This is a new section I thought i would add on to this blog,

Often people find Number system as one the most confusing and tough section.

You can find very good questions on freembastuff but here are some examples which might be helpful.

**Number System : Remainder Theorem**

Three types of solutions for any Remainder problem:

All of us know that remainder run a loop system, i.e a cycle which exactly matches the divisibility cycle.

not going to the basics , consider for example:

53/2^4 which is basically 53/16= rem 3

now what if the problem is

2^1111/17 = 2^3 ( 2^4)^277/17 ( We considered 2^4 because 17 is closest to 2^4 i.e 16 and we get remainder -1 , Why? because as such we know 2^1111 is >>> 17 and the remainder would be +ve, -1 here means 16 but -1^2 =1 and 16^16 /17 = 1 rem both are same thing but -1 keeps things simple)

=>2^3 ( 16)x(16)…./17 = 2^3 (-1)^277/17 =-8/17 or the remainder is 9. [16/17 = -1 remainder or 16 remainder.]

Now above two might be simple one’s third one is the tougher one and based on upper two:

solve 53^111/51

53 is bigger than 51 and 53/51 gives remainder as 2 so 53 ^1111/51 = 2^1111/51

now consider that 51= 3×17

so lets divide 2^1111/51 into two parts

2^1111/3 and 2^1111/17

2^1111/3 = (-1)^1111/3=-1/3=> 2 remainder

Similarly

2^1111/17= 9 remainder

now the solution to above problem is a number which gives 2 mainder from 3 and 9 remainder from 17

or

3K+2=17m+9

try solving by putting values answer is 26.

Ask your doubts!

### Like this:

Like Loading...

*Related*

Posted by Sumedha. on July 29, 2009 at 3:52 am

A simpler method for calculating remainder for 53^111 / 51 would be:

1. 53/51 gives remainder 2

2. Now calculate remainder for 2^111 / 51

3. 2^8 / 51 gives remainder 1

4. so, express 2^111 as 2^8*13 * 2^7

5. Thus remainder is calualted for (2^8*13)*2^7 (Actual remainder becomes R1*R2)

6. R1=1 —->for [ (2^8*13)/51 ] and R2 = 26 —-> for [2^7 /51 ]

7. thus, final remainder = 26

8. This method seems tedious, but once you get a hang of this all sorts of divisiblity can be calulated.

Posted by romesh on September 16, 2009 at 3:04 am

first of all in the above first method it is shown that 53/2^4=3 (how).second thing is that 2^1111/17 gives rem=1then how in the 3rd method it is shown rem=9.

Posted by freembastuff on October 1, 2009 at 1:58 pm

@romesh

2^1111/17 is 2^3 ( 2^4) ^277 /17 => 2^3 /17 x (16)^277 /17

Thus remainder in these cases are

8 /17 x (-1)/17 => -8/17 which is nothing but 9

@Sumedha

and for multiples of 51 is tougher to calculate, though in this case it might have been simpler. The concept is that in large numbers where its tough to calculate multiples of denominator we can divide the denominator in to its miltiples and process as given in the case !