New Batches at TathaGat Delhi & Noida!               Directions to CP centre
numbers...help please
by vivek krishnan - Sunday, 23 September 2012, 12:38 AM
  What is the remainder when 1, 23, 54, 68, 76, 98, 76, 54, 234 is divided by 1001?
(1) 345  (2) 114   (3) 234   (4) 102

Re: numbers...help please
by Mohit Sharma - Sunday, 23 September 2012, 01:17 AM
  a small help... 1001 can be written as 11*13*7
then use chinese remainder theorem
Re: numbers...help please
by vivek krishnan - Monday, 24 September 2012, 02:11 AM
  first of all thank you very much Mohit...
this might be very silly but please bare with me and help me...

i understand,
\begin{align} x &\equiv a_1 \pmod{n_1} \\ x &\equiv a_2 \pmod{n_2} \\ &{}\ \ \vdots \\ x &\equiv a_k \pmod{n_k}
\end{align}, where  a1,a2, …, ak sequence of integers
n1, n2, …, nk are positive integers which are co prime, and x is a common factor (kind of)

but please help me understand how do i fit this concept to the problem stated...(sorry, this is the first time iam dealing with chinese remainder theorem... please help me)




Re: numbers...help please
by Tapesh Kumar - Saturday, 19 January 2013, 04:40 PM
 

I think the answer can be 114. below is the logic I thought of..

1001 - 1000^3+1

by the logic of divisibility, we need to separate the numbers in to group of 3 & alternatively add & subtract.

it will become - 123-546+876-987+654-234 = 114

 

Guys, pls let me know if this is incorrect.