## what is the remeinder of

what is the remeinder of ((25)^625 + 26)/247...???? kan ne1 help me !!!

## Any1 got the answeror the

Any1 got the answer or the question is not correct. The reminder can definitely be calculated but it wld be lengthy and no trick seems to be here.. i tried but could not find any relation ship between 247 and 25 The only close thing came to mind is 247 = 16^2 - 9 and 25 = 16 + 9 but that wont take any where So I feel it shd be 22 or some thing else in place of 247

## Hints for Solving the given Problem

Question: What is the remainder when [{(25)^625} + 26] is divided by 247? Hints: The given problem has to be solved in a different manner. Note that 247 = 13 x 19 So first find out the remainder when (25)^625 is divided by 13 and then find out the remainder when (25)^625 is divided by 19 and then combine both the results to get the remainder when (25)^625 is divided by 247 and then add 26 to that result. If my calculations were correct, then in the remainder when (25)^625 when divided by 247 is 24 and hence the remainder when [{(25)^625} + 26] is divided by 247, the remainder is: 24 + 26 = 50. I am not sure whether my calculations were correct or not as I make a lot of calculation mistakes. So, Please do let me know if 50 is the right answer so that I can post the solution in details. Thank You Ravi Raja

## Hey Ravi its very confusing

Hey Ravi its very confusing . . .

So first find out the remainder when (25)^625 is divided by 13 and then find out the remainder when (25)^625 is divided by 19 and then combine both the results to get the remainder when (25)^625 is divided by 247 and then add 26 to that result.
Are you suggesting that the reminder of N/(a x b) = Reminder of N/a + Reminder of N/b ????

## Note

The remainder of N/(a x b) is NOT equal to remainder of N/a + remainder of N/b Example: Remainder of 25/3 = 1 Remainder of 25/4 = 1 So, remainder of 25/3 + remainder of 25/4 = 1 + 1 = 2 and Remainder of 25/12 = 1 So, it is clear that: Remainder of 25/12 is NOT equal to remainder of 25/3 + remainder of 25/4 That is, The remainder of N/(a x b) is NOT equal to remainder of N/a + remainder of N/b. Actually there is another theory involved in it which obviously is difficult for everyone to understand. Thats why i wanted to confirm my answer. Anyway, I am looking for an alternate method to solve this problem. If I find it out then I will definitely post it and if possible, I will post this theory too. Thank You. Ravi Raja

## Any one solved it ?????

Any one solved it ?????

## well the calculator shows

well the calculator shows that the answer is 220. so keep trying

## ever heard of the Chinese

ever heard of the Chinese remainder theorem ! give it read.

the whole process is too large to explain but i will sure give the solution.

247 = 13 * 19 .clearly 13 and 19 are coprime.

let N = 25^625 + 26 it can be proved easily that :

(i)N = 12 mod(13)................

(ii) N = 11 mod(19)................

now find a number u such that 19u = 1 mod(13) => u = 11 also,find a number v such that 13v = 1 mod(19) => v = 3

now, let r = 12*(19*11) + 11*(13*3) = 2937

NOW, it can be proved that the value of N that satisfies (i) and (ii) will also satisy N = r mod(13*19) now N = 220 mod(247)

HENCE THE REMAINDER IS 220. tada ! If anyone is at all interested in number theory then there is an free book on GOOGLE BOOKS called "The elements of the theory of algebraic numbers". The general process is given on page 72.

## cn i suggest sumthn.....

Solve it by Eulers method......it vil simplify this problm.........

## Solve it by

Solve it by Eulers method......

## I Cudnt understand ur

I Cudnt understand ur method...what is dat "mod" all about???..kan sum1 plz explain it more explicitly !!!

## I am not an expert but if I

I am not an expert but if I get time then I ‘ll write one article on mod with necessary tips and tricks required for solving CAT questions. You can google the word mod to get many stuffs on this.

x = y (mod m)

Means two things (both are same also)

( I ) Both x and y gives the same reminder when divided by m

Ex: 17 = 52  (mod 7)

Both 17 and 52 gives remainder 3 when divided by 7.

(II) Another way of saying the above

y can be written as kx + m where k is a constant

There are few Theorems which are based on above concept and Chinese Remainder Theorem is one of them

#### Chinese Remainder Theorem

This theorem provides a way to combine two equations that use different mod and help in solving many reminder questions

if          x = y (mod m)

x = y (mod n)

and  if m & n are coprime then

x = y (mod mn)

## <cite>.ever heard of the

<blockquote>ever heard of the Chinese

ever heard of the Chinese remainder theorem ! give it read.

the whole process is too large to explain but i will sure give the solution.

247 = 13 * 19 .clearly 13 and 19 are coprime.

let N = 25^625 + 26 it can be proved easily that :

(i)N = 12 mod(13)................

(ii) N = 11 mod(19)................

now find a number u such that 19u = 1 mod(13) => u = 11 also,find a number v such that 13v = 1 mod(19) => v = 3

now, let r = 12*(19*11) + 11*(13*3) = 2937

NOW, it can be proved that the value of N that satisfies (i) and (ii) will also satisy N = r mod(13*19) now N = 220 mod(247)

HENCE THE REMAINDER IS 220. tada ! If anyone is at all interested in number theory then there is an free book on GOOGLE BOOKS called "The elements of the theory of algebraic numbers". The general process is given on page 72. </blockquote>

well i think you application of CRM here is not proper as we are looking for reminders and not the number....

what you say RAJORSHI...

sol: 247=13*19

using euler function for each of them and finding reminder for 13 and 19 separatley..

i.e. (25^625+26)mod 13=12  and (25^625+26) mod 19=11

multiplying both 132 = (25^625+26)mod 247

I know its coming.....and I know i can handle it the best....