@jain4444 said:0 , 1 , 5 , 4
0?
@ScareCrow28 said:Could it be done like this? :R [ 5^5^5^5^5 ] / 10^5 = R [ 5^5k/10]^5 = 03125Hence, 5th digit is = 0??
Am getting the same answer. Longer method. Too lazy to type :)
regards
scrabbler
regards
scrabbler
@scrabbler said:Am getting the same answer. Longer method. Too lazy to type regardsscrabbler
Fir to sahi hai answer 
@ScareCrow28 said:Could it be done like this? :R [ 5^5^5^5^5 ] / 10^5 = R [ 5^5k/10]^5 = 03125Hence, 5th digit is = 0??
Not convinced of your logic though. You are saying that R(x^5/y^5) = [ R(x/y) ] ^5 I think?
Consider remainder of 81 with 49 = 32. This is not equal to [R(9/7)] ^2 = 4.
regards
scrabbler
Consider remainder of 81 with 49 = 32. This is not equal to [R(9/7)] ^2 = 4.
regards
scrabbler
@ScareCrow28 said:Could it be done like this? :R [ 5^5^5^5^5 ] / 10^5 = R [ 5^5k/10]^5 = 03125Hence, 5th digit is = 0??
yeah 0 it is 
5th digit means we need to find remainder by 100000
1,00,000 = 32*3125
remainder by 32 :-
E(32) = 16
E(16) = 8
E(8 ) = 4
E(16) = 8
E(8 ) = 4
5^5 mod 4 = 1
5^1 mod 8 = 5
5^5 mod 16 = 5
5^5 mod 32 = 21
5^5 mod 16 = 5
5^5 mod 32 = 21
remainder by 3125 = 0
3125k = 32n + 21
=> number = 3125 or 03125
3125k = 32n + 21
=> number = 3125 or 03125
so , 5th digit from right = 0
@jain4444 said:0 , 1 , 5 , 4
Then 0.
Effectively we want remainder with 100000 = 3125 * 32. Numerator is anyway divisible by 3125. So we need to find the remainder of 5^5^5^5^5 with 32, which with some contortions I found to be 21.
So basically the number will be of the form 3125k and also of the form 32m+21, and fortunately 3125 itself satisfies this. So the remainder with 100000 (i.e. the last 5 digits) are 03125.
regards
scrabbler
Effectively we want remainder with 100000 = 3125 * 32. Numerator is anyway divisible by 3125. So we need to find the remainder of 5^5^5^5^5 with 32, which with some contortions I found to be 21.
So basically the number will be of the form 3125k and also of the form 32m+21, and fortunately 3125 itself satisfies this. So the remainder with 100000 (i.e. the last 5 digits) are 03125.
regards
scrabbler
@scrabbler said:Then 0.Effectively we want remainder with 100000 = 3125 * 32. Numerator is anyway divisible by 3125. So we need to find the remainder of 5^5^5^5^5 with 32, which with some contortions I found to be 21.So basically the number will be of the form 3125k and also of the form 32m+21, and fortunately 3125 itself satisfies this. So the remainder with 100000 (i.e. the last 5 digits) are 03125.regardsscrabbler
bhai thoda gyaan do :
Suppose if the denominator can be factorized , as in this case 3125*32
Case 1:
both the factors divide the numerator i.e. remainder 0
what if only one of the factors divide the numerator completely. In that case do you just check the remainder on the basis of the other factor. which in this case was 32
other case ;
what if both the factors do not divide the numerator completely.??
If s = {11,111,1111,........... 1111(20 times) } .How many elements of S have odd number of factors ? S is a set.
@saurav205 said:bhai thoda gyaan do :Suppose if the denominator can be factorized , as in this case 3125*32Case 1:both the factors divide the numerator i.e. remainder 0what if only one of the factors divide the numerator completely. In that case do you just check the remainder on the basis of the other factor. which in this case was 32other case ;what if both the factors do not divide the numerator completely.??
Chinese remainder theorem jaante ho? Nai jaante to dekhna ek baar..
Suppose a number N is leaving a remainder r1 with x and r2 with y
Then, N = k*x + r1 = l*y + r2
If both give zero remainder , then N = k*LCM(x,y)
Otherwise we have to calculate accordingly...
@ScareCrow28 said:Chinese remainder theorem jaante ho? Nai jaante to dekhna ek baar..Suppose a number N is leaving a remainder r1 with x and r2 with yThen, N = k*x + r1 = l*y + r2If both give zero remainder , then N = k*LCM(x,y)Otherwise we have to calculate accordingly...
hmm....i know the chinese remainder theorem.
will have to revise it again I guess..
@nole said:If s = {11,111,1111,........... 1111(20 times) } .How many elements of S have odd number of factors ? S is a set.
Only perfect squares have odd number of factors. Set me aisa to koi number lag ni raha jo perfect square ho? 
Ans 0 hai kya?
Ans 0 hai kya?
@ScareCrow28 I also thought the same.will tell u answer in a short while.my frnd texted me,need to get OA from him.
@nole said:@ScareCrow28 I also thought the same.will tell u answer in a short while.my frnd texted me,need to get OA from him.
Matlab it's not 0 kya?
@saurav205 said:bhai thoda gyaan do :Suppose if the denominator can be factorized , as in this case 3125*32
Make sure the factors are co-prime. Important point. Say they are M and N.
@saurav205 said:Case 1:both the factors divide the numerator i.e. remainder 0
Perfect hai.
@saurav205 said:what if only one of the factors divide the numerator completely. In that case do you just
Suppose it divides M completely, then the remainder will also be a multiple of M. And if it leaves a remainder p with N then I look at numbers of the form Nk+p (putting k = 0, 1, 2...) till I find a multiple of M (which is what I did above for the 33 waala)
@saurav205 said:check the remainder on the basis of the other factor. which in this case was 32other case ;what if both the factors do not divide the numerator completely.??
Then find the individual remainders - suppose the numerators are p and q so that we can say they numerator is Mk + p and Nl+q (where k and l are variables). Then:
Note that numbers of the form Mk+p occur every M steps, and Nl+q every N steps. So a number satisfying both will occur every LCM (M, N) steps. And so if we have taken M, N as co-prime, then this will be once every M*N steps. So all we need to do is list out numbers of 1 type (say Mk+p) and see which one also satisfies Nl+q.
For example, is we want to find the remainder of 8^653 by 35, we take 35 = 7 * 5.
Now 8^653 = 7k +1 and also 5l+3 and will have exactly 1 solution from 0 to 34. So I check 7k+1 numbers => 1, 8, 15, 22, 29 and see that 8 satisfies 5l+3 also. And hence answer will be 8.
Hope it helps...
regards
scrabbler
@scrabbler said:
Make sure the factors are co-prime. Important point. Say they are M and N.Perfect hai.Suppose it divides M completely, then the remainder will also be a multiple of M. And if it leaves a remainder p with N then I look at numbers of the form Nk+p (putting k = 0, 1, 2...) till I find a multiple of M (which is what I did above for the 33 waala)Then find the individual remainders - suppose the numerators are p and q so that we can say they numerator is Mk + p and Nl+q (where k and l are variables). Then:Note that numbers of the form Mk+p occur every M steps, and Nl+q every N steps. So a number satisfying both will occur every LCM (M, N) steps. And so if we have taken M, N as co-prime, then this will be once every M*N steps. So all we need to do is list out numbers of 1 type (say Mk+p) and see which one also satisfies Nl+q.For example, is we want to find the remainder of 8^653 by 35, we take 35 = 7 * 5.Now 8^653 = 7k +1 and also 5l+3 and will have exactly 1 solution from 0 to 34. So I check 7k+1 numbers => 1, 8, 15, 22, 29 and see that 8 satisfies 5l+3 also. And hence answer will be 8.Hope it helps...regardsscrabbler
thanks mate....thoda revision was needed of chinese remainder theorem and you saved me a lot of time....
Thanks a lot..
Problem on chinese remainder theorem:
Rem 128^500 when divided by 153
@amresh_maverick said:Problem on chinese remainder theorem:Rem 128^500 when divided by 153
153 equal to 17*9
17x+16=9y+4
gives me 67
P.S
AGAIN COMMITED MISTAKE
EDITED MY ANS