can someone start a thread for approach concept... i mean how to approach a problem and what should we think when we approach a particular number system problem...because thatts the most important thing i suppose...its an urgent necessity as it will make our roots go strong and we can tackle any problem?
plzzz help me......my QA sux........hav always been terrified by QA......... number system particularly................plzz tell me which topics to start with in number systems... im afraid QA will play spoil sport for me in CAT 2011.
hi....... i m getting stuck in this remainder problem..pls show me where i m wrong...
50^51^52 is divided by 11 :what'll be the remainder?
my approach: (50^51)^50*(50^51)^2/11 ->(50^51)^2/11 (using FERMAT'S) ->(50^102)/11 ->50^2/11 ->6^2/11 ->remainder 3............m i wrong?answer given is 6!!!!!!
I think you took 50^51^52 as (50^51)^52 but it is actually (50)^51^52 I mean 2^3^4=2^81 and not 8^4 that is the problem with the solution. Here,Euler will come handy . E(11)=10 51^52=(50+1)^52 so it is in the form 10n+1 Hence 50^(10n+1) mod 11 =50 mod 11=6 (answer) Check the quoted post above of your post for Euler theorem !
hi....... i m getting stuck in this remainder problem..pls show me where i m wrong...
50^51^52 is divided by 11 :what'll be the remainder?
my approach: (50^51)^50*(50^51)^2/11 ->(50^51)^2/11 (using FERMAT'S) ->(50^102)/11 ->50^2/11 ->6^2/11 ->remainder 3............m i wrong?answer given is 6!!!!!!
when u divide 50 by 11 u can write remainder -5, because it is 5 less than next multiple of 11 similarly for 51 = -4 , for 52 it is = -3 now 50*51*52= -5*-4*-3= -60 on dividing by eleven remainder = -5, that means 11-5= 6 :cheerio:
when u divide 50 by 11 u can write remainder -5, because it is 5 less than next multiple of 11 similarly for 51 = -4 , for 52 it is = -3 now 50*51*52= -5*-4*-3= -60 on dividing by eleven remainder = -5, that means 11-5= 6 :cheerio:
=> 5555......93times = 555...9 times (mod 98 ) :drinking::drinking: i dint get it .. euler applies to the exponent of a no. in the form b^n .. but can we use this to no. of digits ??
=> 5555......93times = 555...9 times (mod 98 ) :drinking::drinking: i dint get it .. euler applies to the exponent of a no. in the form b^n .. but can we use this to no. of digits ??
Actually its an extension of the Euler's theorem.
Suppose we have number N having digit 'a' written 'k' times, where k is Euler's number of 'n'. So, when N is divided by n, reminder will be 1. (Note, n should be coprime to 3 and 10)
Proof:- N = aaa...a (k times) = (a/9)(10^k - 1)
Now, since 10 and n are coprime, we can say that (10^k - 1) is divisible by n. Now since n and 9 are also coprime, we can say that
(a/9)(10^k - 1) will be divisible by n
So, actually I made a mistake because 98 and 10 are not coprime. So, we need to break 98 inot 2 and 49. So, clearly using the previous method, we can say 5555...5 (93 times) = 555...9 times (mod 49) = 5*10^8 + 55(10^6 + 10^4 + 10^2 + 1) (mod 49) = 5*16 + 55*(8 + 4 + 2 + 1) (mod 49) = 905 (mod 49) = 23 (mod 49)
so p=1!+2.2! p=5 add 2 thus p+2=7 now divide p by n+1 ie 7/3 remainder is 1
to verify we can take n=3 and divide p+2/n+1 remainder will again be 1
thus we can say that Remainder will be one in any case
Add and subtract 10! 1*1! + 2*2!..................11*10!-10! now 10!= 10*9! 1*1! + 2*2!..................9*9!-10*9!+ 11*10! 1*1! + 2*2!.................8*8!-9*8!+ 11*10! continue this process ultimately you will reach P =-1 + 11*10! P+2=1+11*10! therefore remainder is 1
My first post in this thread for the subscription.
You don't have to post for subscription.
You just have to scroll up, below the page numbers for the thread, you can find Thread Tools and you can find "Subscribe to this thread" if you click on "Thread Tools"
plzz!!!!!