[Pragati Soni]

Originally Posted by thakarsagar

1) 337...
Explanation:
For odd numbers the no gives a remainder of 337....
for even numbers the no gives a remainder of 297....
101 is odd.
-sagar

I think your approach seems to be very logical but I am not aware o it. I do face problem in such questions.

regards,
Pragati.

[thakarsagar]

Originally Posted by ankursanghi

Could any funda of Harmonic Geometric and Arithmetic mean be applied here? I am not able to recall the funda here. Someone please help me recall.. and we could do this question together.

hey ankur,
following your approach....
let M = 1001 + ... + 3001

AM > HM
AM of M = 2001
HM = 2001/S....

=> S > 1
-sagar

[Govi]

Another remainder problem (Pardon me Sagar i know there has been a deluge of remainder problems here )

What is the remainder when all two digit numbers from 19 to 80 inclusive are written down one after the other as a single number N = 192021...7980. and this N is divided by 1980

Hope people ans it with atleast few lines of explanation, although it may not be needed for this prob.

cheerio

SK

[ganeshiyer]

Originally Posted by clsuresh

Friends ,

What Sagar said is absolutely right. I have something different from that. Just have a look at that. But befor going into this I think all of u know the following basic funda.

Let R be the remainder when N is divided by D1.
Now the remainder when N is divided by D2, where D2 is a factor of D1 is either
1. R ( if R< D2) or
2. R1 where R1 is the ramainder obtained when R is divided by D2 ( if R>D2)
Remember the converse is also true.

Let's look at an example here. ( let me give the converse example)

Let 23 be the remainder obtained when a number is N divded by 30.
Now the remainder obtained when the number is divided 90 could be either
23 or 30x1+23 or 30x2+3 i.e
23 or 53 or 63. ( I think this is clear to everyone)


Regards,
Suresh. :smile:

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

[Stefan R]

Originally Posted by Govi

Another remainder problem (Pardon me Sagar i know there has been a deluge of remainder problems here )

What is the remainder when all two digit numbers from 19 to 80 inclusive are written down one after the other as a single number N = 192021...7980. and this N is divided by 1980

Hope people ans it with atleast few lines of explanation, although it may not be needed for this prob.

cheerio

SK

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.

[hitman]

Originally Posted by Govi

What is the remainder when all two digit numbers from 19 to 80 inclusive are written down one after the other as a single number N = 192021...7980. and this N is divided by 1980

Hope people ans it with atleast few lines of explanation, although it may not be needed for this prob.

cheerio

SK

hi,
the answer is 0.
method: (not sure whether its the shortest one)
1980= 4*5*9*11
4,5,9 n 11 are co-prime. now it is clear that N is div by 4 and 5.
after little calculation we get N is also div by 9 and 11. Hope i'm correct
So if N is div by 4,5,9 n 11then it must be div by their product(1980).Rule of co-prime divisibility.


if there is any shorter approach plz metion.

ciao

[ganeshiyer]

Originally Posted by thakarsagar

1) 337...
Explanation:
For odd numbers the no gives a remainder of 337....
for even numbers the no gives a remainder of 297....
101 is odd.
-sagar

hey sagar ,
heres my solution to the q

7777777........7777......101 times
= 777777......7700+77/440
=7777777......7700 % 440+ 77%440
now
777777........7700/440
=77777........770/44......100 times
consider only 7777........777 .......99 times
777.......77% 44 gives 29 for odd values
hence 777.....770%440 gives 290
hence
777.......770+77 % 440
=290+77= 367

please point out where is the mistake????????:huh:

[Wanderlust]

Originally Posted by ganeshiyer

hey sagar ,
heres my solution to the q

7777777........7777......101 times
= 777777......7700+77/440 HERE'S THE ERROR !!!!
=7777777......7700 % 440+ 77%440
now
777777........7700/440
=77777........770/44......100 times
consider only 7777........777 .......99 times
777.......77% 44 gives 29 for odd values
hence 777.....770%440 gives 290
hence
777.......770+77 % 440
=290+77= 367

please point out where is the mistake????????:huh:

hI ganesh

77 is divisible by 11 ,not 777
on the same lines,even no of 7's in a sequence would be divisible by 11 ...
i.e either 777..7770+7 or 777...7000 <100 7's or 98 7's >
but the former wouldn't be divible by 4 ,hence not even by 440
so now we have ..777..7000+777 % 440
=777..7000%440 + 777%440
=0+337....Ans!

[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

[ganeshiyer]

Originally Posted by Wanderlust

hI ganesh

77 is divisible by 11 ,not 777
on the same lines,even no of 7's in a sequence would be divisible by 11 ...
i.e either 777..7770+7 or 777...7000 <100 7's or 98 7's >
but the former wouldn't be divible by 4 ,hence not even by 440
so now we have ..777..7000+777 % 440
=777..7000%440 + 777%440
=0+337....Ans!

thanx ........

got my mistake