@[316144:divishth]
(36^41)/7*11
(36^41) mod 7 = 1
(36^41) mod 11 = 3
so remainder
7x+1 = 11y+3
x=(11y+2)/7
put y = 3
x = 5
so (36^41) mod 77 = 36
(41^36) mod 7*11
(41^36) mod 7 = 1
(41^36) mod 11 = 3
so from here also remainder = 36
now 2*36 mod 77 = 72..correct me if i am wrong
how u come up to this step??
so (36^41) mod 77 = 36
plz explain
can u pls tell me dis method...m nt aware of dis method....wht is dis mod nd ol? i thnk m missing sumthng here bigtym...pls elaborate frm basics....
@allan89 said: E 49 =42now, 6^84 mod 49 = (6^42)^2 mod 49 =1=> 6^83 * 6 mod 49 =1let remainder of 6^83 mod 49 be k=> k* 6 mod 49 =1=> k=41; (hit and trial)again, similarly8^83 md 49 =p=>p* 8 mod 49 =1=>p=43total remainder = (41+43 ) mod 49 =84 mod 49 =35
@Enceladus said: It should be 5.E(7) = 6. 10 mod 6 = 4. 100 mod 6 = 4. 1000 mod 6 = 4. and so on.... So this expression reduced to 10*3^4. => 10*3^4 mod 7 = 5.
expain it pls....
PS: nt aware of fermat or euler's theo...
PS: nt aware of fermat or euler's theo...
remainder when 1234567891011....979899 divided by 1001 and by 101???
@[510729:erm]: even i got 36 as the answer.
@bemad said:expain it pls.... PS: nt aware of fermat or euler's theo...
http://www.pagalguy.com/news/remainders-reloaded-euler-fermat-wilsons-theorems-cat-2011-a-19067
Funda is if m & n are coprime to one other then [n^(E(m))]/ m we get remainder 1(so we use mod. )
Where E(m) is Euler no. of m given as follows :
on prime factorization of m=a^2 * b^5 say
the E(m) = m(1 - 1/a)*(1 - 1/b) .
A very good eg. is given..check the link....
Funda is if m & n are coprime to one other then [n^(E(m))]/ m we get remainder 1(so we use mod. )
Where E(m) is Euler no. of m given as follows :
on prime factorization of m=a^2 * b^5 say
the E(m) = m(1 - 1/a)*(1 - 1/b) .
A very good eg. is given..check the link....
@[597684:Shriram1989]
hey bt explain me how u got this
guys I have a few problems, please explain-
1. find the sum of the series- 1/(3*7)+ 1/(7*11)+ 1/(11*15)+........ answer is 1/3..
2. if x, y, z are any real numbers then find the minimum possible value of x^2+ 2y^2+ z^2+ 2yz subject to x+2y+z =-6.. answer is 12
3. 16
A= [{2n^3 + 2n^2 +n^1/2(n+3)-3}/ {4n*n^1/2(n^1/2+3)+9}] then n necessarily satisfies which of the following? ans is 0.508
4. find the sum of all real values of 'x' that satisfy the equation-
(243^logx base 81)- 2x= (2^logx+2 base 16)-8... answer is 20. please kindly help me with these sums.. I want to solve them in the easiest way possible. TIA.. :).. happy learning..!!
problem1:
if N=192837465564738291
then find the remainder when (N+1)(N+2)(N+3)(N+4)(N+5)(N+6)(N+7) is divided by 11
@bemad said: problem1: if N=192837465564738291 then find the remainder when (N+1)(N+2)(N+3)(N+4)(N+5)(N+6)(N+7) is divided by 11
N is already divisible by 11, so it can be written as 11k
Now, the product will contain 11 is all the terms except the constant term that is 1.2.3.4.5.6.7
So, 1.2.3.4.5.6.7 mod 11 = 2
so that product mod 11 = 2
..bt if v do it dis way....
N= divisible by 11, if v ryt n+1 as a number then the difference btwn odd and even digit is -1 and so on...
so the remainder shud be -1*-2*-3*-4*-5*-6*-7
nd dis gives d ans as 9...pls tell whts wrong here...
@YouMadFellow said:N is already divisible by 11, so it can be written as 11kNow, the product will contain 11 is all the terms except the constant term that is 1.2.3.4.5.6.7So, 1.2.3.4.5.6.7 mod 11 = 2so that product mod 11 = 2
@bemad said: ..bt if v do it dis way.... N= divisible by 11, if v ryt n+1 as a number then the difference btwn odd and even digit is -1 and so on... so the remainder shud be -1*-2*-3*-4*-5*-6*-7nd dis gives d ans as 9...pls tell whts wrong here...
I think you should take the positive difference. i.e. modulus of the difference.
take fr example 120...if u divide it by 11 nd tk modulus of differenc dn it wud gv 1 as remainder...which is ofcorse nt true....
@YouMadFellow said:
I think you should take the positive difference. i.e. modulus of the difference.
@bemad said: take fr example 120...if u divide it by 11 nd tk modulus of differenc dn it wud gv 1 as remainder...which is ofcorse nt true....@YouMadFellow said:
I didn't say that the remainder is the modulus of the difference. If the difference is -1, then take 11-1 as the remainder. But what you are doing is that you are multiplying -1 and -2 and -3.. etc.. which is wrong.. because it means that N+1 gives -1 as the remainder... N+2 gives -2 as the remainder etc... when in reality they give 1 and 2 ..
I meant to say that you need to stick to one definition of difference, say you will always take it as (odd - even) and decide on the basis of that..
In case of 120, odd sum = 1, even sum = 2 , so difference as per above definition is -1, and hence remainder is 11-1 = 10.
In the case of the question, N+1 has odd sum = 46 , even sum = 45 the difference as per above definition comes out to be 1, hence the remainder is 1 and not -1 .
@YouMadFellow said:I didn't say that the remainder is the modulus of the difference. If the difference is -1, then take 11-1 as the remainder. But what you are doing is that you are multiplying -1 and -2 and -3.. etc.. which is wrong.. because it means that N+1 gives -1 as the remainder... N+2 gives -2 as the remainder etc... when in reality they give 1 and 2 ..I meant to say that you need to stick to one definition of difference, say you will always take it as (odd - even) and decide on the basis of that..In case of 120, odd sum = 1, even sum = 2 , so difference as per above definition is -1, and hence remainder is 11-1 = 10.In the case of the question, N+1 has odd sum = 46 , even sum = 45 the difference as per above definition comes out to be 1, hence the remainder is 1 and not -1
k...bt v do tk -1 as a remainder and multiply...at least wen v do binomial expansion...
nd N+1 wud be 192837465564738292 so odd-even is -1 nt 1...so pls clarify....
nd N+1 wud be 192837465564738292 so odd-even is -1 nt 1...so pls clarify....
@bemad said:k...bt v do tk -1 as a remainder and multiply...at least wen v do binomial expansion...nd N+1 wud be 192837465564738292 so odd-even is -1 nt 1...so pls clarify....
In case of N+1, I take 1st digit as the rightmost digit so the odd sum for me is
-> 2 + 2+ 3 + 4+ 5+ 6 +7 +8 + 9 = 46
Even sum = 9+8+7+6+5+4+3+2+1 = 45
so, for my convention, it comes out to be 1
@YouMadFellow said:In case of N+1, I take 1st digit as the rightmost digit so the odd sum for me is-> 2 + 2+ 3 + 4+ 5+ 6 +7 +8 + 9 = 46Even sum = 9+8+7+6+5+4+3+2+1 = 45so, for my convention, it comes out to be 1
hmm...k...got my mistake...was counting frm left...thanks fr d clarification...
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@koyal1990 said: guys I have a few problems, please explain-1. find the sum of the series- 1/(3*7)+ 1/(7*11)+ 1/(11*15)+........ answer is 1/3..2. if x, y, z are any real numbers then find the minimum possible value of x^2+ 2y^2+ z^2+ 2yz subject to x+2y+z =-6.. answer is 123. 16A= [{2n^3 + 2n^2 +n^1/2(n+3)-3}/ {4n*n^1/2(n^1/2+3)+9}] then n necessarily satisfies which of the following? ans is 0.5084. find the sum of all real values of 'x' that satisfy the equation-(243^logx base 81)- 2x= (2^logx+2 base 16)-8... answer is 20. please kindly help me with these sums.. I want to solve them in the easiest way possible. TIA.. .. happy learning..!!
please if any one would help me... please... can't do any of the above sums.. :(
Q) last two digits (201 x 202 x 203 x 204 x 246 x 247 x 248 x 249)^2
pl provide systematic solution that is by dividing 100 :)