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i did read the first part, and i think i understand....woo hoo thanks jane, thanks millions. i am new to this subject, havent got enough basic to do this, and now am reading and studying it , thanks for ur help. GOD bless you...see ya
150 questions!!!
by asking them to be in pair. so we have 50 pairs.
then asking each of them same question about the type of man in front of them.
the worst case is when we had 98 answers of KK (from 49 pairs) and (one group* consists of K and S )1 answer of SS ( already 100 "Questions"), otherwise is easy.
Then ask the other groups about each of the man in the *, again the worst case is if the answer is KK( of course , these answers from the 24 groups of 2 spies), ( already 48 "questions").
Then ask the next group ( must be consists of 2 knights) , then we get the answer : K and S. ( 2 "questions") .
So the total question needed is 150 questions....and that is the maximum questions needed. the minimum is depend on the pair that we could get.
" hope that am right....hehehehe, peace"
hehehehe nice people!!! woo hoo...welcome buddy
thanks a lot, but for the second, i meant there is must be a way to show the works, right master Jane? please forgive my bad math knowledge, can i just give examples by substitute any values of k and n consecutively? then in the end i write ur answer? thx again
thanks miss Jane for the help,thanks a lot, and how to deduce that k = 2^n
hi, its nice to know that i have a new group sharing the same idea, hahaha , i really really....love to be in here. its a very nice site...woo hoo, see ya guys
let p =( 2^k ) + 1, k elements of positive integers, be a prime number and let G be the group of integers 1, 2,..., p-1 under multiplication defined modulo p.
a). by first considering the elements 2^1 , 2^2 , 2^3, ..., 2^k and then the elements 2^(k+1), 2^(k+2), ... show that the order of the element 2 is 2k.
b). deduce that k = 2^n for n elements of N(natural numbers).
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