what is the remeinder of
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
n/a
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
????
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
n/a
well the calculator shows that the answer is 220. so keep trying
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.
Solve it by Eulers method......it vil simplify this problm.........
Solve it by Eulers method......
Could you please explain it in details
I Cudnt understand ur method...what is dat "mod" all about???..kan sum1 plz explain it more explicitly !!!
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