@x2maverickc said: @YouMadFellow @deedeedudu 50^10=1(mod11)=>50^50=1(mod11)=>50^51=6(mod11)=>50^51^51=6^51(mod11) 6^50=1(mod11)6^2=3(mod11)6^52=3(mod11) The remainder should be 3. How come the answer is 6, which is correct.Please clear karein. and isn't 50^51^52 equal to 50^2652 www.wolframalpha.com/input/?i=... www.wolframalpha.com/input/?i=... why is there a difference between the two results Dimaag slow ho gaya hai!! Sona chahiye ab
50^51^52 is not the same as 50^2652...
3^3^4 is not the same as 3 ^ 12 !! I guess this much is enough to know why the answers are different. :)
@[37213:jain_ashu] as u mntnd it dat u hv cmpltd r pursuing ur mba so i need to ask 1 qstn which keepz me bothering dat in my btech i hv secured 68% bt i hd bck in my first yr in a subjct( thnkz to uptu checkng ) bt i cleard in dat respective year nly so will this create any prblm in getting top institutez of mba nd in 12th i got 76% and 10th 86% so whats ur take on dis and till nw i dnt hv a wrk ex bcoz i m 2012 passed out AND plz any1 out dere can help m out in gvng answr to my qstn
@[595194:sahib_sheetu]... I don't think having a back in one subject will be of any concern, had it been for multiple subjects then surely it would have been a concern.
Your low marks in boards can be a hindrance in getting into IIMs now a days bcoz they some how correlate good marks in boards to good performance as managers (never mind that most world's top b-schools beg to differ on that).
You being a fresher won't either be a hindrance nor advantage to you in the b-schools bcoz some sectors prefer freshers while other prefer some work ex and so it will even out.
Figure out which sector you wanna go in and then research.
@Tyger said: Can anyone plz explain how the remainder problems are to be done when the dividend is a factorial.For ex : Find the remainder when 34! is divided by 71.Is Wilson's theorem applicable here?If yes,then how? Plz reply...how to solve remainder problems like these..?
what is the remainder when 9+9^2+9^3............9^(2n+1) is divided by 6.....ans is 3
6=2 *3..all powers of 9 leave remainder 0 and 1 in case of division by 3 and 2 respectively...so remainder should be of form 1+1+1...(2n+1)times...i.e remainder on divin by 2 should be 1...so answer sud b 1...bt answer is 3...??HELP..
@chirag1501 said: u can write it as, n*(n+1)*(n-1)*(5n+2) (n+1)*(n-1) means whatever value of n you take, you will always have one odd and one even number in the expression, hence it will always be divisible by 6
We have any more exceptions for the rule one odd and one even will be divisible by six ....? Primes are to be targetted? 13*14, 19*20 etc ?
@[354462:aditya5921] "this approach will work only with questions with prime numbers as denominators.", its not entirely correct.. fermat says, given 2 co-prime numbers (a,b) , a^ phi(b) mod b=1 .. example:given 2 co-prime numbers :[12,35] phi (35)=24 so, 12^24 mod 35 =1
if u multiply 10!, the last digit u get is 8,(ignoring 0's and taking the nonzero numbers only), then upto 96! we have nine 8's to multiply with 6!(which has last digit 2) of 96! so in totality we have, 8^9*2= will give last digit = 6 :)(u can solve 8^9 using cyclicity of powers) i know this method is not good for bigger numbers if anyone knows to do in a better way, kindly highlight them :)
@Vishal99 said: Find the last non zero digit of 96!a) 2 b) 4 c) 6 d) 8guys plz help me solve this question... need detail solution...
@Tyger said: Can anyone plz explain how the remainder problems are to be done when the dividend is a factorial.For ex : Find the remainder when 34! is divided by 71.Is Wilson's theorem applicable here?If yes,then how?
