysandeep

first of all 9x aint 1.11111.....

1.23456789101112131415.....
-.12345678910101213141....
1.11111110191010918274

so the soln is wrong, though idea was a bit good.....

can anybody tell me how they have solved the problem 17^18+13^18%25...
Please give the elaborate soln ...

yes sandeep, you are right but in such problems you got to approximate...22/7 the value of pi is not exact...it's too an approximation...

As far as the other problem is concerned, i am exhausted...with every pinch of enthusiasm been snapped out....
-sagar

ps: sandeep, that was good thinking...

[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 = 2^6 x 3^3 x 5.
Total factors = (6+1)(3+1)(1+1) = 56.
So total no. ways of writing N as product of two factors is 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.

[clsuresh]

Guys here comes our next question.

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

Regards,
Suresh.